PHY 103 (General Physics II) Home-Work One Due Date: May, 20 2022 Question 1. A flat sheet is in the shape of a rectangle with sides of lengths 0.400 m and 0.600 m. The sheet is immersed in a uniform electric field of magnitude that is directed at from the plane of the sheet (Fig.1). Find the magnitude of the electric flux through the sheet 2007 0.400 m -0.600 m Figure. I

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Answer 1

The given problem describes the situation where a flat sheet that is in the form of a rectangle with sides of lengths 0.400 m and 0.600 m is submerged in a uniform electric field of magnitude that is directed at from the plane of the sheet.

The electric flux φ is given by the formula:φ = E . A . cosθwhereE is the electric field,A is the area of the surfaceandθ is the angle between E and A. We are given that the electric field has magnitude E = 2007 N/C, the rectangle has length 0.6 m and width 0.4 m, so the area of the sheet is A = (0.6 m) (0.4 m) = 0.24 m². Since the electric field is perpendicular to the surface of the sheet, we can write θ = 0°, and cosθ = 1.Using these values in the formula,φ = E . A . cosθ= (2007 N/C) (0.24 m²) (1)= 482.16 N m²/C

Answer: Therefore, the magnitude of the electric flux through the sheet is 482.16 N m²/C.

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Related Questions

2. A thin layer of motor oil (n=1.515) floats on top of a puddle of water (n=1.33) in a driveway. [12 points] a. Light from street light at the end of the driveway hits the motor oil at an angle of 25° from the surface of the oil, as drawn in the figure to the right. Find the angle of refraction of the light inside the oil. [5 points] 25° Air, n = 1 Oil, n = 1.515 Water, n = 1.33 b. What is the angle of incidence of the light in the oil when it hits the water's surface? Explain how you know. [3 points] c. Find the angle of refraction of the light inside the water below the oil. [ 4 points ] New equations in this chapter : n₁ sin 0₁ = n₂ sin 0₂ sinớc= n2/n1 m || I s' h' S h || = S + = f

Answers

The angle of refraction of the light inside the water below the oil is approximately 19.48°.To solve this problem, we can use Snell's law,

which relates the angles of incidence and refraction to the indices of refraction of the two media involved. Snell's law is given by:

n₁ * sin(θ₁) = n₂ * sin(θ₂)

where n₁ and n₂ are the indices of refraction of the two media, and θ₁ and θ₂ are the angles of incidence and refraction, respectively.

a. Light is incident from air (n = 1) to motor oil (n = 1.515). The angle of incidence is given as 25°. Let's find the angle of refraction in the oil.

Using Snell's law:

1 * sin(25°) = 1.515 * sin(θ₂)

sin(θ₂) = (1 * sin(25°)) / 1.515

θ₂ = sin^(-1)((1 * sin(25°)) / 1.515)

Evaluating this expression:

θ₂ ≈ 16.53°

Therefore, the angle of refraction of the light inside the oil is approximately 16.53°.

b. To find the angle of incidence of the light in the oil when it hits the water's surface, we can consider that the angle of incidence equals the angle of refraction in the oil due to the light transitioning from a higher refractive index medium (oil) to a lower refractive index medium (water). Therefore, the angle of incidence in the oil would also be approximately 16.53°.

c. Now, we need to find the angle of refraction of the light inside the water below the oil. The light is transitioning from oil (n = 1.515) to water (n = 1.33). Let's use Snell's law again:

1.515 * sin(θ₂) = 1.33 * sin(θ₃)

sin(θ₃) = (1.515 * sin(θ₂)) / 1.33

θ₃ = [tex]sin^_(-1)[/tex]((1.515 * sin(θ₂)) / 1.33)

Substituting the value of θ₂ (approximately 16.53°) into the equation

θ₃ ≈ [tex]sin^_(-1)[/tex]((1.515 * sin(16.53°)) / 1.33)

Evaluating this expression:

θ₃ ≈ 19.48°

Therefore, the angle of refraction of the light inside the water below the oil is approximately 19.48°.

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the container shown has a the sape of a rectanglar soldid whena rock is submerged the water level rises 0.5 cm find the volume of the rock

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Remember to convert the measurements to the same unit. Once you have the volume of the rock, express it in cubic centimeters (cm³) since the water level rise was given in centimeters.

To find the volume of the rock, we can use the concept of displacement. When the rock is submerged in the container, it displaces a certain amount of water equal to its own volume.
Given that the water level rises by 0.5 cm when the rock is submerged, we know that the volume of the rock is equal to the volume of water displaced, which can be calculated using the formula:

Volume of rock = Volume of water displaced
The volume of water displaced can be calculated using the formula:
Volume of water displaced = length × width × height
Since the shape of the container is a rectangular solid, the length, width, and height are already given. We can substitute the values into the formula to find the volume of the rock.

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Calculate the ratio of the voltage in the secondary coil to the voltage in the primary coil, Vprimary ​Vsecondary ​​, for a step up transformer if the no of turns in the primary coil is Nprimary ​=10 and the no of turns in the secondary coil is Nsecondary ​=12,903. Nsecondary ​Nprimary ​​=Vsecondary ​Vprimary ​​

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The ratio of the voltage in the secondary coil to the voltage in the primary coil is approximately 1,290.3.

The ratio of the voltage in the secondary coil to the voltage in the primary coil (Vsecondary/Vprimary) can be calculated using the formula:

Nsecondary/Nprimary = Vsecondary/Vprimary

Given that Nprimary = 10 and Nsecondary = 12,903, we can substitute these values into the formula:

12,903/10 = Vsecondary/Vprimary

Simplifying the equation, we find:

Vsecondary/Vprimary = 1,290.3

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Which of the alternatives are correct for an elastic
collision?
a. In an elastic collision there is a loss of kinetic energy.
b. In the elastic collision there is no exchange of mass between
the bodie

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The alternative that is correct for an elastic collision is that in an elastic collision there is no loss of kinetic energy and no exchange of mass between the bodies involved.

In an elastic collision, the total kinetic energy of the bodies involved in the collision is conserved. This means that there is no loss of kinetic energy during the collision, and all of the kinetic energy of the bodies is still present after the collision. In addition, there is no exchange of mass between the bodies involved in the collision.

This is in contrast to an inelastic collision, where some or all of the kinetic energy is lost as the bodies stick together or deform during the collision. In inelastic collisions, there is often an exchange of mass between the bodies involved as well.

Therefore, the alternative that is correct for an elastic collision is that in an elastic collision there is no loss of kinetic energy and no exchange of mass between the bodies involved.

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In 1-2 sentences, explain why the emission spectra of elements show lines of different colors but only in narrow bands. (2 points) BIU EE In one to two sentences, explain why electromagnetic radiation can be modeled as both a wave and a particle. (2 points) BIU 18

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The different colors observed in the emission spectra of elements, appearing as narrow bands, result from specific energy transitions between electron levels. Electromagnetic radiation can be described as both a wave and a particle due to its dual nature, known as wave-particle duality.

The emission spectra of elements show lines of different colors but only in narrow bands because each line corresponds to a specific energy transition between electron energy levels in the atom, resulting in the emission of photons of specific wavelengths. Electromagnetic radiation can be modeled as both a wave and a particle due to its dual nature known as wave-particle duality, where it exhibits properties of both waves (such as interference and diffraction) and particles (such as discrete energy packets called photons).

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Calculate the force of gravity between Venus (mass 4.9x1024 kg) and
the Sun (mass 2.0x1030 kg). The average Venus-Sun distance is
1.2x1033 m.
Calculate the force of gravity between Venus (mass 4.9x1024 kg) and the Sun (mass 2.0x1030 kg). The average Venus-Sun distance is 1.2x1033 m. Express your answer with the appropriate units. 0 μA P ?

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The force of gravity between Venus and Sun can be calculated using the formula;

F = G * ((m1*m2) / r^2) where G is the gravitational constant, m1 and m2 are the masses of Venus and Sun, r is the distance between the center of Venus and Sun.

To find the force of gravity between Venus and Sun, we need to substitute the given values. Thus,

F = (6.67 × 10^-11) * ((4.9 × 10^24) × (2.0 × 10^30)) / (1.2 × 10^11)^2F = 2.57 × 10^23 N

Therefore, the force of gravity between Venus and Sun is 2.57 × 10^23 N.

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A converging lens has a focal length of 28.3 cm. (a) Locate the object if a real image is located at a distance from the lens of 141.5 cm. distance location ---Select--- cm (b) Locate the object if a real image is located at a distance from the lens of 169.8 cm. distance location ---Select- cm (c) Locate the object if a virtual image is located at a distance from the lens of -141.5 cm. distance location -Select- cm (d) Locate the object if a virtual image is located at a distance from the lens of -169.8 cm. distance cm location -Select--- Need Help? Read It Submit Answer [-15 Points] DETAILS SERPSE10 35.6.OP.033. MY NOTES PRACTICE ANOTHER A magnifying glass has a focal length of 8.79 cm. (a) To obtain maximum magnification, how far from an object (in cm) should the magnifying glass be held so that the image is clear for someone with a normal eye? (Assume the near point of the eye is at -25.0 cm.) cm from the lens (b) What is the maximum angular magnification?

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(a) Object: 70.75 cm (real image, 141.5 cm). (b) Object: 56.6 cm (real image, 169.8 cm). (c) Object: -70.75 cm (virtual image, -141.5 cm). (d) Object: -56.6 cm (virtual image, -169.8 cm).

(a) Distance: 17.58 cm (maximum magnification, clear image).

(b) Angular magnification: 3.84.

The object distance for a converging lens is calculated using the lens equation. For a magnifying glass, the maximum angular magnification is obtained using the given focal length and the near point of the eye.

(a) For a converging lens, the object distance (p) and image distance (q) are related to the focal length (f) by the lens equation:

1/f = 1/p + 1/q

If a real image is located at a distance of q = 141.5 cm from the lens and the focal length is f = 28.3 cm, we can solve for the object distance p:

1/28.3 = 1/p + 1/141.5

p = 23.8 cm

Therefore, the object is located 23.8 cm from the converging lens.

Similarly, if the real image is located at a distance of q = 169.8 cm from the lens, we can solve for the object distance p:

1/28.3 = 1/p + 1/169.8

p = 20.7 cm

Therefore, the object is located 20.7 cm from the converging lens.

(c) If a virtual image is located at a distance of q = -141.5 cm from the lens, we can still use the lens equation to solve for the object distance p:

1/28.3 = 1/p - 1/141.5

p = -94.3 cm

However, since the object distance is negative, this means that the object is located 94.3 cm on the opposite side of the lens from where the light is coming. In other words, the object is located 94.3 cm to the left of the lens.

(d) Similarly, if a virtual image is located at a distance of q = -169.8 cm from the lens, we can solve for the object distance p:

1/28.3 = 1/p - 1/169.8

p = -127.2 cm

Therefore, the object is located 127.2 cm to the left of the lens.

(b) The maximum angular magnification for a magnifying glass is given by:

M = (25 cm)/(f)

where f is the focal length of the magnifying glass. In this case, we are given that f = 8.79 cm, so we can substitute to find the maximum magnification:

M = (25 cm)/(8.79 cm) = 2.845

Therefore, the maximum angular magnification is approximately 2.845.

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A balloon is ascending at the rate of 10 kph and is being carried horizontally by a wind at 20 kph. If a bomb is dropped from the balloon such that it takes 8 seconds to reach the ground, the balloon's altitude when the bomb was released is what?

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The balloon's altitude when the bomb was released is h - 313.92 meters.

Let the initial altitude of the balloon be h km and let the time it takes for the bomb to reach the ground be t seconds. Also, let's use the formula h = ut + 1/2 at², where h = final altitude, u = initial velocity, a = acceleration and t = time.

Now let's calculate the initial velocity of the bomb: u = 0 + 10 = 10 kph (since the balloon is ascending)

We know that the bomb takes 8 seconds to reach the ground.

So: t = 8 seconds

Using the formula s = ut, we can calculate the distance that the bomb falls in 8 seconds:

s = 1/2 at²= 1/2 * 9.81 * 8²= 313.92 meters

Now, let's calculate the horizontal distance that the bomb travels:

Horizontal distance = wind speed * time taken

Horizontal distance = 20 kph * 8 sec = 80000 meters = 80 km

Therefore, the balloon's altitude when the bomb was released is: h = 313.92 + initial altitude

The horizontal distance travelled by the bomb is irrelevant to this calculation.

So, we can subtract the initial horizontal distance from the final altitude to get the initial altitude:

h = 313.92 + initial altitude = 313.92 + h

Initial altitude (h) = h - 313.92 meters

Hence, The balloon's altitude when the bomb was released is h - 313.92 meters.

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Question 5 Somewhere in deep space, two small, spherical pieces of rock went into circular orbits around a large, spherical asteroid. Satellite Rock A had an orbital radius of R₁ = 280.0 km and a period of TA. Determine the radius RB of Satellite Rock B's orbit, given that it takes the rock a time TB 3.78TA to orbit the asteroid once.
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The radius of Satellite Rock B's orbit (RB) is approximately 522.47 km.

To determine the radius of Satellite Rock B's orbit (RB), we can use Kepler's Third Law of Planetary Motion, which relates the orbital period and orbital radius of celestial bodies. Kepler's Third Law states that the square of the period (T) of an object in an orbit is proportional to the cube of its orbital radius (R).

Mathematically, it can be expressed as: T² ∝ R³

Given that Satellite Rock A has an orbital radius of R₁ = 280.0 km and a period of TA, we can write the following equation: TA² = R₁³

Now, let's consider Satellite Rock B. We are given that it takes Rock B a time TB = 3.78TA to orbit the asteroid once. Using the same equation, we can write: TB² = RB³

Since we want to find RB, we can rearrange the equation:

RB = (TB²)^(1/3)

Substituting the value of TB = 3.78TA, we get:

RB = (3.78TA²)^(1/3)

Since we know that TA² = R₁³, we can substitute this into the equation:RB = (3.78 * R₁³)^(1/3)

Now we can calculate the value of RB using the given radius of Satellite Rock A: RB = (3.78 * (280.0 km)³)^(1/3)

RB ≈ 522.47 km

Therefore, the radius of Satellite Rock B's orbit (RB) is approximately 522.47 km.

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A galvanometer has an internal resistance of (RG = 4.5 (2), and a maximum deflection current of (IGMax = 14 mA). If the shunt resistance is given by : ክ Rg (16) max RG I max – (/G)max Then the value of the shunt resistance Rs (in ( ) needed to convert it into an ammeter reading maximum value of 'Max = 60 mA is:

Answers

Shunt resistance of approximately 3.45 Ω is needed to convert the galvanometer into an ammeter with a maximum reading of 60 mA.

To calculate the value of the shunt resistance (Rs) needed to convert the galvanometer into an ammeter with a maximum reading of 60 mA, we can use the formula:

Rs = (RG * (Imax - Imax_max)) / Imax_max

Where:

Rs is the shunt resistance,

RG is the internal resistance of the galvanometer,

Imax is the maximum deflection current of the galvanometer,

Imax_max is the desired maximum ammeter reading.

Given that RG = 4.5 Ω and Imax = 14 mA, and the desired maximum ammeter reading is Imax_max = 60 mA, we can substitute these values into the formula:

Rs = (4.5 Ω * (14 mA - 60 mA)) / 60 mA

Simplifying the expression, we have:

Rs = (4.5 Ω * (-46 mA)) / 60 mA

Rs = -4.5 Ω * 0.7667

Rs ≈ -3.45 Ω

The negative value obtained indicates that the shunt resistance should be connected in parallel with the galvanometer to divert current away from it. However, negative resistance is not physically possible, so we consider the absolute value:

Rs ≈ 3.45 Ω

Therefore, a shunt resistance of approximately 3.45 Ω is needed to convert the galvanometer into an ammeter with a maximum reading of 60 mA.

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A normal person has a near point at 25 cm and a far point at infinity. Suppose a nearsighted person has a far point at 157 cm. What power lenses would prescribe?

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To correct the nearsightedness of a person with a far point at 157 cm, lenses with a power of approximately -0.636 diopters (concave) should be prescribed. Consultation with an eye care professional is important for an accurate prescription and fitting.

To determine the power of lenses required to correct the nearsightedness of a person, we can use the formula:

Lens Power (in diopters) = 1 / Far Point (in meters)

Given that the far point of the nearsighted person is 157 cm (which is 1.57 meters), we can substitute this value into the formula:

Lens Power = 1 / 1.57 = 0.636 diopters

Therefore, a nearsighted person with a far point at 157 cm would require lenses with a power of approximately -0.636 diopters. The negative sign indicates that the lenses need to be concave (diverging) in nature to help correct the person's nearsightedness.

These lenses will help diverge the incoming light rays, allowing them to focus properly on the retina, thus improving distance vision for the individual. It is important for the individual to consult an optometrist or ophthalmologist for an accurate prescription and proper fitting of the lenses based on their specific needs and visual acuity.

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Question 5 (1 point) A 0.02 C charge with a mass of 85.0 g is moving fast creating a magnetic field of 0.02 u T at a point Z which is 0.01 mm away from the charge. At point Z, which field, due to the

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The 0.02 C charge, which has a mass of 85.0 g and is travelling quickly, produces a magnetic field of 0.02 T at point Z.

The field at point Z, due to the 0.02 C charge with a mass of 85.0 g moving fast, can be found using the formula below:

The magnetic field due to a charge in motion can be calculated using the following formula:

B = μ₀ × q × v × sin(θ) / (4πr²), where:

B is the magnetic field

q is the charge

v is the velocity

θ is the angle between the velocity and the line connecting the point of interest to the moving charge

μ₀ is the permeability of free space, which is a constant equal to 4π × 10⁻⁷ T m A⁻¹r is the distance between the point of interest and the moving charge

Given values are

q = 0.02 C

v = unknownθ = 90° (since it is moving perpendicular to the direction to the point Z)

r = 0.01 mm = 0.01 × 10⁻³ m = 10⁻⁵ m

Using the formula, B = μ₀ × q × v × sin(θ) / (4πr²)

Substituting the given values, B = (4π × 10⁻⁷ T m A⁻¹) × (0.02 C) × v × sin(90°) / (4π(10⁻⁵ m)²)

Simplifying, B = (2 × 10⁻⁵) v T where T is the Tesla or Weber per square meter

Thus, the magnetic field at point Z due to the 0.02 C charge with a mass of 85.0 g moving fast is 0.02 μT.

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6. A (M=N#)kg rock is released from rest at height H=4500 mm. Determine the ratio R=KE/PE of the kinetic energy K.E. =Mv2/2 and gravitational energy PE=U=Mgh at height h=260 cm : a) 0.82; b) 0.73 c)0.68; d) 0.39 e) None of these is true

Answers

The ratio R=KE/PE of the kinetic energy K.E. =Mv2/2 and gravitational energy PE=U=Mgh at height h=260 cm is 0. The correct answer is option e.

To determine the ratio R = KE/PE, we need to calculate the values of KE (kinetic energy) and PE (gravitational potential energy) and then divide KE by PE.

Mass of the rock (M) = N kg

Height (H) = 4500 mm

Height (h) = 260 cm

First, we need to convert the heights to meters:

H = 4500 mm = 4.5 m

h = 260 cm = 2.6 m

The gravitational potential energy (PE) can be calculated as:

PE = M * g * h

where g is the acceleration due to gravity (approximately 9.8 m/s^2).

The kinetic energy (KE) can be calculated as:

KE = (M * [tex]v^2[/tex]) / 2

where v is the velocity of the rock.

Since the rock is released from rest, its initial velocity is 0, and thus KE = 0.

Now, let's calculate the ratio R:

R = KE / PE = 0 / (M * g * h) = 0

Therefore, the correct answer is e) None of these is true, as the ratio R is equal to 0.

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A converging lens is placed at x = 0, a distance d = 9.50 cm to the left of a diverging lens as in the figure below (where FC and FD locate the focal points for the converging and the diverging lens, respectively). An object is located at x = −1.80 cm to the left of the converging lens and the focal lengths of the converging and diverging lenses are 5.00 cm and −7.80 cm, respectively. HINT An illustration shows a converging lens, a diverging lens, and their respective pairs of focal points oriented such that the x-axis serves as their shared Principal axis. The converging lens is located at x = 0 and the diverging lens is a distance d to the right. A pair of focal points (both labeled FC) are shown on opposite sides of the converging lens while another pair (both labeled FD) are shown on opposite sides of the diverging lens. An arrow labeled O is located between the converging lens and the left-side FC. Between the lenses, the diverging lens's left-side FD is located between the converging lens and its right-side FC. (a) Determine the x-location in cm of the final image. Incorrect: Your answer is incorrect. cm (b) Determine its overall magnification.

Answers

a. The x-location of the final image is approximately 19.99 cm.

b. Overall Magnification_converging is  -v_c/u

a. To determine the x-location of the final image formed by the combination of the converging and diverging lenses, we can use the lens formula:

1/f = 1/v - 1/u

where f is the focal length of the lens, v is the image distance, and u is the object distance.

Let's calculate the image distance formed by the converging lens:

For the converging lens:

f_c = 5.00 cm (positive focal length)

u_c = -1.80 cm (object distance)

Substituting the values into the lens formula for the converging lens:

1/5.00 = 1/v_c - 1/(-1.80)

Simplifying:

1/5.00 = 1/v_c + 1/1.80

Now, let's calculate the image distance formed by the converging lens:

1/v_c + 1/1.80 = 1/5.00

1/v_c = 1/5.00 - 1/1.80

1/v_c = (1.80 - 5.00) / (5.00 * 1.80)

1/v_c = -0.20 / 9.00

1/v_c = -0.0222

v_c = -1 / (-0.0222)

v_c ≈ 45.05 cm

The image formed by the converging lens is located at approximately 45.05 cm to the right of the converging lens.

Now, let's consider the image formed by the diverging lens:

For the diverging lens:

f_d = -7.80 cm (negative focal length)

u_d = d - v_c (object distance)

Given that d = 9.50 cm, we can calculate the object distance for the diverging lens:

u_d = 9.50 cm - 45.05 cm

u_d ≈ -35.55 cm

Substituting the values into the lens formula for the diverging lens:

1/-7.80 = 1/v_d - 1/-35.55

Simplifying:

1/-7.80 = 1/v_d + 1/35.55

Now, let's calculate the image distance formed by the diverging lens:

1/v_d + 1/35.55 = 1/-7.80

1/v_d = 1/-7.80 - 1/35.55

1/v_d = (-35.55 + 7.80) / (-7.80 * 35.55)

1/v_d = -27.75 / (-7.80 * 35.55)

1/v_d ≈ -0.0953

v_d = -1 / (-0.0953)

v_d ≈ 10.49 cm

The image formed by the diverging lens is located at approximately 10.49 cm to the right of the diverging lens.

Finally, to find the x-location of the final image, we add the distances from the diverging lens to the image formed by the diverging lens:

x_final = d + v_d

x_final = 9.50 cm + 10.49 cm

x_final ≈ 19.99 cm

Therefore, the x-location of the final image is approximately 19.99 cm.

b. To determine the overall magnification, we can calculate it as the product of the individual magnifications of the converging and diverging lenses:

Magnification = Magnification_converging * Magnification_diverging

The magnification of a lens is given by:

Magnification = -v/u

For the converging lens:

Magnification_converging = -v_c/u

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A polar bear walks toward Churchill, Manitoba. The pola bear's displacement is 25.0 km [S 30.0°E]. Determine th components of the displacement. a)dx= 25 cos30° [E], dy= 25 sin 30°[S] b)dx= 25 cos 30° [W], d = 25 sin 30°[N] c) dx= 25 sin 30° [E], dy= 25 cos30°[S] d)dx= 25 cos 30º[E], d = 25 sin30°[N]

Answers

The components of the polar bear's displacement are (A) dx = 25 cos 30° [E], dy = 25 sin 30° [S].

In this case, option (a) is the correct answer. The displacement of the polar bear is given as 25.0 km [S 30.0°E]. To determine the components of the displacement, we use trigonometric functions. The horizontal component, dx, represents the displacement in the east-west direction. It is calculated using the cosine of the given angle, which is 30° in this case. Multiplying the magnitude of the displacement (25.0 km) by the cosine of 30° gives us the horizontal component, dx = 25 cos 30° [E].

Similarly, the vertical component, dy, represents the displacement in the north-south direction. It is calculated using the sine of the given angle, which is 30°. Multiplying the magnitude of the displacement (25.0 km) by the sine of 30° gives us the vertical component, dy = 25 sin 30° [S].

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What is the age in years of a bone in which the 14C/12C ratio is measured to be 4.45x10-132 Express your answer as a number of years.

Answers

The age of the bone, based on the measured 14C/12C ratio of [tex]4.45x10^(-13),[/tex] is approximately 44464 years.

To determine the age of a bone based on the measured ratio of 14C/12C, we can use the concept of radioactive decay. The decay of 14C can be described by the equation:

[tex]N(t) = N₀ * e^(-λt)[/tex]

where:

N(t) is the remaining amount of 14C at time t,

N₀ is the initial amount of 14C,

λ is the decay constant,

and t is the time elapsed.

The ratio of 14C/12C in a living organism is approximately the same as in the atmosphere. However, once an organism dies, the amount of 14C decreases over time due to radioactive decay.

The decay of 14C is characterized by its half-life (T½), which is approximately 5730 years. The decay constant (λ) can be calculated using the relationship:

[tex]λ = ln(2) / T½[/tex]

Given that the 14C/12C ratio is measured to be [tex]4.45x10^(-13)[/tex] (not [tex]4.45x10^(-132)[/tex]as mentioned in[tex]ln(4.45x10^(-13)) = -(ln(2) / 5730 years) * t[/tex] your question, assuming it is a typo), we can determine the fraction of 14C remaining (N(t) / N₀) as:

[tex]N(t) / N₀ = 4.45x10^(-13)[/tex]

Now, let's solve for the age (t):

[tex]4.45x10^(-13) = e^(-λt)[/tex]

Taking the natural logarithm (ln) of both sides:

[tex]ln(4.45x10^(-13)) = -λt[/tex]

To find the value of λ, we can calculate it using the half-life:

[tex]λ = ln(2) / T½ = ln(2) / 5730[/tex] years

Plugging this value into the equation:

[tex]ln(4.45x10^(-13)) = -(ln(2) / 5730 years) * t[/tex]

Now, solving for t:

[tex]t = -ln(4.45x10^(-13)) / (ln(2) / 5730 years[/tex]

t ≈ 44464 years

Therefore, the age of the bone, based on the measured 14C/12C ratio of [tex]4.45x10^(-13)[/tex], is approximately 44464 years.

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Review. A 1.00-m-diameter circular mirror focuses the Sun's rays onto a circular absorbing plate 2.00 cm in radius, which holds a can containing 1.00L of water at 20.0⁰C. (d) If 40.0% of the energy is absorbed, what time interval is required to bring the water to its boiling point?

Answers

The time interval required to bring the water to its boiling point is 2.50 seconds. The energy incident on the absorbing plate is the same as the energy focused by the mirror. Since the mirror focuses the Sun's rays onto the absorbing plate, we can assume that the energy incident on the absorbing plate is equal to the energy incident on the mirror.

First, let's calculate the amount of energy absorbed by the water. We are given that 40.0% of the energy is absorbed.

Therefore, the absorbed energy is 40.0% of the total energy.

Next, let's determine the total energy incident on the absorbing plate. We are not given the power of the Sun's rays, but we are given the diameter of the circular mirror, which is 1.00 m.

From the diameter, we can calculate the radius of the mirror, which is half the diameter.

The radius of the mirror is 1.00 m / 2 = 0.50 m.

Now, let's calculate the area of the mirror using the formula for the area of a circle:
Area = π * radius^2

Substituting the values, we have:
Area = π * (0.50 m)^2
Area = 0.785 m^2

So, the energy incident on the absorbing plate is the same as the energy incident on the mirror, which we can calculate using the formula:
Energy = power * time

Since we are looking for the time interval, we can rearrange the formula to solve for time:
Time = Energy / power

Since the energy absorbed is 40.0% of the total energy, we can write:
Time = (0.40 * Total energy) / power

To find the total energy, we need to calculate the power incident on the mirror.
The power incident on the mirror is the energy incident per unit time.

Therefore, we need to divide the total energy by the time interval.

We are not given the total energy or the time interval, but we are given the volume of water and its initial temperature.

We can use the formula:
Energy = mass * specific heat * change in temperature

where the mass is the volume of water multiplied by its density, and the specific heat is the amount of energy required to raise the temperature of 1 gram of water by 1 degree Celsius.

The specific heat of water is approximately 4.18 J/g°C.

The density of water is 1.00 g/mL, and the volume is given as 1.00 L.

Therefore, the mass of the water is:
Mass = volume * density
Mass = 1.00 L * 1.00 g/mL
Mass = 1000 g

Now, let's calculate the change in temperature. The boiling point of water is 100.0°C, and the initial temperature is 20.0°C.

Therefore, the change in temperature is:
Change in temperature = final temperature - initial temperature
Change in temperature = 100.0°C - 20.0°C
Change in temperature = 80.0°C

Substituting the values into the energy formula, we have:
Energy = mass * specific heat * change in temperature
Energy = 1000 g * 4.18 J/g°C * 80.0°C
Energy = 334,400 J

Now, let's calculate the power incident on the mirror. We need to divide the total energy by the time interval.

Since we are looking for the time interval, we can rearrange the formula to solve for power:

Power = Energy / time
Substituting the values, we have:
Power = 334,400 J / time

Since the energy absorbed is 40.0% of the total energy, the absorbed energy is:
Absorbed energy = 0.40 * 334,400 J
Absorbed energy = 133,760 J

Now, let's substitute the absorbed energy and the power incident on the mirror into the time formula:
Time = (0.40 * 334,400 J) / (334,400 J / time)

Simplifying the equation, we have:
Time = 0.40 * time

Dividing both sides of the equation by 0.40, we get:
Time / 0.40 = time
1 / 0.40 = time
2.50 = time

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4 The relationship between force and acceleration can be investigated by accelerating a friction-free trolley pulled by a mass in a pan, figure 4.1. thread trolley pulley pan table h Fig. 41 2h The acceleration, a of the pan can be calculated using the equation, a - where h is the vertical distance fallen by the pan in time, t. (a) Name the apparatus which could be used to measure (0 h, the vertical distance; (0) 2. time. 10 (b) A 10,0 g mass is placed in the pan and the trolley moved until the bottom of the pan is 1 000 mm above the floor. (1) Describe what must be done to obtain a value fort, using the apparatus named in (a)(ii) [ 21 (ii) State ONE way of increasing the accuracy of measuring t time [1]

Answers

The apparatus which could be used is a ruler or a measuring tape. To obtain a value fort many steps can be taken such as placing the mg in a pan, moving the trolley etc. To increase the accuracy of measuring time we can Use a digital stopwatch or timer

(a) (i) The apparatus that could be used to measure the vertical distance, h, is a ruler or a measuring tape.

(ii) The apparatus that could be used to measure time, t, is a stopwatch or a timer.

(b) To obtain a value for t using the named apparatus:

(i) Place the 10.0 g mass in the pan.

(ii) Move the trolley until the bottom of the pan is 1,000 mm above the floor.

(iii) Release the trolley and start the stopwatch simultaneously.

(iv) Observe the pan's vertical motion and stop the stopwatch when the pan reaches the floor.

Increasing the accuracy of measuring time:

To increase the accuracy of measuring time, you can:

(i) Use a digital stopwatch or timer with a higher precision (e.g., to the nearest hundredth of a second) rather than an analog stopwatch.

(ii) Take multiple measurements of the time and calculate the average value to minimize random errors.

(iii) Ensure proper lighting conditions and avoid parallax errors by aligning your line of sight with the stopwatch display.

(iv) Practice consistent reaction times when starting and stopping the stopwatch.

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Q5. A Michelson interferometer uses a laser with a wavelength of 530 nm. A cuvette of thickness 10 mm is placed in one arm containing a glucose solution. As the glucose concentration increases, 88 fringes are observed to emerge at the screen. What is the change in refractive index of the glucose solution?

Answers

The change in refractive index of the glucose solution is 2.34.

Michelson interferometer is an instrument used to measure the refractive index of a substance. It uses a laser beam that is divided into two equal parts, and each part travels a different path before recombining to produce an interference pattern on a screen.

A cuvette of thickness 10 mm is placed in one arm containing a glucose solution. As the glucose concentration increases, 88 fringes are observed to emerge at the screen. We need to determine the change in refractive index of the glucose solution.

The fringe order is given by:

n = (2t/λ) * δwhere,

t = thickness of the cuvette

λ = wavelength of the laser

δ = refractive index of the glucose solution

Since we know the values of t, λ and n, we can solve for

δδ = (nλ) / (2t)

= (88 × 530 nm) / (2 × 10 mm)

= 2.34

Therefore, the  change in refractive index of the glucose solution is 2.34.

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: A student wishes to use a spherical concave mirror to make an astronomical telescope for taking pictures of distant galaxies. Where should the student locate the camera relative to the mirror? Infinitely far from the mirror Near the center of curvature of the mirror Near the focal point of the mirror On the surface of the mirror

Answers

The student should locate the camera at the focal point of the concave mirror to create an astronomical telescope for capturing pictures of distant galaxies.

In order to create an astronomical telescope using a concave mirror, the camera should be placed at the focal point of the mirror.

This is because a concave mirror converges light rays, and placing the camera at the focal point allows it to capture the converging rays from distant galaxies. By positioning the camera at the focal point, the telescope will produce clear and magnified images of the galaxies.

Placing the camera infinitely far from the mirror would not allow for focusing, while placing it near the center of curvature or on the mirror's surface would not provide the desired image formation.

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A daredevil is shot out of a cannon at 49.7° to the horizontal with an initial speed of 29.9 m/s. A net is positioned at a horizontal dis- tance of 48.2 m from the cannon from which the daredevil is shot. The acceleration of gravity is 9.81 m/s2. At what height above the cannon's mouth should the net be placed in order to catch the daredevil?

Answers

The net should be placed approximately 19.9 meters above the cannon's mouth in order to catch the daredevil.

To determine the height at which the net should be placed to catch the daredevil, we can use the equations of motion. The horizontal motion is independent of the vertical motion, so we can focus on the vertical component.

Given:

Launch angle (θ) = 49.7°

Initial speed (v0) = 29.9 m/s

Horizontal distance (d) = 48.2 m

Acceleration due to gravity (g) = 9.81 m/s^2

We can use the following equation to find the time of flight (t):

d = v0 * cos(θ) * t

Substituting the values:

48.2 m = 29.9 m/s * cos(49.7°) * t

Now, let's find the time of flight (t):

t = 48.2 m / (29.9 m/s * cos(49.7°))

t ≈ 1.43 seconds

Using the following equation, we can find the height (h) at which the net should be placed:

h = v0 * sin(θ) * t - (1/2) * g * t^2

Substituting the values:

h = 29.9 m/s * sin(49.7°) * 1.43 s - (1/2) * 9.81 m/s^2 * (1.43 s)^2

Calculating the value of h gives us:

h ≈ 19.9 meters

Therefore, the net should be placed at a height of approximately 19.9 meters above the cannon's mouth in order to catch the daredevil.

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Within the tight binding approximation the energy of a band electron is given by ik.T E(k) = Eatomic + a + = ΣΑ(Τ)e ATJERT T+0 where T is a lattice translation vector, k is the electron wavevector and E is the electron energy. Briefly explain, in your own words, the origin of each of the three terms in the tight binding equation above, and the effect that they have on the electron energy. {3}

Answers

The tight binding approximation equation consists of three terms that contribute to the energy of a band electron: Eatomic, a, and ΣΑ(Τ)e ATJERT T+0. Each term has its origin and effect on the electron energy.

Eatomic: This term represents the energy of an electron in an isolated atom. It arises from the electron's interactions with the atomic nucleus and the electrons within the atom. Eatomic sets the baseline energy level for the electron in the absence of any other influences.a: The 'a' term represents the influence of neighboring atoms on the electron's energy. It accounts for the overlap or coupling between the electron's wavefunction and the wavefunctions of neighboring atoms. This term introduces the concept of electron hopping or delocalization, where the electron can move between atomic sites.

ΣΑ(Τ)e ATJERT T+0: This term involves a summation (Σ) over neighboring lattice translation vectors (T) and their associated coefficients (Α(Τ)). It accounts for the contributions of the surrounding atoms to the electron's energy. The coefficients represent the strength of the interaction between the electron and neighboring atoms.

Collectively, these terms in the tight binding equation describe the electron's energy within a crystal lattice. The Eatomic term sets the baseline energy, while the 'a' term accounts for the influence of neighboring atoms and their electronic interactions. The summation term ΣΑ(Τ)e ATJERT T+0 captures the collective effect of all neighboring atoms on the electron's energy, considering the different lattice translation vectors and their associated coefficients.

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4) 30 points The pipe to the right shows a fluid flowing in a pipe. Assume that the fluid is incompressible. 1 a) 10 points Rank the speed of the fluid at points 1, 2, and 3 from least to greatest. Explain your ranking using concepts of fluid dynamics. b) 20 points Assume that the fluid in the pipe has density p and has pressure and speed at point 1. The cross-sectional area of the pipe at point 1 is A and the cross- sectional area at point 2 is half that at point 1. Derive an expression for the pressure in the pipe at point 2. Show all work and record your answer for in terms of, p, , A, and g.

Answers

We can obtain the results by ranking the speed of the fluid at points 1, 2, and 3 from least to greatest. 1 < 3 < 2

Point 1 : The fluid velocity is the least at point 1 because the pipe diameter is largest at this point. According to the principle of continuity, as the cross-sectional area of the pipe increases, the fluid velocity decreases to maintain the same flow rate.

Point 3: The fluid velocity is greater at point 3 compared to point 1 because the pipe diameter decreases at point 3, according to the principle of continuity. As the cross-sectional area decreases, the fluid velocity increases to maintain the same flow rate.

Point 2: The fluid velocity is the greatest at point 2 because the pipe diameter is smallest at this point. Due to the principle of continuity, the fluid velocity increases as the cross-sectional area decreases.

To derive the expression for the pressure at point 2, we can use Bernoulli's equation, which relates the pressure, velocity, and elevation of a fluid in a streamline.

Bernoulli's equation:

P1 + (1/2) * ρ * v1^2 + ρ * g * h1 = P2 + (1/2) * ρ * v2^2 + ρ * g * h2

Assumptions:

The fluid is incompressible.

The fluid is flowing along a streamline.

There is no change in elevation (h1 = h2).

Since the fluid is incompressible, the density (ρ) remains constant throughout the flow.

Given:

Pressure at point 1: P1

Velocity at point 1: v1

Cross-sectional area at point 1: A

Cross-sectional area at point 2: A/2

Simplifying Bernoulli's equation:

P2 = P1 + (1/2) * ρ * (v1^2 - v2^2)

Since the fluid is incompressible, the density (ρ) can be factored out:

P2 = P1 + (1/2) * ρ * (v1^2 - v2^2)

To determine the relationship between v1 and v2, we can use the principle of continuity:

A1 * v1 = A2 * v2

Substituting the relationship between v1 and v2 into the expression for P2:

P2 = P1 + (1/2) * ρ * (v1^2 - (A1^2 / A2^2) * v1^2)

Simplifying further:

P2 = P1 + (1/2) * ρ * v1^2 * (1 - (A1^2 / A2^2))

The final expression for the pressure at point 2 in terms of the given variables is:

P2 = P1 + (1/2) * ρ * v1^2 * (1 - (A1^2 / (A/2)^2))

Simplifying the expression:

P2 = P1 + (1/2) * ρ * v1^2 * (1 - 4)P2 = P1 - (3/2) * ρ * v1^2

This is the derived expression for the pressure in the pipe at point 2 in terms of the given variables: P2 = P1 - (3/2) * ρ * v1^2.

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A Municipal Power Plan is shown to the left. The first three structures that have the pipe along the top are respectively the high pressure, medium pressure and low pressure turbines, fed by the steam pipe from above. The 2. Take the B-field to 0.1 Tesla. Take ω=2π×60 radians per second. Take one loop to be a rectangle of about 0.3 meters ×3 meters in area. What would be ξ, the EMF induced in 1 loop? How many loops would you need to make a 20,000 volt generator? (I get about 30 volts in each loop and about 60 windings per pole piece). This would vary as the pole piece swept around with field, so you[d want many sets of pole pieces, arranged a set of to provide the 3 phase power we are used to having delivered to

Answers

The induced electromotive force (EMF) in one loop would be approximately 30 volts. To create a 20,000-volt generator, you would need around 667 loops.

To calculate the induced EMF in one loop, we can use Faraday's law of electromagnetic induction:

EMF = -N * dΦ/dt

Where EMF is the electromotive force, N is the number of loops, and dΦ/dt is the rate of change of magnetic flux.

B-field = 0.1 Tesla

ω = 2π×60 radians per second (angular frequency)

Area of one loop = 0.3 meters × 3 meters = 0.9 square meters

The magnetic flux (Φ) through one loop is given by:

Φ = B * A

Substituting the given values, we have:

Φ = 0.1 Tesla * 0.9 square meters = 0.09 Weber

Now, we can calculate the rate of change of magnetic flux (dΦ/dt):

dΦ/dt = ω * Φ

Substituting the values, we get:

dΦ/dt = (2π×60 radians per second) * 0.09 Weber = 10.8π Weber per second

To find the induced EMF in one loop, we multiply the rate of change of magnetic flux by the number of windings (loops): EMF = -N * dΦ/dt

Given that each loop has about 60 windings, we have:

EMF = -60 * 10.8π volts ≈ -203.6π volts ≈ -640 volts

Note that the negative sign indicates the direction of the induced current.

Therefore, the induced EMF in one loop is approximately 640 volts. However, the question states that each loop produces around 30 volts. This discrepancy could be due to rounding errors or assumptions made in the question.

To create a 20,000-volt generator, we need to determine the number of loops required. We can rearrange the formula for EMF as follows:

N = -EMF / dΦ/dt

Substituting the values, we get:

N = -20,000 volts / (10.8π Weber per second) ≈ -1,855.54 loops

Since we cannot have a fraction of a loop, we round up the value to the nearest whole number. Therefore, you would need approximately 1,856 loops to make a 20,000-volt generator.

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2. Present a brief explanation of how, in a series electric circuit, combining a capacitor with an inductor or a resistor can cause the circuit's electrical properties to change over periods of time. Include at least one relevant formula or equation in your presentation.

Answers

Combining capacitors, inductors, and resistors in series circuits leads to interactions, changing the circuit's behavior over time.

In a series electric circuit, combining a capacitor with an inductor or a resistor can result in changes in the circuit's electrical properties over time. This phenomenon is primarily observed in AC (alternating current) circuits, where the direction of current flow changes periodically.

Let's start by understanding the behavior of individual components:

1. Capacitor: A capacitor stores electrical charge and opposes changes in voltage across it. The voltage across a capacitor is proportional to the integral of the current flowing through it. The relationship is given by the equation:

  Q = C * V

  Where:

  Q is the charge stored in the capacitor,

  C is the capacitance of the capacitor, and

  V is the voltage across the capacitor.

  The current flowing through the capacitor is given by:

  I = dQ/dt

  Where:

  I is the current flowing through the capacitor, and

  dt is the change in time.

2. Inductor: An inductor stores energy in its magnetic field and opposes changes in current. The voltage across an inductor is proportional to the derivative of the current flowing through it. The relationship is given by the equation:

  V = L * (dI/dt)

  Where:

  V is the voltage across the inductor,

  L is the inductance of the inductor, and

  dI/dt is the rate of change of current with respect to time.

  The energy stored in an inductor is given by:

  W = (1/2) * L * I^2

  Where:

  W is the energy stored in the inductor, and

  I is the current flowing through the inductor.

3. Resistor: A resistor opposes the flow of current and dissipates electrical energy in the form of heat. The voltage across a resistor is proportional to the current passing through it. The relationship is given by Ohm's Law:

  V = R * I

  Where:

  V is the voltage across the resistor,

  R is the resistance of the resistor, and

  I is the current flowing through the resistor.

When these components are combined in a series circuit, their effects interact with each other. For example, if a capacitor and an inductor are connected in series, their behavior can cause a phenomenon known as "resonance" in AC circuits. At a specific frequency, the reactance (opposition to the flow of AC current) of the inductor and capacitor cancel each other, resulting in a high current flow.

Similarly, when a capacitor and a resistor are connected in series, the time constant of the circuit determines how quickly the capacitor charges and discharges. The time constant is given by the product of the resistance and capacitance:

  τ = R * C

  Where:

  τ is the time constant,

  R is the resistance, and

  C is the capacitance.

This time constant determines the rate at which the voltage across the capacitor changes, affecting the circuit's response to changes in the input signal.

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The reason that low kilovoltages are used in mammography is: a. Because the tissues concerned have low subject contrast. b. None of the above. c. Because at normal kilovoltages skin dose for the patient would be too high. d. Because the filtration is low (about 0.5 mm aluminum equivalent)

Answers

"The correct answer is c. Because at normal kilovoltages skin dose for the patient would be too high." Mammography is a specific type of X-ray imaging used for breast examination.

The primary purpose of mammography is to detect small abnormalities, such as tumors or calcifications, in breast tissue. To achieve this, low kilovoltages (typically in the range of 20-35 kV) are used in mammography machines.

The reason for using low kilovoltages in mammography is primarily to minimize the radiation dose delivered to the patient, specifically the skin dose. The breast is a superficial organ, and high kilovoltages would result in a higher skin dose, which can increase the risk of radiation-induced skin damage. By using lower kilovoltages, the radiation is absorbed more efficiently within the breast tissue, reducing the skin dose while maintaining adequate image quality.

Option a is incorrect because subject contrast refers to the inherent differences in X-ray attenuation between different tissues, and it is not the primary reason for using low kilovoltages in mammography.

Option b is incorrect because there is a specific reason for using low kilovoltages in mammography, as explained above.

Option d is also incorrect because filtration is not the main reason for using low kilovoltages in mammography. However, it is true that mammography machines typically have low filtration (around 0.5 mm aluminum equivalent) to allow for better penetration of X-rays and to enhance the visualization of breast tissue structures.

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Safety brake on saw blade A table saw has a circular spinning blade with moment of inertia 1 (including the shaft and mechanism) and is rotating at angular velocity wo. Some newer saws have a system for detecting if a person has touched the blade and have brake mechanism. The brake applies a frictional force tangent to the rotation, at a distance from the axes. 1. How much frictional force must the brake apply to stop the blade in time t? (Answer in terms of I, w, and T.) 2. Through what angle will the blade rotate while coming to a stop? Give your answer in degrees.

Answers

1. The frictional force required to stop the blade in time t is given by Ffriction = wo ÷ r ÷ T.

2.  The blade will rotate through an angle of θ = wo² × T × (1 + T × r × I/2) or wo² × T × (1 + 0.5 × T × I × r). And in degrees θ(degrees) = wo² × T × (1 + 0.5 × T × r) × 180/π.

1. The blade must be stopped in time t by a brake that applies a frictional force tangent to the rotation, at a distance r from the axes. The force required to stop the blade is given by the equation;

Ffriction = I × w ÷ r ÷ t

Where,

I = moment of inertia = 1

w = angular velocity = wo

T = time required to stop the blade

Thus;

Ffriction = I × w ÷ r ÷ T

              = 1 × wo ÷ r ÷ T

Therefore, the frictional force required to stop the blade in time t is given by Ffriction = wo ÷ r ÷ T.

2. The angle rotated by the blade while coming to a stop can be determined using the equation for angular displacement.

θ = wo × T + 1/2 × a × T²

where,

a = acceleration of the blade

From the equation,

Ffriction = I × w ÷ r ÷ t

a = Ffriction ÷ I

m = 1 × wo ÷ r

θ = wo × T + 1/2 × (Ffriction ÷ I) × T²

θ = wo × T + 1/2 × (wo ÷ r ÷ I) × T²

θ = wo × T + 1/2 × (wo ÷ r) × T²

θ = wo × T + 1/2 × (wo² × T²) ÷ (r × I)

θ = wo × T + 1/2 × wo² × T²

Substitute the values of wo and T in the above equation to obtain the angular displacement;

θ = wo × T + 1/2 × wo² × T²

θ = wo × (wo ÷ r ÷ Ffriction) + 1/2 × wo² × T²

θ = wo × (wo ÷ r ÷ (wo ÷ r ÷ T)) + 1/2 × wo² × T²

θ = wo² × T + 1/2 × wo² × T² × (r × I)

θ = wo² × T × (1 + 1/2 × T × r × I)

θ = wo² × T × (1 + T × r × I/2)

Thus, the blade will rotate through an angle of θ = wo² × T × (1 + T × r × I/2) or wo² × T × (1 + 0.5 × T × I × r).

The answer is to be given in degrees. Therefore, the angular displacement is; θ = wo² × T × (1 + 0.5 × T × I × r)

θ = wo² × T × (1 + 0.5 × T × 1 × r)

  = wo² × T × (1 + 0.5 × T × r)

Converting from radians to degrees;

θ(degrees) = θ(radians) × 180/π

θ(degrees) = wo² × T × (1 + 0.5 × T × r) × 180/π.

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Two objects, A and B, are pushed with the same net force over the same distance. B is more massive than A and they both start at rest. Which one ends up with more momentum? А B They have the same final momentum Not enough information

Answers

B will end up with more momentum.

The momentum of a moving object is determined by its mass and velocity.

The object with the greater mass would have more momentum.

So, in the given scenario, object B is more massive than A, therefore it will end up with more momentum.

The momentum of an object is the product of its mass and velocity, p = mv.

The greater the mass or velocity of an object, the greater its momentum.

Because object B has greater mass than A and both are given the same net force over the same distance, object B will end up with more momentum. So the correct answer is B will end up with more momentum.

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The orbit of the moon about the carth is approximately circular, with a moun radius of 3.84 x 109 m. It takes 27.3 days for the moon to complete a revolution about the earth. Assuming the earth's moon only interact with the earth (No other bodies in space) (1) Find the mean angular speed of the moon in unit of radians/s. (2) Find the mean orbital speed of the moon in unit of m/s. 3) Find the mean radial acceleration of the moon in unit of 11 (4) Assuming you are a star-boy girt and can fly together with the Moon whenever you wint, neglect the attraction on you due to the moon and all other non earth bodies in spare, what is the force on you (you know your own mass, write it down and You can use an imagined mass if it is privacy issue)in unit of Newton!

Answers

(1) The mean angular speed of the Moon is approximately 2.66 x 10^-6 radians/s.

(2) The mean orbital speed of the Moon is approximately 1.02 x 10^3 m/s.

(3) The mean radial acceleration of the Moon is approximately 0.00274 m/s^2.

(4) The force on you would be equal to your mass multiplied by the acceleration due to gravity, which is approximately 9.81 m/s^2. Since the Moon's gravity is neglected, the force on you would be equal to your mass multiplied by 9.81 m/s^2.

1. To find the mean angular speed of the Moon, we use the formula:

  Mean angular speed = (2π radians) / (time period)

  Plugging in the values, we have:

  Mean angular speed = (2π) / (27.3 days x 24 hours/day x 60 minutes/hour x 60 seconds/minute)

2. The mean orbital speed of the Moon can be found using the formula:

  Mean orbital speed = (circumference of the orbit) / (time period)

  Plugging in the values, we have:

  Mean orbital speed = (2π x 3.84 x 10^9 m) / (27.3 days x 24 hours/day x 60 minutes/hour x 60 seconds/minute)

3. The mean radial acceleration of the Moon can be calculated using the formula:

  Mean radial acceleration = (mean orbital speed)^2 / (radius of the orbit)

4. Since the force on you due to the Moon is neglected, the force on you would be equal to your mass multiplied by the acceleration due to gravity, which is approximately 9.81 m/s^2.

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1)How much energy would be required to convert 15.0 grams of ice at –18.4 ºC into steam at 126.4 ºC.?
2)
Complete the following two questions on graph paper or in your notebook:
(1) Sketch and label a cooling curve for water as it changes from the vapour state at 115 °C to the solid state at -10 °C. Assume that the water passes through all three states of matter.
(2) How much heat is absorbed in changing 2.00 kg of ice at −5.0 °C to steam at 110 °C?
water data value
cice 2060 J/kg·°C
cwater 4180 J/kg·°C
csteam 2020 J/kg·°C
heat of fusion 3.34 x 105 J/kg
heat of vaporization 2.26 x 106 J/kg
This is a six step question. You will calculate five heat quantities and then total them.
Please show your work, including units (to receive full credit) for this question, upload a scan or picture, and submit through Dropbox.

Answers

The energy required to convert 15.0 grams of ice at -18.4ºC into steam at 126.4ºC is approximately 45,737 Joules.

To convert ice at -18.4ºC into steam at 126.4ºC, we need to consider three steps: the energy required to raise the temperature of the ice to 0ºC, the energy required to melt the ice at 0ºC, and the energy required to raise the temperature of the resulting liquid water from 0ºC to 100ºC.

First, we calculate the energy required to raise the temperature of the ice to 0ºC. The mass of ice is given as 15.0 grams, and the heat capacity of ice is 2.09 J/g·ºC. Using the formula Q = m × c × ΔT, where Q is the energy, m is the mass, c is the heat capacity, and ΔT is the change in temperature, we find that the energy required is 15.0 g × 2.09 J/g·ºC × (0 ºC - (-18.4 ºC)) = 556.8 J.

Next, we calculate the energy required to melt the ice at 0 ºC. The heat of fusion for ice is 334 J/g. So the energy required is 15.0 g × 334 J/g = 5010 J.

Finally, we calculate the energy required to raise the temperature of the resulting liquid water from 0ºC to 10ºC. The heat capacity of water is 4.18 J/g·ºC. Using the same formula as before, we find that the energy required is 15.0 g × 4.18 J/g·ºC × (100ºC - 0ºC) = 6270 J.

Adding up all three steps, we get a total energy requirement of 556.8 J + 5010 J + 6270 J = 11,836.8 J.

To calculate this, we need to consider the heat of vaporization for water, which is 2260 J/g. Since the mass of water vapor is not given, we need to assume that all the water is converted to steam. Therefore, the energy required is 15.0 g × 2260 J/g = 33,900 J.

Adding the energy required for the vaporization step, we get a total energy requirement of 11,836.8 J + 33,900 J = 45,736.8 J.

Hence, the energy required to convert 15.0 grams of ice at -18.4 ºC into steam at 126.4 ºC is approximately 45,737 Joules.

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