The electric potential in a particular region of space varies only as a function of y-position and is given by the function V(y)1.69y2 +15.6y 52.5) Volts. Calculate the magnitude of the electric field at the position y - 22.5 meters

Complete Question

The distance between the objective and eyepiece lenses in a microscope is 19 cm . The objective lens has a focal length of 5.5 mm .

What eyepiece focal length will give the microscope an overall angular magnification of 300?

Answer:

The  eyepiece focal length is  [tex]f_e = 0.027 \ m[/tex]

Explanation:

From the question we are told that

    The focal length is  [tex]f_o = 5.5 \ mm = -0.0055 \ m[/tex]

This negative sign shows the the microscope is diverging light

     The  angular magnification is [tex]m = 300[/tex]

     The  distance between the objective and the eyepieces lenses is  [tex]Z = 19 \ cm = 0.19 \ m[/tex]

Generally the magnification is mathematically represented as

        [tex]m = [\frac{Z - f_e }{f_e}] [\frac{0.25}{f_0} ][/tex]

Where [tex]f_e[/tex] is the eyepiece focal length of the microscope

  Now  making [tex]f_e[/tex] the subject  of the formula

         [tex]f_e = \frac{Z}{1 - [\frac{M * f_o }{0.25}] }[/tex]

substituting values

        [tex]f_e = \frac{ 0.19 }{1 - [\frac{300 * -0.0055 }{0.25}] }[/tex]

         [tex]f_e = 0.027 \ m[/tex]

Answer:

A. Her total angular momentum has decreased

Explanation:

Total angular momentum is the product of her moment of inertia and angular velocity. In this scenario it doesn’t decrease but rather remains constant as the movement of the arms doesn’t have any effect on the total angular momentum.

The movement of the arm under certain conditions however has varying effects and changes on parameters such as the moment of inertia and the angular speed.

Answer:

(a) Distance between deer and car = 5 m

(b) Vmax = 21.92 m/s

Explanation:

a.

First we calculate distance covered during response time:

s₁ = vt   --------- equation 1

where,

s₁ = distance covered during response time = ?

v = speed of car = 20 m/s

t = response time = 0.5 s

Therefore,

s₁ = (20 m/s)(0.5 s)

s₁ = 10 m

Now, we calculate the distance covered by the car during deceleration. Using 3rd equation of motion:

2as₂ = Vf² - Vi²

s₂ = (Vf² - Vi²)/2a ------ eqation 2

where,

a = deceleration = - 10 m/s²

s₂ = Distance covered during deceleration = ?

Vf = Final Velocity = 0 m/s (since car finally stops)

Vi = Initial Velocity = 20 m/s

Therefore,

s₂ = [(0 m/s)² - (20 m/s)²]/2(-10 m/s²)

s₂ = (400 m²/s²)/(20 m/s²)

s₂ = 20 m

thus, the total distance covered by the car before coming to rest is given as:

s = s₁ + s₂

s = 10 m + 20 m

s = 30 m

Now, the distance between deer and car, when it comes to rest, can be calculated as:

Distance between deer and car = 35 m - s = 35 m - 30 m

Distance between deer and car = 5 m

b.

Since, the distance covered by the car in total must be equal to 35 m at maximum. Therefore,

s₁ + s₂ = 35 m

using equation 1 and equation 2 from previous part:

Vi t + (Vf² - Vi²)/2a = 35 m

Vi(0.5 s) + [(0 m/s)² - Vi²]/2(-10 m/s²) = 35 m

0.5 Vi + 0.05 Vi² = 35

0.05 Vi² + 0.5 Vi - 35 = 0

solving this quadratic equation, we get:

Vi = - 31.92 m/s  (OR)  Vi = 21.92 m/s

For maximum velocity:

Vmax = 21.92 m/s

Answer:

The maximum speed of the mass is 4.437 m/s.

Explanation:

Given;

length of pendulum, L = 1.62 m

attached mass, m = 117 g

angle of inclination, θ = 38°

This mass was raised to a height of

h = 1.62 - cos38° = 1.0043 m

Apply the principle of conservation of mechanical energy

PE = KE

mgh = ¹/₂mv²

v  = √(2gh)

v = √(2 * 9.8 * 1.0043)

v = 4.437 m/s.

Therefore, the maximum speed of the mass is 4.437 m/s.

Answer:

7.63 x 10^-8ohmm

Explanation:

resistivity of the wire = 7.63 x 10^-8ohmm

Explanation:

a) How much work is done by gravity?

b) How much work is done by tension?

w = f x d

w = 950 x 0.05 x 5.5 = 261.25j

Answer:

60.8 cm²

Explanation:

The charge density, σ on the surface is σ = Q/A where q = charge = 87.6 pC = 87.6 × 10⁻¹² C and A = area = 65.2 cm² = 65.2 × 10⁻⁴ m².

σ = Q/A = 87.6 × 10⁻¹² C/65.2 × 10⁻⁴ m² = 1.34 × 10⁻⁸ C/m²

Now, the charge through the Gaussian surface is q = σA' where A' is the charge in the Gaussian surface.

Since the flux, Ф = 9.20 Nm²/C and Ф = q/ε₀ for a closed Gaussian surface

So, q = ε₀Ф = σA'

ε₀Ф = σA'

making A' the area of the Gaussian surface the subject of the formula, we have

A' = ε₀Ф/σ

A' = 8.854 × 10⁻¹² F/m × 9.20 Nm²/C ÷ 1.34 × 10⁻⁸ C/m²

A' = 81.4568/1.34 × 10⁻⁴ m²

A' = 60.79 × 10⁻⁴ m²

A' ≅ 60.8 cm²

The flux through the Gaussian surface is 9.20 N⋅m^2/C then the surface area of the Gaussian Sheet is 60.76 square cm.

A certain amount of electrons in excess or defect is called a charge. Charge density is the amount of charge distributed over per unit of volume.

Given that, for a thin sheet of insulating material, the charge Q is 87.6 pC and surface area A is 65.2 square cm. Then the charge density for a thin sheet is given below.

[tex]\sigma = \dfrac {Q}{A}[/tex]

[tex]\sigma = \dfrac {87.6\times 10^{-12}}{65;.2\times 10^{-4}}[/tex]

[tex]\sigma = 1.34\times 10^{-8} \;\rm C/m^2[/tex]

Thus the charge density for a thin sheet of insulating material is [tex]1.34\times 10^{-8} \;\rm C/m^2[/tex].

Now, the flux through the Gaussian surface is 9.20 N⋅m^2/C. The charge over the Gaussian Surface is given as below.

[tex]Q' = \sigma A'[/tex]

Where Q' is the charge at the Gaussian Surface, A' is the surface area of the Gaussian surface and [tex]\sigma[/tex] is the charge density.

For the closed Gaussian Surface, Flux is given below.

[tex]\phi = \dfrac {Q'}{\epsilon_\circ}[/tex]

Hence the charge can be written as,

[tex]Q' = \phi\epsilon_\circ[/tex]

So the charge can be given as below.

[tex]Q' = \phi\epsilon_\circ = \sigma A'[/tex]

Then the surface area of the Gaussian surface is given below.

[tex]A' = \dfrac {\phi\epsilon_\circ}{\sigma}[/tex]

Substituting the values in the above equation,

[tex]A' = \dfrac {9.20 \times 8.85\times 10^{-12}}{1.38\times 10^{-8}}[/tex]

[tex]A' =0.006076\;\rm m^2[/tex]

[tex]A' = 60.76 \;\rm cm^2[/tex]

Hence we can conclude that the area of the Gaussian Surface is 60.76 square cm.

To know more about the charge and charge density, follow the link given below.

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Answer:

A)   E = 0N/C

B)   0i + 0^^j

C)   F = 0N

D)   0^i  + 0^j

Explanation:

You assume that the rings are in the zy plane but in different positions.

Furthermore, you can consider that the origin of coordinates is at the midway between the rings.

A) In order to calculate the magnitude of the electric field at the middle of the rings, you take into account that the electric field produced by each ring at the origin is opposite to each other and parallel to the x axis.

You use the following formula for the electric field produced by a charge ring at a perpendicular distance of r:

[tex]E=k\frac{rQ}{(r+R^2)^{3/2}}[/tex]               (1)

k: Coulomb's constant = 8.98*10^9Nm^2/C

Q: charge of the ring

r: perpendicular distance to the center of the ring

R: radius of the ring

You use the equation (1) to calculate the net electric field at the midpoint between the rings:

[tex]E=k\frac{rQ}{(r^2+R^2)^{3/2}}-k\frac{rQ}{(r^2+R^2)^{3/2}}=0\frac{N}{C}[/tex]

The electric field produced by each ring has the same magnitude but opposite direction, then, the net electric field is zero.

B) The direction of the electric field is 0^i + 0^j

C) The magnitude of the force on a proton at the midpoint between the rings is:

[tex]F=qE=q(0N/C)=0N[/tex]

D) The direction of the force is 0^i + 0^j

Part A: The magnitude of the electric field generated at the midpoint between two rings is equal that is 0 N/C.

Part B: The direction of the electric field at the midpoint is opposite.

Part C: The magnitude of the force generated on a proton at the midpoint between two rings is equal that is 0 N.

Part D: The direction of the force on a proton at the midpoint is opposite.

An electric field is defined as the region that surrounds electrically charged particles and exerts a force on all other charged particles within the region, either attracting or repelling them.

Given that diameter of the ring is 10 m and they are 25 m apart from each other. The charge on the left ring is -25nC and on the right ring is 25nC. The electric field can be given as below.

[tex]E = \dfrac {kQ}{(r+R)^2}\\ [/tex]

Where Q is the charge, r is the radius of the ring, R is the mid-point distance and k is the constant.

Part A

The electric field at the mid-point will be the sum of the electric field generated by both the rings. Substituting the values in the above equation,

[tex]E = \dfrac {8.9\times 10^9\times 25}{(10 +12.5)^2}+\dfrac {8.9\times 10^9\times (-25)}{(10 +12.5)^2}[/tex]

[tex]E = \dfrac {222.5\times 10^9}{506.25} - \dfrac {222.5\times 10^9}{506.25}\\ [/tex]

[tex]E = 0\;\rm N/C[/tex]

Hence we can conclude that both the rings generate the electric field with the same magnitude but they are opposite in direction.

Part B

The electric field at the mid-point is 0 N/C. In the vector form, the electric field can be given as below.

[tex]E = 0i+0j[/tex]

The vector form shows that the electric field at the mid-point between the two rings has the same magnitude but is opposite in direction.

Part C

The force can be given as below.

[tex]F = qE[/tex]

[tex]F = 0 \;\rm N[/tex]

If the electric field at the mid-point is zero, then the force at the mid-point will be zero.

Part D

The vector form of the force at the midpoint is given below.

[tex]F = 0i+0j[/tex]

Hence we can conclude that at the midpoint of two rings, the electric field generates an equal force on the proton but in opposite direction. Hence the net force will be zero.

To know more about the electric field, follow the link given below.

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Answer:

The initial velocity of spaceship P was u₁ = 5.06 m/s

Explanation:

In an elastic collision between two bodies the expression for the final velocity of the second body is given as follows:

[tex]V_{2} = \frac{(m_{2}-m_{1}) }{(m_{1}+m_{2})}u_{2} + \frac{2m_{1} }{(m_{1}+m_{2})}u_{1}[/tex]

Here, subscript 1 is used for spaceship P and subscript 2 is used for spaceship T. In this equation:

V₂ = Final Speed of Spaceship T = 8.1 m/s

m₁ = mass of spaceship P = 4 M

m₂ = mass of spaceship T = M

u₁ = Initial Speed of Spaceship P = ?

u₂ = Initial Speed of Spaceship T = 0 m/s

Using these values in the given equation, we get:

[tex]8.1 m/s = \frac{M-4M }{4M+M}(0 m/s) + \frac{2(4M) }{4M+M}u_{1}[/tex]

8.1 m/s = (8 M/5 M)u₁

u₁ = (5/8)(8.1 m/s)

u₁ = 5.06 m/s

Answer:

Ф = 28.9°

Explanation:

given:

radius (r) = 117m

velocity (v) = 25.1 m/s

required: angle Ф

Ф = inv tan (v² / (r * g))      we know that g = 9.8

Ф = inv tan (25.1² / (117 * 9.8))

Ф = 28.9°

Answer:

I believe it is they will weigh the same

Explanation:

Center of gravity is the axis on which the mass rotates evenly if I remember correctly from AP Physics

The head side is heavier than the handle side. - this will be discovered.

Theoretically, the body's center of gravity is where all of the weight is believed to be concentrated. Knowing the centre of gravity is crucial because it may be used to forecast how a moving object will behave when subjected to the effects of gravity. In designing immobile constructions like buildings and bridges, it is also helpful.

We know that center of gravity is  close to some particular point refers the mass of the point is greater then others. It is given that: The center of gravity of an ax is on the centerline of the handle, close to the head.

So, we can conclude that the head side of the ax is heavier than the handle side of it.

Learn more about center of gravity here:

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Answer:

Tension in the cable = 200.655 × 10^(18) N

Explanation:

The tension in the cable will be the force on the string which is caused by when it holds against the centripetal acceleration.

Now, the formula for centripetal acceleration is given by;

a = v²/R

Now, let's find the velocity.

The circumference is 2πR

We are given that R = 3.85 x 10^(8) m

So, Circumference = 2 × π × 3.85 x 10^(8) = 24.19 × 10^(8) m

We are told the period of the moon orbit is 27.3 days

Distance travelled per day = Circumference/ period

Distance travelled per day = (24.19 × 10^(8))/27.3 = 88608058.608 m/day

Now, a day has 24 hours = 24 × 60 × 60 = 86400 seconds

Thus,

Distance travelled per seconds =

88608058.608/86400 = 1025.556 m/sec

So, from a_c = v²/r,

a_c = 1025.556²/3.85x10^8

a_c = 0.00273 m/s²

Now, Force = Mass x Acceleration = ma

We are given mass = 7.35 x 10^(22) kg

Plugging in the relevant values, we have;

F = 0.00273 x 7.35 x 10^(22)

F = 200.655 × 10^(18) N

Answer:

The wavelength will be 33.9 cm

Explanation:

Given;

frequency of the wave, F = 1200 Hz

Tension on the wire, T = 800 N

wavelength, λ = 39.1 cm

[tex]F = \frac{ \sqrt{\frac{T}{\mu} }}{\lambda}[/tex]

Where;

F is the frequency of the wave

T is tension on the string

μ is mass per unit length of the string

λ is wavelength

[tex]\sqrt{\frac{T}{\mu} } = F \lambda\\\\\frac{T}{\mu} = F^2\lambda^2\\\\\mu = \frac{T}{F^2\lambda^2} \\\\\frac{T_1}{F^2\lambda _1^2} = \frac{T_2}{F^2\lambda _2^2} \\\\\frac{T_1}{\lambda _1^2} = \frac{T_2}{\lambda _2^2}\\\\T_1 \lambda _2^2 = T_2\lambda _1^2\\\\[/tex]

when the tension is decreased to 600 N, that is T₂ = 600 N

[tex]T_1 \lambda _2^2 = T_2\lambda _1^2\\\\\lambda _2^2 = \frac{T_2\lambda _1^2}{T_1} \\\\\lambda _2 = \sqrt{\frac{T_2\lambda _1^2}{T_1}} \\\\\lambda _2 = \sqrt{\frac{600* 0.391^2}{800}}\\\\\lambda _2 = \sqrt{0.11466} \\\\\lambda _2 =0.339 \ m\\\\\lambda _2 =33.9 \ cm[/tex]

Therefore, the wavelength will be 33.9 cm

Answer:

Explanation:

a = F/m = 16.8t/45 = 0.373*t

v = ∫a*dt = ∫(0.373*t)dt = 0.1865*t²

x = ∫v*dt = 0.1865∫(t²)dt = 0.062*t³

For t = 5.0 sec,

x = 0.062*(5)^3 = 7.75 m

Answer:

the right answer is option C

Explanation:

at the horizon

The position of rainbows depends on the some factors like wavelength of light and position of sun. So, it can occur at the horizon or above the horizon at the 40 - 42 Degrees. Hence, option (A) and (C) are correct.

The given problem is based on the concepts and fundamentals of Rainbow. The rainbow is the band of seven of seven colors namely, violet, indigo, blue, green, yellow, orange and red.

Thus, we can conclude that the position of rainbows depends on the some factors like wavelength of light and position of sun. So, it can occur at the horizon or above the horizon at the 40 - 42 Degrees. Hence, option (A) and (C) are correct.

Learn more about the Rainbows here:

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Answer:

Explanation:

Let the common  acceleration be a .

Considering motion of mL

20.5g - TL = 20.5 a

Considering motion of mR

TR - 11.7g = 11.7 a

TR-TL + 20.5 g- 11.7 g = (20.5 + 11.7 ) a

TL-TR = 8.8 g - 32.2 a

Considering motion of wheel

( TL - TR ) x rp = 1/2 x mp x rp² x α where α is angular acceleration of wheel

=  ( TL - TR )  = 1/2 x mp x a      (  a = α x rp )

Putting the values from equation above

8.8g - 32.2a = .5 mp a

8.8g = 32.2 a + .5 x 5.33 a

8.8 g = 34.865 a

a = 2.47 m /s²

20.5 g - TL = 20.5 a

TL = 20.5 x 9.8 - 20.5 x 2.47

= 150.26 N

TR - 11.7g = 11.7 a

TR - 11.7 x 9.8  = 11.7 x 2.47

TR = 143.56 N  

Answer:

superscript

Explanation:

When looking at the chemical symbol, the charge of the ion is displayed as the Superscript. This is because the charge of ions is usually written up on the chemical symbol while the atom/molecule is usually written down the chemical symbol. The superscript refers to what is written up on the formula while the subscript is written down on the formula.

An example is H2O . The 2 present represents two molecule of oxygen and its written as the subscript while Fe2+ in which the 2+ is written up is known as the superscript.

Answer:

superscript

Explanation:

Answer:

6 atm

Explanation:

PV = PV

(4 atm) (3 L) = P (2 L)

P = 6 atm

Answer:

the correct answer is d Beats

Explanation:

when two sound waves interfere time has different frequencies, the result is the sum of the waves is

       y = 2A cos 2π (f₁-f₂)/2    cos 2π (f₁ + f₂)/2

where in this expression the first part represents the envelope and the second part represents the pulse or beatings of the wave.

When examining the correct answer is d Beats

Answer:

the image is virtual and erect and the lens divergent; therefore the correct answer is C

Explanation:

In a thin lens the magnification given by

      m = h '/ h = - q / p

where h ’is the height of the image, h is the height of the object, q is the distance to the image and p is the distance to the object.

It indicates that the object is straight and is placed at a distance p> f

analyze the situation tells us that the magnification is positive so the distance to the image must be negative, that is, that the image is on the same side as the object.

Consequently the lens must be divergent

The magnification value is

          0.4 = h ’/ h

          h ’= 0.4 h

therefore the erect images

therefore the image is virtual and erect and the lens divergent; therefore the correct answer is C

Answer:

The apparent weight of the pilot is 3311 N

Explanation:

Given;

radius of the vertical circle, r = 600 m

speed of the plane, v = 150 m/s

mass of the pilot, m = 70 kg

Weight of the pilot due to his circular motion;

[tex]W= F_v\\\\F_v = \frac{mv^2}{r} \\\\F_v = \frac{70*150^2}{600} \\\\F_v = 2625 \ N[/tex]

Real weight of the pilot;

[tex]W_R = mg\\\\W_R = 70 *9.8\\\\W_R = 686 \ N[/tex]

Apparent weight - Real weight of pilot = weight due to centripetal force

[tex]F_N - mg = \frac{mv^2}{r} \\\\F_N = \frac{mv^2}{r} + mg\\\\F_N = 2625 \ N + 686 \ N\\\\F_N = 3311\ N[/tex]

Therefore, the apparent weight of the pilot is 3311 N

Answer:

he correct answers are a, b

Explanation:

In the two-slit interference phenomenon, the expression for interference is

          d sin θ= m λ                       constructive interference

          d sin θ = (m + ½) λ             destructive interference

in general this phenomenon occurs for small angles, for which we can write

           tanθ = y / L

           tan te = sin tea / cos tea = sin tea

           sin θ = y / La

un

derestimate the first two equations.

Let's do the calculation for constructive interference

         d y / L = m λ

the distance between maximum clos is and

         y = (me / d) λ

this is the position of each maximum, the distance between two consecutive maximums

         y₂-y₁ = (L   2/d) λ - (L 1 / d) λ₁          y₂ -y₁ = L / d λ

examining this equation if the wavelength decreases the value of y also decreases

the same calculation for destructive interference

         d y / L = (m + ½) κ

         y = [(m + ½) L / d] λ

again when it decreases the decrease the distance

the correct answers are a, b

Answer:

3) Transmitted intensity of light if unpolarized light passes through a single polarizing filter = 40 W/m²

- Transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup described = 7.5 W/m²

Explanation:

Complete Question

3) What is the transmitted intensity of light if unpolarized light passes through a single polarizing filter and the initial intensity is 80 W/m²?

- What is the transmitted intensity of light if an additional polarizer is added perpendicular to the first polarizer in the setup described in Question 3 (the setup)? Show all work in your answer.

The image of this setup attached to this question as obtained from online is attached to this solution.

Solution

3) When unpolarized light passes through a single polarizer, the intensity of the light is cut in half.

Hence, if the initial intensity of unpolarized light is I₀ = 80 W/m²

The intensity of the light rays thay pass through the first single polarizer = I₁ = (I₀/2) = (80/2) = 40 W/m²

- According to Malus' law, the intensity of transmitted light through a polarizer is related to the intensity of the incident light and the angle at which the polarizer is placed with respect to the major axis of the polarizer before the current polarizer of concern.

I₂ = I₁ cos² θ

where

I₂ = intensity of light that passes through the second polarizer = ?

I₁ = Intensity of light from the first polarizer that is incident upon the second polarizer = 40 W/m²

θ = angle between the major axis of the first and second polarizer = 30°

I₂ = 40 (cos² 30°) = 40 (0.8660)² = 30 W/m²

In the same vein, the intensity of light that passes through the third/additional polarizer is related to the intensity of light that passes through the second polarizer and is incident upon this third/additional polarizer through

I₃ = I₂ cos² θ

I₃ = intensity of light that passes through the third/additional polarizer = ?

I₂ = Intensity of light from the second polarizer that is incident upon the third/additional polarizer = 30 W/m²

θ = angle between the major axis of the second and third/additional polarizer = 60° (although, it is 90° with respect to the first polarizer, it is the angle it makes with the major axis of the second polarizer, 60°, that matters)

I₃ = 30 (cos² 60°) = 30 (0.5)² = 7.5 W/m²

Hope this Helps!!!

Answer:

[tex]F=-2A[\frac{1}{x^3}\hat{i}+\frac{1}{y^3}\hat{j}][/tex]

Explanation:

You have the following potential energy function:

[tex]U(x,y)=A[\frac{1}{x^2}+\frac{1}{y^2}}][/tex]           (1)

A > 0 constant

In order to find the force in terms of the unit vectors, you use the gradient of the potential function:

[tex]\vec{F}=\bigtriangledown U(x,y)=\frac{\partial}{\partial x}U\hat{i}+\frac{\partial}{\partial y}U\hat{j}[/tex]         (2)

Then, you replace the expression (1) into the expression (2) and calculate the partial derivatives:

[tex]\vec{F}=A\frac{\partial}{\partial x}[\frac{1}{x^2}+\frac{1}{y^2}]} \hat{i}+A\frac{\partial}{\partial x}[\frac{1}{x^2}+\frac{1}{y^2}]\hat{j}\\\\\vec{F}=A(-2x^{-3})\hat{i}+A(-2y^{-3})\hat{j}\\\\F=-2A[\frac{1}{x^3}\hat{i}+\frac{1}{y^3}\hat{j}][/tex](3)

The result obtained in (3) is the force expressed in terms of the unit vectors, for the potential energy function U(x,y).

Answer:

Frequency, [tex]f=2.18\times 10^9\ Hz[/tex]

Explanation:

We have,

Speed of radio waves is [tex]3\times 10^8\ m/s[/tex]

Wavelength of radio waves is [tex]\lambda=0.137\ m[/tex]

It is required to find the frequency of the radio waves. The speed of a wave is given by :

[tex]v=f\lambda\\\\f=\dfrac{v}{\lambda}\\\\f=\dfrac{3\times 10^8}{0.137}\\\\f=2.18\times 10^9\ Hz[/tex]

So, the frequency of the radio wave is [tex]2.18\times 10^9\ Hz[/tex].

Answer:

  K_{f} / K₀ =1.12

Explanation:

This problem must work using the conservation of angular momentum (L), so that the moment is conserved in the system all the forces must be internal and therefore the torque is internal and the moment is conserved.

Initial moment. With arms outstretched

         L₀ = I₀ w₀

the wo value is 5.0 rad / s

final moment. After he shrugs his arms

         [tex]L_{f}[/tex] = I_{f}  w_{f}

indicate that the moment of inertia decreases by 11%

        I_{f} = I₀ - 0.11 I₀ = 0.89 I₀

        L_{f} = L₀

        I_{f} w_{f}  = I₀ w₀

        w_{f} = I₀ /I_{f}    w₀

let's calculate

        w_{f} = I₀ / 0.89 I₀   5.0

        w_{f} = 5.62 rad / s

Having these values ​​we can calculate the change in kinetic energy

         [tex]K_{f}[/tex] / K₀ = ½ I_{f} w_{f}² (½ I₀ w₀²)

         K_{f} / K₀ = 0.89 I₀ / I₀ (5.62 / 5)²

         K_{f} / K₀ =1.12

Answer:

16 years.

Explanation:

Using Kepler's third Law.

P2=D^3

P=√d^3

Where P is the orbital period and d is the distance from the sun.

From the question the semi major axis of the asteroid is 4 AU= distance. The distance is always express in astronomical units.

P=?

P= √4^3

P= √256

P= 16 years.

Orbital period is 16 years.

Answer:

the fraction of submerged volume is equal to the ratio of the densities of the body between the density of the fluid.

Explanation:

This is a fluid mechanics problem, where as the boat is in equilibrium with the pushing force we can write Newton's second law

                  B- W = 0

                  B = W

the thrust force is equal to the weight of the liquid that is dislodged

                  B = ρ g V

we substitute

             ρ g V = m g

             V = m /ρ_fluid          1

we can write the mass of the pot as a function of its density

             ρ_body = m / V_body

            m = ρ_body  V_body

             V_fluid / V_body = ρ_body / ρ _fluid         2

Equations 1 and 2 are similar, although 2 is easier to analyze, the fraction of submerged volume is equal to the ratio of the densities of the body between the density of the fluid.

The effect appears the pot as if it had a lower apparent weight

Answer:

the speed of the block when it reaches point B is 14 m/s

Explanation:

Given that:

mass of the block slides = 1.5 - kg

height = 10 m

Force constant  = 200 N/m

distance of rough surface patch = 20 m

coefficient of kinetic friction = 0.15

In order to determine the speed of the block when it reaches point B.

We consider the equation for the energy conservation in the system which can be represented by:

[tex]\dfrac{1}{2}mv^2=mgh[/tex]

[tex]\dfrac{1}{2}v^2=gh[/tex]

[tex]v^2=2 \times g \times h[/tex]

[tex]v^2=2 \times 9.8 \times 10[/tex]

[tex]v=\sqrt{2 \times 9.8 \times 10[/tex]

[tex]v=\sqrt{196[/tex]

v = 14 m/s

Thus; the speed of the block when it reaches point B is 14 m/s

Answer:

the answers are provided in the attachments below

Explanation:

Gauss law state that the net electric field coming out of a closed surface is directly proportional to the charge enclosed inside the closed surface

Applying Gauss law to the long solid cylinder

A) E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex]

B) E = 2K λ / r

C) Answers from parts a and b are the same

D) attached below

Applying Gauss's law which states that the net electric field in an enclosed surface is directly ∝ to the charge found in the enclosed surface.

A ) The expression for the electric field inside the volume at a distance r

Gauss law :  E. A = [tex]\frac{q}{e_{0} }[/tex]  ----- ( 1 )

where : A = surface area = 2πrL ,  q = p(πr²L)

back to equation ( 1 )

E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex]

B) Electric field at point Outside the volume in terms of charge per unit length  λ

Given that:  linear charge density = area * volume charge density

                                            λ    =  πR²P

from Gauss's law : E ( 2πrL) = [tex]\frac{q}{e_{0} }[/tex]

∴ E = [tex]\frac{\pi R^{2}P }{2e_{0}r\pi }[/tex]    ----- ( 2 )

where : πR²P = λ

Back to equation ( 2 )

E = λ  / 2e₀π*r              where : k = 1 / 4πe₀

∴ The electric field ( E ) at point outside the volume in terms of charge per unit Length λ

E = 2K λ / r

C) Comparing answers A and B

Answers to part A and B are similar

Hence we can conclude that Applying Gauss law to the long solid cylinder

E ( electric field ) = p*r / 2 * [tex]e_{0}[/tex], E = 2K λ / r also Answers from parts a and b are the same.

Learn more about Gauss's Law : https://brainly.com/question/15175106