An emf is induced by rotating a 1000 turn, 19 cm diameter coil in the Earth’s 5.00 x 10-5 T magnetic field. Randomized Variables d= 19 cm What average emf is induced, given the plane of the coil is originally perpendicular to the Earth’s field and is rotated to be parallel to the field in 8 ms?

The moment of inertia (I) of a solid cylinder about its geometrical axis can be calculated using the formula:

I = (1/2) * m * r^2

Where:

m = mass of the cylinder

r = radius of the cylinder

Given:

Mass of the cylinder (m) = 20 kg

Radius of the cylinder (r) = 0.2 m

Substituting the given values into the formula:

I = (1/2) * 20 kg * (0.2 m)^2

I = (1/2) * 20 kg * 0.04 m^2

I = 0.4 kg·m^2

Therefore, the moment of inertia of the solid cylinder about its geometrical axis is 0.4 kg·m^2.

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i) The static sensitivity at 283K is approximately -0.0926g^2.

ii) The static sensitivity at 350K is approximately -0.0576g^2.

A thermistor's static sensitivity is defined as the change in resistance per unit change in temperature. It can be stated mathematically as follows:

S = dR/dT

Given the thermistor calibration curve, we have:

0.5e(-inTg2) = R(T).

Taking the derivative with respect to T, we obtain:

dR/dT = -0.5 inTg2 e(-inTg2).

(i) We have the following at 283K:

-0.5in(283)g2 e(-in(283)g2) S = dR/dT

S ≈ -0.0926g^2

At 283K, the static sensitivity is roughly -0.0926g2.

(ii) We have the following at 350K:

[tex]-0.5in(350)g2 e(-in(350)g2) S = dR/dT[/tex]

S ≈ -0.0576g^2

At 350K, the static sensitivity is roughly -0.0576g2.

As a result, as the temperature rises, the thermistor's static sensitivity diminishes.

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When a double slit is illuminated by a 416-nm blue laser, the spacing of the slits in the double-slit experiment is approximately 1703.3 nm.

To calculate the spacing of the slits in a double-slit interference pattern, we can use the formula:

sin(θ) = (mλ) / d

where θ is the angle of the bright fringe, m is the order of the fringe (m=1 for the first bright fringe), λ is the wavelength of the light, and d is the spacing between the slits. We are given the angle (14.0°) and the wavelength (416 nm), so we can solve for d:

sin(14.0°) = (1 * 416 nm) / d

To isolate d, we can rearrange the formula:

d = (1 * 416 nm) / sin(14.0°)

Now we can plug in the values and calculate the spacing of the slits:

d ≈ (416 nm) / sin(14.0°) ≈ 1703.3 nm

Therefore, the spacing of the slits in the double-slit experiment is approximately 1703.3 nm.

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The spacing of the slits if the first bright fringe of an interference pattern occurs at an angle of 14.0° from the central fringe when a double slit is illuminated by a 416-nm blue laser is approximately 1.7 × 10⁻⁶ meters.

To find the spacing of the slits when the first bright fringe of an interference pattern occurs at an angle of 14.0° from the central fringe and is illuminated by a 416-nm blue laser, follow these steps:

1. Use the double-slit interference formula: sin(θ) = (mλ) / d, where θ is the angle of the fringe, m is the order of the fringe (m = 1 for the first bright fringe), λ is the wavelength of the laser, and d is the spacing between the slits.

2. Plug in the known values: sin(14.0°) = (1 × 416 × 10⁻⁹ m) / d.

3. Solve for d: d = (1 × 416 × 10⁻⁹  m) / sin(14.0°).

4. Calculate the result: d ≈ 1.7 × 10⁻⁶ m.

Thus, the spacing of the slits is approximately 1.7 × 10⁻⁶ meters.

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The electric potential at a distance of 45.0 cm from the center of the sphere is 100 volts. The electric potential at a distance of 30.0 cm from the center of the sphere is 150 volts.

The electric potential due to a uniformly charged sphere at a point outside the sphere can be found using the following formula:

V = k * Q / r

where V is the electric potential at a distance r from the center of the sphere, k is the Coulomb constant , and Q is the total charge on the sphere.

1. At a distance of 45.0 cm from the center of the sphere, the electric potential is:

V = k * Q / r

V = (9.0 x [tex]10^9 N*m^2/C^2[/tex]) * (5.00 x [tex]10^-9 C[/tex]) / (0.450 m)

V = 100 V

Therefore, the electric potential at a distance of 45.0 cm from the center of the sphere is 100 volts.

2. At a distance of 30.0 cm from the center of the sphere, the electric potential is:

V = k * Q / r

V = (9.0 x [tex]10^9 N*m^2/C^2[/tex]) * (5.00 x [tex]10^-9[/tex]C) / (0.300 m)

V = 150 V

Therefore, the electric potential at a distance of 30.0 cm from the center of the sphere is 150 volts.

3. At a distance of 16.0 cm from the center of the sphere, the electric potential is:

V = k * Q / r

V = (9.0 x [tex]10^9 N*m^2/C^2[/tex]) * (5.00 x [tex]10^{-9[/tex] C) / (0.160 m)

V = 281.25 V

Therefore, the electric potential at a distance of 16.0 cm from the center of the sphere is 281.25 volts.

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The emissive power of the sun  is 8.21x10²¹ W

The sun’s surface temperature is 5760 K

At 504 nm emissive power of the sun a maximum.

The model used here assumes a black body surface for the Earth and does not take into account the effects of the atmosphere.

(a) The energy flux is given as 1353 W/m². The surface area of the sun is A = πr² = π(0.5 x 1.39x10⁹)² = 6.07x10¹⁸ m². Therefore, the total power output or emissive power of the sun is

P = E.A

  = (1353 W/m²)(6.07x10¹⁸ m²)

  = 8.21x10²¹ W.

(b) Using the Stefan-Boltzmann law, the emissive power of a black body is given by P = σAT⁴, where σ is the Stefan-Boltzmann constant (5.67x10⁻⁸ W/m²K⁴). Rearranging the equation, we get

T = (P/σA)¹∕⁴.

Substituting the values, we get

T = [(8.21x10²¹ W)/(5.67x10⁻⁸ W/m²K⁴)(6.07x10¹⁸ m²)]¹∕⁴

  = 5760 K.

(c) The maximum spectral emissive power occurs at the wavelength where the derivative of the Planck's law with respect to wavelength is zero. The wavelength corresponding to the maximum spectral emissive power can be calculated using Wien's displacement law, which states that

λmaxT = b,

where b is the Wien's displacement constant (2.90x10⁻³ mK). Therefore, λmax = b/T

         = (2.90x10⁻³ mK)/(5760 K)

         = 5.04x10⁻⁷ m or 504 nm.

(d) The power received by the Earth is given by P = E.A(d/D)², where d is the diameter of the Earth, D is the distance between the Earth and the sun, and A is the cross-sectional area of the Earth. Substituting the values, we get

P = (1353 W/m²)(π(0.5x1.29x10⁷)²)(1.5x10¹¹ m/1.5x10¹¹ m)²

  = 1.74x10¹⁷ W. The power absorbed by the Earth is given by Pabs = εP, where ε is the absorptivity of the Earth (0.7). Therefore,

Pabs = (0.7)(1.74x10¹⁷ W)

        = 1.22x10¹⁷ W.

Using the Stefan-Boltzmann law, the temperature of the Earth can be calculated as

T = (Pabs/σA)¹∕⁴

  = [(1.22x10¹⁷ W)/(5.67x10⁻⁸ W/m²K⁴)(π(0.5x1.29x10⁷)²)]¹∕⁴

  = 253 K.

The actual average temperature of the Earth is higher than the predicted temperature (288 K vs 253 K) because the Earth's atmosphere plays a significant role in trapping the incoming solar radiation, leading to a greenhouse effect that increases the temperature of the Earth's surface.

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The integrating factor u(t) is given by the exponential of the integral of the coefficient of y, which is (4/t):
u(t) = e^(∫(4/t)dt) = e^(4ln(t)) = t^4 and y(t) = (3/5)t + 37/(5t^4).

To solve the initial value problem t(dy/dt) + 4y = 3t with y(1) = 8, first, we need to find the integrating factor u(t). The equation can be written as a first-order linear ordinary differential equation (ODE): (dy/dt) + (4/t)y = 3
The integrating factor u(t) is given by the exponential of the integral of the coefficient of y, which is (4/t):
u(t) = e^(∫(4/t)dt) = e^(4ln(t)) = t^4 Now, multiply the ODE by u(t):
t^4(dy/dt) + 4t^3y = 3t^4 The left side of the equation is now an exact differential:
d/dt(t^4y) = 3t^4 Integrate both sides with respect to t: ∫(d/dt(t^4y))dt = ∫3t^4 dt   t^4y = (3/5)t^5 + C
To find the constant C, use the initial condition y(1) = 8: (1)^4 * 8 = (3/5)(1)^5 + C  C = 40/5 - 3/5 = 37/5
Now, solve for y(t): y(t) = (1/t^4) * ((3/5)t^5 + 37/5) y(t) = (3/5)t + 37/(5t^4)

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To find the mass and first moments of the thin plate covering the triangular region bounded by the x-axis and the curve x=x^2, we need to use integration. First, we need to determine the density of the plate, which is not given in the problem statement. Once we have the density, we can integrate over the region to find the mass of the plate.

Let's assume that the density of the plate is constant and equal to ρ. Then the mass of the plate can be found using the following integral:

m = ∫∫ρdA

where dA is an infinitesimal element of area and the integral is taken over the triangular region. Using polar coordinates, we can write:

m = ∫0^1∫0^r ρrdrdθ

Evaluating this integral, we get:

m = ρ/6

Now, to find the first moments of the plate about the x- and y-axes, we need to use the following integrals:

M_x = ∫∫yρdA
M_y = ∫∫xρdA

where M_x and M_y are the first moments about the x- and y-axes, respectively. Using polar coordinates again, we get:

M_x = ∫0^1∫0^r ρr^3sinθdrdθ = ρ/20
M_y = ∫0^1∫0^r ρr^4cosθdrdθ = ρ/15

Therefore, the mass of the plate is ρ/6 and its first moments about the x- and y-axes are ρ/20 and ρ/15, respectively. Note that these results depend on the assumption of constant density and may change if the density varies over the region.

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A) The torque on the system after the bug lands on the left sphere is 3MgL, where g is the acceleration due to gravity.

B) The angular acceleration of the rod-sphere-bug system immediately after the bug lands when the rod is vertical is (3g/5L).

C) The angular speed of the bug is (3g/5L)(L/2) = (3g/10), where L/2 is the distance from the axis of rotation to the bug.

D) The angular momentum of the system is conserved, so the initial angular momentum is zero and the final angular momentum is (3MgL)(2L) = 6MgL².

E) The force that must be exerted on the bug by the sphere to keep the bug from being thrown off the sphere is equal in magnitude but opposite in direction to the force exerted on the sphere by the bug. This force can be found using Newton's second law, which states that force equals mass times acceleration.

The acceleration of the bug is the same as the acceleration of the sphere to which it is attached, so the force on the bug is (3M)(3g/5) = (9Mg/5) and it is directed towards the center of the sphere. Therefore, the force exerted on the sphere by the bug is also (9Mg/5) and is directed away from the center of the sphere.

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Answer:

When heat is applied, the liquid expands moderately

Explanation:

Reason: Particles move around each other faster where the force of attraction between these particles is less than solids, which makes liquids expand more than solids.

The electric potential energy of the charge is 2.7 J. The formula to calculate electric potential energy is U = q × V, where U is the potential energy, q is the charge, and V is the electric potential. Plugging in the given values, U = (4.5 × 10^-5 C) × (2.0 × 10^4 N/C) × (0.030 m) = 2.7 J.

The electric potential energy (U) of a charged object in an electric field is given by the formula U = q × V, where q is the charge of the object and V is the electric potential at the location of the object.

In this case, the charge (q) is 4.5 × 10^-5 C, and the electric field strength (V) is 2.0 × 10^4 N/C. The distance of the charge from the source of the electric field is given as 0.030 m.

Plugging in the values into the formula, we have U = (4.5 × 10^-5 C) × (2.0 × 10^4 N/C) × (0.030 m). Simplifying the expression, we get U = 2.7 J.

Therefore, the electric potential energy of the charge is 2.7 Joules.

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The peak electric field strength for this wave is approximately 1.23 x 10^3 V/m.

To find the peak electric field strength (E) in an electromagnetic wave, you can use the relationship between the magnetic field (B) and the electric field, which is given by the formula:

E = c * B

where c is the speed of light in a vacuum (approximately 3.0 x 10^8 m/s).

In this case, the peak magnetic field strength (B) is given as 4.1 μT (4.1 x 10^-6 T). Plug the values into the formula:

E = (3.0 x 10^8 m/s) * (4.1 x 10^-6 T)

E ≈ 1.23 x 10^3 V/m

So, the peak electric field strength for this wave is approximately 1.23 x 10^3 V/m.

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The load factor of a hash table is defined as the ratio of the number of elements stored in the hash table to the number of buckets in the hash table. In this case, the hash table has 20 buckets and 12 elements, so the load factor is: Load factor = number of elements / number of buckets
Load factor = 12 / 20
Load factor = 0.6

Therefore, the answer is d) 0.6.

To calculate the load factor of a hash table, you can use the formula: load factor = number of elements / number of buckets. In this case, the hash table has 20 buckets and 12 elements.

Your question is: If a hash table has 20 buckets and 12 elements, what will the load factor be?

Step 1: Identify the number of elements and buckets.
- Number of elements: 12
- Number of buckets: 20

Step 2: Apply the formula.
- Load factor = number of elements / number of buckets
- Load factor = 12 / 20

Step 3: Calculate the result.
- Load factor = 0.6

So, the load factor of the hash table is 0.6, which corresponds to option d) 0.6.

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Where f is the focal length of the lens, do is the distance between the object and the lens, and di is the distance between the lens and the image.

The prescription of 1.25 D indicates the power of the contact lens. It tells us how much the lens will bend the light that enters it. Using the formula 1/f = 1/do + 1/di, we can calculate the distance between the lens and the image (di) by knowing the distance between the object and the lens (do) and the focal length of the lens (f).

The near point is the closest distance at which an object can be brought into focus. For normal human vision, this distance is 25.0 cm. By calculating the distance between the lens and the image using the prescription and the formula, we can determine the child's near point.

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The required gap δ is approximately 6.936 mm so the rails just touch one another when the temperature is increased from t1 = -14 ∘f to t2 = 90 ∘f.

The required gap δ can be determined by using the formula: δ = αL(t2 - t1), where α is the coefficient of linear expansion, L is the length of the rails, and t1 and t2 are the initial and final temperatures, respectively.

When the temperature increases from t1 = -14 ∘f to t2 = 90 ∘f, the change in temperature is Δt = t2 - t1 = 90 - (-14) = 104 ∘f. To find the coefficient of linear expansion α, we need to know the material of the rails.

Assuming the rails are made of steel, the coefficient of linear expansion is α = 1.2 x 10^-5 / ∘C. Converting the temperature difference to ∘C, we have Δt = 57.8 ∘C.

The length of the rails is not given, so let's assume it is 10 meters. Using the formula, we can now calculate the required gap:

δ = αLΔt = (1.2 x 10^-5 / ∘C) x (10 m) x (57.8 ∘C) = 6.936 mm

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To determine the elastic characteristics of the aortal material, the surgeon must understand how it responds to force and deformation. The test results on the 16.0 cm strip of donated aorta reveal that it stretches 3.75 cm when a 1.50 N pull is exerted on it. This indicates that the material has an elastic modulus of 2.50 N/cm.

Now, if the maximum distance the aorta will be able to stretch when it replaces the damaged one is 1.14 cm, the surgeon needs to calculate the greatest force it will be able to exert there. This can be done using the formula:

F = kx

Where F is the force, k is the elastic modulus, and x is the distance stretched.

Substituting the values, we get:

F = (2.50 N/cm) x (1.14 cm) = 2.85 N

Therefore, the greatest force the aortal material will be able to exert on the damaged heart is 2.85 N. It is important for the surgeon to know this information to ensure that the material is strong enough to withstand the physiological stresses and strains of the heart's pumping action. By using this information, the surgeon can make informed decisions about the materials and techniques to be used during the repair procedure.

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The greatest force the material will be able to exert in the damaged heart is 0.456 N.The force constant of the strip of aortal material can be calculated using the formula:

force constant = force applied / extension

Substituting the given values, we get:

force constant = 1.50 N / 3.75 cm
force constant = 0.4 N/cm

Therefore, the force constant of the strip of aortal material is 0.4 N/cm.

To find the greatest force the material can exert when it replaces the damaged aorta, we can use the same formula but rearrange it to solve for force applied:

force applied = force constant x extension

Substituting the given values, we get:

force applied = 0.4 N/cm x 1.14 cm
force applied = 0.456 N

Therefore, the greatest force the material will be able to exert in the damaged heart is 0.456 N. This information is important for the surgeon to ensure that the material can handle the stress and strain of the patient's heart.

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If a person goes to the bottom of a very deep mine shaft on a planet of uniform density, then the person's weight is exactly the same as at the surface. Option(A) is true.

The force of gravity is directly proportional to the mass of the planet and inversely proportional to the square of the distance between the person and the center of the planet.

Gravity is a fundamental force that governs the motion of objects in the universe. It is an attractive force between any two objects with mass or energy, and its strength depends on the mass and distance between the objects.

Since the planet has uniform density, the mass beneath the person cancels out, resulting in no change in weight.

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The angular acceleration of the fan is 0.969 rad/s²  and it takes 20.25 s for the fan to stop rotating.

To determine the angular acceleration of the fan, we need to use the formula:
angular acceleration = (final angular velocity - initial angular velocity) / time
Since the final angular velocity is 0 (the fan comes to a stop), and the initial angular velocity is 19 rad/s, we can substitute these values into the formula to get:
angular acceleration = (0 - 19 rad/s) / time
To find time, we need to use the fact that the fan rotates through an angle of 7.3 rad while slowing down. We can use the formula:
angle = (initial angular velocity x time) + (0.5 x angular acceleration x time²)
Substituting the given values, we get:
7.3 rad = (19 rad/s x time) + (0.5 x angular acceleration x time²)
Simplifying this equation, we get a quadratic equation:
0.5 x angular acceleration x time² + 19 rad/s x time - 7.3 rad = 0
Solving for time using the quadratic formula, we get:
time = (-19 rad/s ± sqrt((19 rad/s)² - 4 x 0.5 x (-7.3 rad) ) ) / (2 x 0.5 x angular acceleration)
time = (-19 rad/s ± sqrt(361.69 + 7.3) ) / angular acceleration
time = (-19 rad/s ± 19.6 ) / angular acceleration
We can ignore the negative root since time cannot be negative. So, we get:
time = (19.6 rad/s) / angular acceleration
Now, we can substitute this value of time into the equation for angular acceleration to get:
angular acceleration = -19 rad/s / ((19.6 rad/s) / angular acceleration)
Simplifying, we get:
angular acceleration = -0.969 rad/s²
Therefore, the angular acceleration of the fan is 0.969 rad/s² (magnitude only, since it's negative).
To find the time it takes for the fan to stop rotating, we can use the equation we derived earlier:
7.3 rad = (19 rad/s x time) + (0.5 x (-0.969 rad/s²) x time²)
Simplifying, we get another quadratic equation:
0.4845 x time² + 19 rad/s x time - 7.3 rad = 0
Solving for time using the quadratic formula, we get:
time = (-19 rad/s ± sqrt((19 rad/s)² - 4 x 0.4845 x (-7.3 rad) ) ) / (2 x 0.4845)
time = (-19 rad/s ± sqrt(361.69 + 14.1) ) / 0.969
We can ignore the negative root again, so we get:
time = (19.6 rad/s) / 0.969
time = 20.25 s
Therefore, it takes 20.25 s for the fan to stop rotating.

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Answer:

c

Explanation:

to get a kelvin from degrees u add 273

To convert Celsius to Kelvin, we need to add 273.15 to the Celsius temperature. Therefore, the temperature in Kelvin would be 423 K, which is answer choice c.

To explain this further, the Kelvin scale is an absolute temperature scale where 0 Kelvin represents the theoretical lowest possible temperature, also known as absolute zero. On the other hand, the Celsius scale is a relative temperature scale where 0 °C represents the freezing point of water at sea level.
So, when we convert a temperature from Celsius to Kelvin, we add 273.15 to the Celsius temperature to obtain the corresponding Kelvin temperature. In this case, 150 °C + 273.15 = 423.15 K, which we can round down to 423 K.
Therefore, the correct answer to the question is c) 423 K.
The correct answer for converting 150 °C to Kelvin is a) 523 K. To convert a temperature in Celsius to Kelvin, you simply add 273.15. In this case, 150 °C + 273.15 = 523.15 K. Since we are rounding to whole numbers, the temperature is approximately 523 K.

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The correct option is A) increases with increasing values of n.

In the Bohr model of the hydrogen atom, the electron is assumed to move in circular orbits around the nucleus. These orbits are characterized by a principal quantum number n, where n = 1, 2, 3, and so on. The value of n determines the energy of the electron and the size of the orbit.

The radius of the nth orbit in the Bohr model is given by the equation:

rn = n^2 * h^2 / (4 * π^2 * me * ke^2)

where rn is the radius of the nth orbit, h is Planck's constant, me is the mass of the electron, ke is Coulomb's constant, and π is a mathematical constant.

As we can see from the equation, the radius of the nth orbit is directly proportional to [tex]n^2[/tex]. This means that the distance between adjacent orbit radii, which is the difference between the radii of two adjacent orbits, increases with increasing values of n.

Therefore, option A) is the correct answer.

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The angular magnification produced by the large reflecting telescope with a 10.0m radius of curvature objective mirror and a 3.00m focal length eyepiece is not provided in the question.

The angular magnification of a telescope can be calculated using the formula:

M = - fo/fe

Where M is the angular magnification, fo is the focal length of the objective mirror and fe is the focal length of the eyepiece.

In this case, fo = 2R = 20.0m (since the radius of curvature is 10.0m) and fe = 3.00m. Substituting these values in the above formula, we get:

M = - (20.0m) / (3.00m) = -6.67

Therefore, the angular magnification produced by the large reflecting telescope is -6.67. A negative value indicates that the image produced by the telescope is inverted. The sketch below shows how the telescope produces an inverted image of the object being viewed.

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The corresponding absolute pressure would be the sum of the gauge pressure and the atmospheric pressure at sea level. The atmospheric pressure at sea level is approximately 101,325 Pa. Therefore, the absolute pressure at the bottom of the tank would be:
Absolute pressure = 301,325 Pa

The corresponding absolute pressure at the bottom of the tank would be 301,325 Pa. The absolute pressure at the bottom of the tank can be calculated using the formula:
Absolute Pressure = Gauge Pressure + Atmospheric Pressure

Given the gauge pressure is 200,000 Pa, and the atmospheric pressure at sea level is approximately 101,325 Pa, we can find the absolute pressure:Absolute Pressure = 200,000 Pa + 101,325 Pa = 301,325 Pa

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To determine the tangential stress in the air cylinder, we can use the formula for hoop stress in a cylindrical vessel:

Hoop stress (σ_h) = Pressure (P) * Internal radius (r_i) / Wall thickness (t)

Given:

Pressure (P) = 2 MPa

Internal diameter (D) = 800 mm

Internal radius (r_i) = D / 2 = 400 mm

Plate thickness (t) = 15 mm

Substituting the values into the formula, we have:

σ_h = (2 MPa) * (400 mm) / (15 mm)

Converting the radius and thickness to meters to maintain consistent units:

σ_h = (2 MPa) * (0.4 m) / (0.015 m)

Calculating:

σ_h ≈ 53.33 MPa

Therefore, the approximate tangential stress in the air cylinder is 53.33 MPa.

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The work done by the force of 30 lb applied at an angle of 60° to pull a handcart 10 ft across the ground is approximately 150 foot-pounds.

To calculate the work done by the force, we need to find the displacement of the handcart and the component of the force in the direction of displacement.

The displacement is 10 ft in the direction of the force, so we can use the formula:

Work = force x distance x cos(theta)

where theta is the angle between the force and displacement.

In this case, the force is 30 lb and theta is 60 degrees. So:

Work = 30 lb x 10 ft x cos(60°) = 150 ft-lb

Therefore, the work done by the force is 150 foot-pounds.

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A 0.25-kg ball to the floor from a height of 2.1 m  and it bounces to a height of 1.2 m. The magnitude of the change in its momentum as a result of the bounce is 2.387 Ns.

To find the magnitude of the change in momentum of the ball as a result of the bounce, we can use the principle of conservation of momentum. The momentum of an object is given by the product of its mass and velocity. Since the ball is dropped vertically and bounces back, we consider the change in momentum in the vertical direction.

Initially, when the ball is dropped, its velocity is purely downward, so the initial momentum is:

p_initial = m * v_initial

where m is the mass of the ball and v_initial is the initial velocity.

When the ball bounces back, its velocity changes direction and becomes purely upward. The final momentum is:

p_final = m * v_final

where v_final is the final velocity.

According to the principle of conservation of momentum, the change in momentum is:

Δp = p_final - p_initial

Substituting the given values:

m = 0.25 kg

v_initial = -√(2gh)   (negative because it is downward)

v_final = √(2gh)     (positive because it is upward)

h = 2.1 m (initial height)

h = 1.2 m (final height)

g = 9.8 m/s² (acceleration due to gravity)

v_initial = -√(2 * 9.8 * 2.1) ≈ -6.132 m/s

v_final = √(2 * 9.8 * 1.2) ≈ 3.416 m/s

Δp = (0.25 kg * 3.416 m/s) - (0.25 kg * -6.132 m/s)

=>Δp = 0.854 Ns + 1.533 Ns

=>Δp ≈ 2.387 Ns

The magnitude of the change in momentum is approximately 2.387 Ns.

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The angle from the vertical of the axis of the second polarizing filter is approximately 45.94°.

Note: If the two polarizing filters are not ideal or if their polarization axes are not perpendicular to each other, the equation for the intensity of the emerging light will be more complex, and the angle between the polarization axes may not be the same as the angle from the vertical.

Using Malus's Law, we can determine the angle from the vertical of the axis of the second polarizing filter. Malus's Law states that the intensity of light after passing through two polarizing filters is given by:
I = I₀ * cos²θ
where I is the final intensity (121 W/m²), I₀ is the initial intensity (350 W/m²), and θ is the angle between the axes of the two filters. Rearranging the equation to find the angle θ:
cos²θ = I / I₀
cos²θ = 121 / 350
Taking the square root: cosθ = sqrt(121 / 350)
Now, we find the inverse cosine to get the angle:
θ = arccos(sqrt(121 / 350))
θ ≈ 45.94°

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The entry fee for a child at the amusement park is $65.

To find the entry fee for a child at the amusement park, we need to determine the difference in entry fees between the two families and divide it by the difference in the number of children between the two families.

Entry fee difference: $155 - $90 = $65

The difference in number of children: 3 - 2 = 1

To find the entry fee for a child, we divide the entry fee difference ($65) by the difference in the number of children (1):

Entry fee for a child = Entry fee difference / Difference in number of children

Entry fee for a child = $65 / 1 = $65

Therefore, the entry fee for a child at the amusement park is $65.

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A gazelle is running at 9.09 m/s. he hears a lion and accelerates at 3.80 m/s²; the gazelle has traveled approximately 25.14 meters after 2.16 seconds since hearing the lion.

To find the total distance traveled by the gazelle, we'll use the formula d = v0t + 0.5at^2, where d is the distance, v0 is the initial velocity, t is the time, and a is the acceleration. Given the initial velocity of 9.09 m/s, acceleration of 3.80 m/s², and time of 2.16 seconds:
1. Calculate the distance covered during the initial velocity: d1 = v0 * t = 9.09 m/s * 2.16 s = 19.6344 m
2. Calculate the distance covered during acceleration: d2 = 0.5 * a * t^2 = 0.5 * 3.80 m/s² * (2.16 s)^2 = 5.50896 m
3. Add the distances to find the total distance: d = d1 + d2 = 19.6344 m + 5.50896 m ≈ 25.14 m
The gazelle has traveled approximately 25.14 meters after 2.16 seconds since hearing the lion.

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To calculate the potential (v) through which an electron has been accelerated to reach a speed of 9.11 x 10^6 m/s, we can use the equation for the kinetic energy of the electron:

KE = 1/2mv^2

Where KE is the kinetic energy of the electron, m is the mass of the electron (9.11 x 10^-31 kg), and v is the speed of the electron.

Since the electron is accelerated through a potential, it gains potential energy (PE) which is then converted into kinetic energy as it accelerates. The potential energy gained by the electron is equal to the potential difference (v) multiplied by the charge of the electron (e = 1.6 x 10^-19 C):

PE = eV

Setting the initial potential energy of the electron equal to its final kinetic energy:

eV = 1/2mv^2

Solving for v:

v = sqrt(2eV/m)

Substituting the given values:

v = sqrt(2 x 1.6 x 10^-19 x v / 9.11 x 10^-31)

v = sqrt(3.2 x 10^-12 x v)

v = 1.79 x 10^6 sqrt(v) m/s

To find the value of v that would result in a speed of 9.11 x 10^6 m/s:

9.11 x 10^6 = 1.79 x 10^6 sqrt(v)

Solving for v:

v = (9.11 x 10^6 / 1.79 x 10^6)^2

v = 25 V

Therefore, the potential through which the electron has been accelerated is 25 volts.

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In astronomy, an eon refers to a period of one billion years. This timescale is often used to describe the age of the universe, the lifespan of a star, or the evolution of a galaxy.

Astronomers use the term eon to describe a very long period of time in the history of the universe, typically one billion years. This timescale is often used when discussing topics such as the age of the universe or the lifespan of stars. For example, the current age of the universe is estimated to be around 13.8 billion years, which is equivalent to 13.8 eons. Similarly, the lifespan of a star can range from a few million to trillions of years, depending on its mass. By using the eon as a unit of time, astronomers can more easily discuss and compare these vast timescales.

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The average pressure from the girl's weight exerted on the horizontal surface is 16558.3 Pa.

(a) The volume of the girl's body can be calculated using the formula:

volume = mass/density

Substituting the given values, we get:

volume = 23.6 kg / 987 kg/m3 = 0.0239 m3

Therefore, the volume of the girl's body is 0.0239 m3.

(b) The weight of the girl is given by:

weight = mass x gravity

where the acceleration due to gravity, g = 9.81 m/s2

Substituting the given values, we get:

weight = 23.6 kg x 9.81 m/s2 = 231.816 N

The pressure exerted by the girl's weight on the horizontal surface is given by:

pressure = weight / area

Substituting the given values, we get:

pressure = 231.816 N / 1.40 ✕ [tex]10^{-2} m^2[/tex] = 16558.3 Pa

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