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Svetradugi [14.3K]
2 years ago
15

⦁ Determining the magnetic flux, A rectangular piece of stiff paper measures 10 cm x 5 cm. You hold the piece of paper in a unif

orm magnetic field that has a magnitude of 4.0 x10-3 T For each situation below, - sketch a diagram showing the magnetic field and the paper, - determine the magnitude of the magnetic flux through the paper, when the magnitude of the flux is (a) maximized (b) minimized (c) halfway between its maximum and minimum value, when the angle between the magnetic field lines and the area normal vector is 300
Physics
1 answer:
MArishka [77]2 years ago
4 0

Answer:

jguewjdofe

Explanation:

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A hollow sphere of radius 0.200 m, with rotational inertia I = 0.0484 kg·m2 about a line through its center of mass, rolls witho
d1i1m1o1n [39]

Answer:

Part a)

KE_r = 8 J

Part b)

v = 3.64 m/s

Part c)

KE_f = 12.7 J

Part d)

v = 2.9 m/s

Explanation:

As we know that moment of inertia of hollow sphere is given as

I = \frac{2}{3}mR^2

here we know that

I = 0.0484 kg m^2

R = 0.200 m

now we have

0.0484 = \frac{2}{3}m(0.200)^2

m = 1.815 kg

now we know that total Kinetic energy is given as

KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2

KE = \frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{R})^2

20 = \frac{1}{2}(1.815)v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2

20 = 1.5125 v^2

v = 3.64 m/s

Part a)

Now initial rotational kinetic energy is given as

KE_r = \frac{1}{2}I(\frac{v}{R})^2

KE_r = \frac{1}{2}(0.0484)(\frac{3.64}{0.200})^2

KE_r = 8 J

Part b)

speed of the sphere is given as

v = 3.64 m/s

Part c)

By energy conservation of the rolling sphere we can say

mgh = (KE_i) - KE_f

1.815(9.8)(0.900sin27.1) = 20- KE_f

7.30 = 20 - KE_f

KE_f = 12.7 J

Part d)

Now we know that

\frac{1}{2}mv^2 + \frac{1}{2}I(\frac{v}{r})^2 = 12.7

\frac{1}{2}(1.815) v^2 + \frac{1}{2}(0.0484)(\frac{v}{0.200})^2 = 12.7

1.5125 v^2 = 12.7

v = 2.9 m/s

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Help awnser need experts​
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Label and describe what is happening in this picture
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