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3 years ago

a boy is enjoying a ride on giant wheel which is moving with a constant speed of 10 metre per second it implies a he is

Physics

1 answer:

Sloan [31]3 years ago

6 0

Answer:

in accelerated motion

Explanation: tbh I just guessed if im wrong sorry

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V and R are ___________ proportional.

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B- V and R are directly proportional to eachother

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4 years ago

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Question 7 (4 points) (01.02 MC) The gravitational strength at the poles is greater than the gravitational strength at the equat

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I think it’s A but not really sure

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4 years ago

Block A, with a mass of 6.0 kg, is sliding across a frictionless track at 65 m/s when it collides with Block B ( 9.0 kg) which i

Mars2501 [29]

The correct answer is a I hope that helped enjoy the rest of your weekend

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3 years ago

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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3 years ago

The earth's magnetic moment is 8 x 1022 a-m2. If that moment were created by a loop of wire going around the earth (r = 6378 km)

Elena L [17]

Answer:

M = I A definition of magnetic moment - current * area

A = π R^2 = π * (6.4E6)^2 = 1.3E14 m^2

I = 8E22 A-m^2 / 1.3E14 m^2 = 6.2E8 amperes

I = 620,000,000 amps

4 0

2 years ago