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Ber [7]
3 years ago
10

A 0.8 m length of wire is formed into a single turn, square loop in which there is a current of 12 A. The loop is placed in magn

etic field of 0.12 T, which is perpendicular to a normal to the loop plane. What is the maximum torque that the loop can experience?
Physics
1 answer:
rewona [7]3 years ago
8 0

Answer:

0.073 N-m

Explanation:

i = 12 A, l = 0.8 m, B = 0.12 T

The circumference of the loop is 0.8 m.

Let r be the radius of the loop.

2 x 3.14 x r = 0.8

r = 0.127 m

Maximum Torque = i x A x B

Maximum Torque = 12 x 3.14 x 0.127 x 0.127 x 0.12 = 0.073 N-m

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A fish swims 12.0 m in 5.0 s. It swims the first 4.0 m in 2.0 s, the next 3.0 m in 1.2 s, and the last 5.0 m in 1.8 s. What is t
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A room with 3.1-m-high ceilings has a metal plate on the floor with V = 0V and a separate metal plate on the ceiling. A 1.1g gla
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Answer:

The ball traveled 0.827 m

Explanation:

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distance between the metal plates of the room, d = 3.1 m

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charge on the glass, q = 4.7 nC

speed of the glass ball, v = 4.8 m/s

voltage of the ceiling, V = +3.0 x 10⁶ V

The repulsive force experienced by the ball when shot to the ceiling with positive voltage, can be calculated using Coulomb's law;

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|F| = (4.7 x 10⁻⁹ x 3 x  10⁶) / (3.1)

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F = - 4.548 x 10⁻³ N

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F_{net} = F_c - mg\\\\F_{net} = -4.548 *10^{-3} - (1.1*10^{-3} * 9.8)\\\\F_{net} = -15.328*10^{-3} \ N

The work done between the ends of the plate is equal to product of the  magnitude of net force on the ball and the distance traveled by the ball.

W = F_{net} *h\\\\W = 15.328 *10^{-3} *  h

W = K.E

15.328*10^{-3} *h = \frac{1}{2}mv^2\\\\ 15.328*10^{-3} *h = \frac{1}{2}(1.1*10^{-3})(4.8)^2\\\\ 15.328*10^{-3} *h =0.0127\\\\h = \frac{0.0127}{15.328*10^{-3}}\\\\ h = 0.827 \ m

Therefore, the ball traveled 0.827 m

4 0
3 years ago
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