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Tcecarenko [31]

3 years ago

7

A coil of 1000 turns of wire has a radius of 12 cm and carries a counterclockwise current of 15A. If it is lying flat on the gro

und, and the Earth's magnetic field points due north, has a magnitude of 5.8 x 10-5 T, and makes a downward angle of 25Â° with the vertical, what is the torque on the loop?

1 answer:

grin007 [14]3 years ago

5 0

Answer:

torque is 1.7 * $10^{-2}$ Nm

Explanation:

Given data

turns n = 1000 turns

radius r = 12 cm

current I = 15A

magnitude B = 5.8 x 10^-5 T

angle θ = 25°

to find out

the torque on the loop

solution

we know that torque on the loop is

torque = N* I* A*B* sinθ

here area A = πr² = π(0.12)²

put all value

torque = N* I* A*B* sinθ

torque = 1000* 15* π(0.12)² *5.8 x 10-5 * sin25

torque = 0.0166 N m

torque is 1.7 * $10^{-2}$ Nm

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What is the internal energy of 3.00mol of N2 gas at 25c?

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Answer:

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2 years ago

Two 2.0 kg bodies, A and B, collide. The velocities before the collision are ~vA = (15ˆi + 30ˆj) m/s and ~vB = (−10ˆi + 5.0ˆj) m

AleksandrR [38]

Answer:

Part a)

$10\hat i + 15\hat j = \vec v$

Part b)

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Explanation:

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$5\hat i + 35\hat j = (-5\hat i + 20\hat j) +\vec v$

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Part b)

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magnitude of final speed of B = $\sqrt{10^2 + 15^2} = 18.03 m/s$

Now change in total kinetic energy is given as

$\Delta K = (\frac{1}{2}m_1v_{1i}^2 + \frac{1}{2}m_2v_{2i}^2) - (\frac{1}{2}m_1v_{1f}^2 + \frac{1}{2}m_2v_{2f}^2)$

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$\Delta K = 500 J$

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3 years ago

Which describes how a simple machine can make work easier ?

sveta [45]

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2 years ago

Tennis balls traveling at greater than 100 mph routinely bounce off tennis rackets. At some sufficiently high speed, however, th

Kipish [7]

Answer:

Probability of tunneling is $10^{- 1.17\times 10^{32}}$

Solution:

As per the question:

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Mass of the tennis ball, m = 100 g = 0.1 kg

Thickness of the tennis ball, t = 2.0 mm = $2.0\times 10^{- 3}\ m$

Max velocity of the tennis ball, $v_{m} = 200\ mph$ = 89 m/s

Now,

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$KE = \frac{1}{2}mv_{m}^{2} = \frac{1}{2}\times 0.1\times 89^{2} = 396.05\ J$

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$\eta = \frac{\bar{h}}{\sqrt{2m(KE - KE')}}$

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Thus

$\eta = \frac{1.0545\times 10^{- 34}}{\sqrt{2\times 0.1(396.05 - 154.8)}}$

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$\eta = 1.52\times 10^{-35}\ m$

Now,

We can calculate the tunneling probability as:

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$P(t) = e^{\frac{- 2\times 2.0\times 10^{- 3}}{1.52\times 10^{-35}}} = e^{-2.63\times 10^{32}}$

$P(t) = e^{-2.63\times 10^{32}}$

Taking log on both the sides:

$logP(t) = -2.63\times 10^{32} loge$

$P(t) = 10^{- 1.17\times 10^{32}}$

6 0

2 years ago