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Nadya [2.5K]
3 years ago
7

Consider a 150 turn square loop of wire 17.5 cm on a side that carries a 42 A current in a 1.7 T. a) What is the maximum torque

on the loop of wire in N.m? b) What is the magnitude of the torque in N·m when the angle between the field and the normal to the plane of the loop θ is 14°?
Physics
1 answer:
ankoles [38]3 years ago
7 0

Answer:

(a) 328 Nm

(b) 79.35 Nm

Explanation:

N = =150, side = 17.5 cm = 0.175 m, i = 42 A, B = 1.7 T

A = side^2 = 0.175^2 = 0.030625 m^2

(a) Torque = N x i x A x B x Sinθ

For maximum torque, θ = 90 degree

Torque = 150 x 42 x 0.030625 x 1.7 x Sin 90

Torque = 328 Nm

(b) θ = 14 degree

Torque =  150 x 42 x 0.030625 x 1.7 x Sin 14

Torque = 79.35 Nm

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Answer:

<h2> $1.50</h2>

Explanation:

Given data

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3 years ago
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Answer:

C. Up, equal to the can's weight

Explanation:

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C. Up, equal to the can's weight

D. Not enough information is given

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13 Henry Wadsworth Longfellow, John Greenleaf Whittier, Oliver Wendell Holmes, and James Russel Lowell were known as​
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3 years ago
A tank containing 200 L of hydrogen gas at 0.0 Celsius is kept at 10 kPa. The pressure is raised to 95C, and the volume is decre
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Answer:

The new pressure of the gas is 15.40 kPa.

Explanation:

Gay-Lussac's law indicates that when there is a constant volume, as the temperature increases, the pressure of the gas increases. And when the temperature is decreased, the pressure of the gas decreases. Mathematically this law indicates that the quotient between pressure and temperature is constant:

\frac{P}{T}=k

On the other hand, Boyle's law says that the volume occupied by a certain gaseous mass at constant temperature is inversely proportional to the pressure. This law is expressed mathematically as:

P*V=k

Finally, Charles's law indicates that as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases. Mathematically, this law says that when the amount of gas and pressure are kept constant, the quotient that exists between the volume and the temperature will always have the same value:

\frac{V}{T}=k

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\frac{P*V}{T}=k

Studying an initial state 1 and a final state 2, it is fulfilled:

\frac{P1*V1}{T1}=\frac{P2*V2}{T2}

In this case:

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Replacing:

\frac{10 kPa*200 L}{273 K}=\frac{P2*175 L}{368 K}

Solving:

P2=\frac{368 K}{175 L} *\frac{10 kPa*200 L}{273 K}

P2= 15.40 kPa

<u><em>The new pressure of the gas is 15.40 kPa.</em></u>

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