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3 years ago

The rectangular coils in a 345-turn generator are 12 cm by 12 cm. Part A

Physics

1 answer:

Likurg_2 [28]3 years ago

4 0

Answer:

118.166 volt

Explanation:

We have given number of turns N =345

Sides of the rectangular coils is 12 cm =0.12 m

So area A =0.12×0.12=0.0144 $m^2$

Magnetic field B =0.45 T

Angular speed =505 rpm

Speed in rad/sec $\omega =\frac{2\pi N}{60}=\frac{2\times 3.14\times 505}{60}=52.8566rad/sec$

The emf is given by $E=NAB\omega SIN\omega t$

For maximum emf sinwt =1

So $E=345\times 0.0144\times 0.45\times 52.8566=118.166volt$

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In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400

liq [111]

Complete Question:

In the "before" part of Fig. 9-60, car A (mass 1100 kg) is stopped at a traffic light when it is rear-ended by car B (mass 1400 kg). Both cars then slide with locked wheels until the frictional force from the slick road (with a low ?k of 0.15) stops them, at distances dA = 6.1 m and dB = 4.4 m. What are the speeds of (a) car A and (b) car B at the start of the sliding, just after the collision? (c) Assuming that linear momentum is conserved during the collision, find the speed of car B just before the collision.

Answer:

a) Speed of car A at the start of sliding = 4.23 m/s

b) speed of car B at the start of sliding = 3.957 m/s

c) Speed of car B before the collision = 7.28 m/s

Explanation:

NB: The figure is not provided but all the parameters needed to solve the question have been given.

Let the frictional force acting on car A, $f_{ra} = \mu mg\\$ ............(1)

Since frictional force is a type of force, we are safe to say $f_{ra} = ma$ .......(2)

Equating (1) and (2)

$ma = \mu mg\\a = \mu g\\\mu = 0.15\\a = 0.15 * 9.8 = 1.47 m/s^{2}$

a) Speed of A at the start of the sliding

$d_{A} = 6.1 m\\Speed of A at the start of sliding, v_{A} = \sqrt{2ad_{A} }\\ v_{A} = \sqrt{2*1.47*6.1 } \\v_{A} = \sqrt{17.934 } \\v_{A} = 4.23 m/s$

b) Speed of B at the start of the sliding

$d_{A} = 4.4 m\\Speed of A at the start of sliding, v_{B} = \sqrt{2ad_{B} }\\ v_{B} = \sqrt{2*1.47*4.4 } \\v_{B} = \sqrt{12.936 } \\v_{B} = 3.957 m/s$

Let the speed of car B before collision = $v_{B1}$

Momentum of car B before collision = $m_{B} v_{B1}$

Momentum after collision = $m_{A} v_{A} + m_{B} v_{B2}$

Applying the law of conservation of momentum:

$m_{B} v_{B1} = m_{A} v_{A} +m_{B} v_{B2}$

$m_{A} = 1100 kg\\m_{B} = 1400 kg$

$(1400*v_{B1} ) = (1100 * 4.23) + ( 1400 * 3.957)\\(1400*v_{B1} ) = 10192.8\\v_{B1} = 10192.8/1400\\v_{B1 = 7.28 m/s$

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1. A car is stopped at a red light. When the light turns green, the car starts

Lubov Fominskaja [6]

Answer:

Because of inertia (Newton's First Law of motion)

Explanation:

According to Newton's First Law of motion:

"An object at rest (or in motion at constant velocity) will tend to stay at rest (or tend to keep moving with same velocity) unless acted upon an unbalanced force"

In this problem, the object we are analyzing is the coffee cup.

At the beginning, the cup is at rest, together with the car.

Later, the car starts moving when the light turns green.

If we apply Newton's First Law of motion to the cup, we see that the coffee cup tends to keep its state of rest: for this reason, as the car moves forward, the coffee in the cup will spill backward, into the rear seat. This property of an object to mantain its state of motion is also called as inertia.

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Which part of an atom is mostly empty space?

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Answer:

The electron cloud is mostly empty space

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