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OverLord2011 [107]
3 years ago
6

An engine creat a 2230N force that pushes a car forward 58.3m in 9.4s

Physics
1 answer:
Scilla [17]3 years ago
5 0

Answer:

13830.74 W

Explanation:

Applying

P = W/t.................... Equation 1

Where P = Power produce in the engine, W = Work done by the engine, t = time.

But,

W = F×d................... Equation 2

Where F = force, d = distance.

Substitute equation 2 into equation 1

P = (F×d)/t...................... Equation 3

From the question,

Given: F = 2230 N, d = 58.3 m, t = 9.4 s

Substitute these values into equation 3

P = (2230×58.3)/9.4

P = 130009/9.4

P = 13830.74 W

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the answer is c. The induced current will increase

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Ivan

Answer: B.

Explanation:

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A block mass m (0.25 kg) is pressed against (but is not attached to) an ideal spring of force constant k (100 N/m) and negligibl
VMariaS [17]

Answer:

d. All the above choices are correct.

Explanation:

When a spring of spring constant k is compressed by distance x , the potential energy stored in it is equal to

E = 1/2 k x²

If spring constant is 2 k , potential energy stored

E = 1/2 2k x²

= k x²

which is twice the earlier potential energy.

In the first case , the energy of spring is imparted to box . The energy given to box is spent by frictional force due to which box comes to rest.

So energy of box acquired from spring = work done by frictional force.

So energy of box acquired from spring =  F X d  , F is frictional force , d is displacement .

In the second case ,

energy acquired by box becomes  two times

Work done by frictional force will also become two times to put box at rest

So displacement will be two times ( because frictional force is constant )

so option a is correct .

option b is also correct .

Because kinetic energy  of box will be twice as explained above .

So option d will be correct.

5 0
3 years ago
The angular momentum of a flywheel having a rotational inertia of 0.200 kg · m2 about its central axis decreases from 3.80 to 0.
Bingel [31]

Answer:

14.3065 rad

-36.777 J

-28.29 W

Explanation:

L_f = Final angular momentum = 0.6 kgm²/s

L_i = Initial angular momentum = 3 kgm²/s

I = Moment of inertia = 0.2 kgm²

Torque is given by

\tau=\dfrac{L_f-L_i}{t}\\\Rightarrow \tau=\frac{0.6-3.8}{1.3}\\\Rightarrow \tau=-2.46\ Nm

\theta=\omega_it+\dfrac{1}{2}\alpha t^2

Initial angular speed is given by

\omega_i=\dfrac{L_i}{I}

Angular acceleration is given by

\alpha=\dfrac{\tau}{I}

\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\Rightarrow \theta=\dfrac{L_it+\dfrac{1}{2}\tau t^2}{I}\\\Rightarrow \theta=\dfrac{3.8\times 1.3+\dfrac{1}{2}\times -2.46\times 1.3^2}{0.2}\\\Rightarrow \theta=14.3065\ rad

The angle is 14.3065 rad

Work done is given by

W=\tau \theta\\\Rightarrow W=-2.46\times 14.95\\\Rightarrow W=-36.777\ J

The work done on the wheel is -36.777 J

Power is given by

P=\dfrac{W}{t}\\\Rightarrow P=\dfrac{-36.777}{1.3}\\\Rightarrow P=-28.29\ W

The power is -28.29 W

6 0
3 years ago
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