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Semenov [28]
4 years ago
8

two idential train engines are parked on parallel track. one engine has ten cars attached and the other has no cars attached. ea

ch engine exerts the same net force. which allow sthe train to overcome static friction. which engine will be able to accelerate more
Physics
1 answer:
nikklg [1K]4 years ago
6 0

Answer:

The second engine with no cars will accelerate more.

Explanation:

Since the engines are identical so they hacve the same mass. The first engine has 10 cars attached which adds more mass to it whereas second engine has no cars attached.

Therefore second engine is lighter in mass than the first one.

<u>We know from the Newton's second law of motion:</u>

F=m.a

\therefore a=\frac{F}{m}

Now, according to question both the engines exert same force for moving.

We see that :

a\propto\frac{1}{m}

So more is the mass, lesser is the acceleration, hence the second engine with no cars will accelerate more.

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A wheel with moment of inertia 25 kg. m2 and angular velocity 10 rad/s begins to speed up, with angular acceleration 15 rad/sec2
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Answer:

(A) Angular speed 40 rad/sec

Rotation = 50 rad

(b) 37812.5 J

Explanation:

We have given moment of inertia of the wheel I=25kgm^2

Initial angular velocity of the wheel \omega _0=10rad/sec

Angular acceleration \alpha =15rad/sec^2

(a) We know that \omega =\omega _0+\alpha t

We have given t = 2 sec

So \omega =10+15\times  2=40rad/sec

Now \Theta =\omega _0t+\frac{1}{2}\alpha t^2=10\times 2+\frac{1}{2}\times 15\times 2^2=50rad

(b) After 3 sec \omega =10+15\times 3=55rad/sec

We know that kinetic energy is given by Ke=\frac{1}{2}I\omega ^2=\frac{1}{2}\times 25\times 55^2=37812.5J

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3 years ago
What is the difference between each distance traveled and displacement travled
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8 0
3 years ago
A coil of 40 turns is wrapped around a long solenoid of cross-sectional area 7.5×10−3m2. The solenoid is 0.50 m long and has 500
defon

To solve this problem it is necessary to apply the concepts related to mutual inductance in a solenoid.

This definition is described in the following equation as,

M = \frac{\mu_0 N_1 N_2A_1}{l_1}

Where,

\mu =permeability of free space

N_1 = Number of turns in solenoid 1

N_2 = Number of turns in solenoid 2

A_1= Cross sectional area of solenoid

l = Length of the solenoid

Part A )

Our values are given as,

\mu_0 = 4\pi *10^{-7}H/m

N_1 = 500

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Substituting,

M = \frac{\mu_0 N_1 N_2A_2}{l_1}

M = \frac{(4\pi *10^{-7})(500)(40)(7.5*10^{-4})}{0.5}

M = 3.77*10^{-4}H

PART B) Considering that many of the variables remain unchanged in the second solenoid, such as the increase in the radius or magnetic field, we can conclude that mutual inducantia will appear the same.

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I hope I helped.
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If a car driving down the road doubles its speed what happens to its kinetic energy?
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