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n200080 [17]
3 years ago
12

"3.) By Newton’s 3rd Law, a train pulls backward on its engine exactly as hard as the engine pulls forward on the train. Since N

ewton’s 3rd Law is correct, the force of the engine on the train should be equal and opposite to the force of the train on the engine. Therefore, the train shouldn’t go anywhere. Carefully explain why this analysis is wrong."
Physics
1 answer:
dem82 [27]3 years ago
3 0

Answer:

The force is applicable according to the newton's third law of motion but the force on the engine is compensated in the form of stess on the fixed parts and rigid links whereas the wheel is free to roll.

Explanation:

This interpretation that the engine applies a force on the train and the train also applies an equal force on the engine and hence the train should not move is wrong because the engine imparts a rotational force in the form of torque on the train and the train imparts an equal force on the engine in the opposite direction but the engine is fixed like a structure on the chassis of the train and consists of rigid links which resist the motion and deformation as compared to the relative motion between the wheels and the rail tracks.

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Which planet does the sun have the strongest gravitational pull
Irina18 [472]

saturn is the planet that the sun has the strongest gravitational pull

4 0
3 years ago
If a 4 ohms resister, a 7 ohms resistor and a 12 ohms resistor are connected in a parallel, which resistor has the most current
Margaret [11]

Each resistor has the same voltage across it, and Current=(voltage)/(resistance).

The lowest resistance has the highest current through it.  That's the 4-ohm unit.

7 0
4 years ago
A foul ball is hit straight up into the air with a speed of about 25 m/s. ? how high does it go? (b) how long is it in the air?
nikdorinn [45]

To solve this problem, we make use of the equations of motion:

vf^2 = v0^2 – 2 g h

vf = v0 – g t

h = v0 t – 0.5 g t^2

 

A. Finding for max height:

Max height occurs when vf = 0, therefore:

0^2 = 25^2 – 2 (9.8) h

h = 31.89 m

 

B. Finding for t:

t = t(up) + t(down)

 

vf = v0 – g t(up)

0 = 25 – 9.8 t(up)

t(up) = 2.55 s

 

h = v0 t(down) – 0.5 g t(down)^2

31.89 = 0.5 (9.8) t(down)^2

t(down)^2 = 6.51

t(down) = 2.55 s

 

so,

<span>t = 5.1 s</span>

5 0
4 years ago
A 3000 kg freight train traveling at a rate of 2 m/s strikes a stationery 3000 kg freight car and couples up. What is the speed
Contact [7]

Answer:

1 m/s

Explanation:

Momentum before the collision = momentum after the collision

m₁ u₁ + m₂ u₂ = m₁ v₁ + m₂ v₂

(3000 kg) (2 m/s) + (3000 kg) (0 m/s) = (3000 kg) v + (3000 kg) v

6000 kgm/s = 6000 kg v

v = 1 m/s

5 0
4 years ago
Justin drives his unicycle 7.0 m/s northward. How long will it take him (in seconds) to travel 12 km?
Aleksandr-060686 [28]

Answer:

The time required is 28.5 [min]

Explanation:

We can find the time using the kinematic equation, but first, we must convert the distance to travel in meters.

x = distance = 12 [km] = 12000 [m]

velocity = 7 [m/s]

Tt = \frac{x}{v} \\t=\frac{12000}{7} \\t=1714 [s] = 28.5 [min]

7 0
3 years ago
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