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Travka [436]
3 years ago
14

If a train is traveling eastward with a constant velocity of 15 m/s, what is the net external force action on it? There is not e

nough information, so it cannot be solved. There is no net force acting on the train. the force from the engines
Physics
2 answers:
sammy [17]3 years ago
6 0

Answer:

There is no net force acting on the train.

Explanation:

The net force equals mass times acceleration:

∑F = ma

The train moves at constant velocity, so its acceleration is 0.  Therefore, the net force is 0.

ExtremeBDS [4]3 years ago
3 0

If a train is travelling eastward with a constant velocity of 15 m/s, there is no net force acting on the train.

Answer: Option B

<u>Explanation: </u>

According to Newton 2nd law, the external force is proportionate to the rates of change of momentum which states that force will be indirectly proportional to the rates of variation in velocity. So, as the velocity is constant for the train, the rate of variation in velocity or the acceleration will be zero. Thus the total external forces acting on the train will be zero. But the train will be moving continuously due to the energy obtained from the engine.

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Explanation:

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Which items are matter?
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Without being provided a list of items, I would have to generally say that everything around you is matter. There are a few exceptions to this list, but a general rule of thumb is anything you can touch, taste, smell or hold would be considered matter. Sound, light, time (Dr. Who may disagree) and heat would be considered non-matter items. 
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3 years ago
An observer stands on the side of the front of a stationary train. When the train starts moving with constant acceleration, the
Zarrin [17]

To solve this problem we will use the linear motion description kinematic equations. We will proceed to analyze the general case by which the analysis is taken for the second car and the tenth. So we have to:

x = v_0 t \frac{1}{2} at^2

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x= Displacement

v_0 = Initial velocity

a = Acceleration

t = time

Since there is no initial velocity, the same equation can be transformed in terms of length and time as:

L = \frac{1}{2} a t_1 ^2

For the second cart

2L \frac{1}{2} at_2^2

When the tenth car is aligned the length will be 9 times the initial therefore:

9L = \frac{1}{2} at_3^2

When the tenth car has passed the length will be 10 times the initial therefore:

10L = \frac{1}{2}at_4^2

The difference in time taken from the second car to pass it is 5 seconds, therefore:

t_2-t_1 = 5s

From the first equation replacing it in the second one we will have that the relationship of the two times is equivalent to:

\frac{1}{2} = (\frac{t_1}{t_2})^2

t_1 = \frac{t_2}{\sqrt{2}}

From the relationship when the car has passed and the time difference we will have to:

(t_2-\frac{t_2}{\sqrt{2}}) = 5

t_2 (\sqrt{2}-1) = 3\sqrt{2}

t_2= (\frac{5\sqrt{2}}{\sqrt{2}-1})^2

Replacing the value found in the equation given for the second car equation we have to:

\frac{L}{a} = \frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2

Finally we will have the time when the cars are aligned is

18 \frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2 = t_3^2

t_3 = 36.213s

The time when you have passed it would be:

20\frac{1}{4} (\frac{5\sqrt{2}}{\sqrt{2}-1})^2 = t_4^2

t_4 = 38.172

The difference between the two times would be:

t_4-t_3 = 38.172-36.213 \approx 2s

Therefore the correct answer is C.

4 0
3 years ago
A car rounds a 75-m radius curve at a constant speed of 18 m/s. a ball is suspended by a string from the ceiling the car and mov
Svetlanka [38]
The angle between the string and the vertical is computed by:
Centripetal acceleration Ac = V²/R = 18²/75 = 4.32 m/s² horizontal 
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Θ = arctan(Ac/g)
= arctan (4.32m/s^2 / 9.8 m/s^2)
= 23.79° is the answer.
8 0
3 years ago
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