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Wewaii [24]
2 years ago
9

Suppose that a ball is released from the window of a train that is moving with constant velocity.  The path of the ball, as obse

rved from the train window, will be a horizontal straight line.
true or false
please need a answer​
Physics
1 answer:
barxatty [35]2 years ago
7 0

True, the path of the ball, as observed from the train window, will be a horizontal straight line.

An object projected from a certain height has a parabolic path when observed from a fixed point.

However, if the reference point is moving at the same velocity as the object, the path of the object's motion appears to be a straight line.

When the ball is released from the window of the train, it will move at the same constant velocity as the train, and the path of the ball's motion observed from the train window will be a straight line.

Thus, we can conclude that the given statement is true. The path of the ball, as observed from the train window, will be a horizontal straight line.

Learn more about path of motion of objects here: brainly.com/question/82610

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Which of the following types of particles can be categorized into different kind of elements?
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When making a turn, do not have the steering wheel turned in the direction of the turn before beginning the turning maneuver.a)
RideAnS [48]

Answer:

a) True.

Explanation:

If you turn the wheel in the direction of the turn before beginning the turning maneuver then it's possible that there might be not enough space available for turning and also if you are waiting for the traffic to get clear with rear  ended then it will get pushed forward onto the coming traffic.

7 0
3 years ago
Consider the two moving boxcars in Example 5. Car 1 has a mass of m1 = 65000 kg and a velocity of v01 = +0.80 m/s. Car 2 has a m
Amiraneli [1.4K]

Answer:

1.034m/s

Explanation:

We define the two moments to develop the problem. The first before the collision will be determined by the center of velocity mass, while the second by the momentum preservation. Our values are given by,

m_1 = 65000kg\\v_1 = 0.8m/s\\m_2 = 92000kg\\v_2 = 1.2m/s

<em>Part A)</em> We apply the center of mass for velocity in this case, the equation is given by,

V_{cm} = \frac{m_1v_1+m_2v_2}{m_1+m_2}

Substituting,

V_{cm} = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

V_{cm} = 1.034m/s

Part B)

For the Part B we need to apply conserving momentum equation, this formula is given by,

m_1v_1+m_2v_2 = (m_1+m_2)v_f

Where here v_f is the velocity after the collision.

v_f = \frac{m_1v_1+m_2v_2}{m_1+m_2}

v_f = \frac{(65000*0.8)+(92000*1.2)}{92000+65000}

v_f = 1.034m/s

8 0
3 years ago
A car accelerates at a rate of 13m/s^2[S]. If the car's initial velocity is 120km/h[N]. What will its final velocity be in m/s,
Delvig [45]

Answer:

the final velocity of the car is 59.33 m/s [N]

Explanation:

Given;

acceleration of the car, a = 13 m/s²

initial velocity of the car, u = 120 km/h = 33.33 m/s

duration of the car motion, t = 2 s

The final velocity of the car in the same direction is calculated as follows;

v = u + at

where;

v is the final velocity of the car

v = 33.33 + (13 x 2)

v = 59.33 m/s [N]

Therefore, the final velocity of the car is 59.33 m/s [N]

6 0
2 years ago
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