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
Answer:
Initial velocity, U = 4.5m/s
Explanation:
Given the following data;
Final velocity, v = 12m/s
Time, t = 5 seconds
Acceleration, a = 1.5m/s²
To find the initial velocity, we would use the first equation of motion.
Where;
V is the final velocity.
U is the initial velocity.
a is the acceleration.
t is the time measured in seconds.
Substituting into the equation, we have;
12 = U + 1.5*5
12 = U + 7.5
U = 12 - 7.5
Initial velocity, U = 4.5m/s
Dinosaurs but I need the whole groups yo tell you ;)
Answer:
Explanation:
The vehicle is experiencing a large force created by the concrete wall.
Equation
vf^2 = vi^2 + 2*a * d
Givens
vf = 0 The car eventually does stop.
vi = 72 km/hr * [ 1000 m/ km] * [1 hour / 3600 seconds]
vi = 20 meters / second
a = ?
m = 850 kg
Solution
vf^2 = vi^2 + 2a*d
0 = 20 m/s + 2* 2 *a
-20 m/s = 4a
-20/4 = a
a = - 5 m/s^2 The minus sign tells you the vehicle is slowing down. It sure should be.
Force = m * a
F = - 850 * (-5)
F = - 4250 N
The car provides a 4250 N force on it going east to west and a 4250 N force going from west to east provided by the concrete wall.
