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:
The rock's speed after 5 seconds is 98 m/s.
Explanation:
A rock is dropped off a cliff.
It had an initial velocity of 0 m/s. And now it is moving downwards under the influence of gravitational force with the gravitational acceleration of 9.8 m/s².
Speed after 5 seconds = V
We know that acceleration = average speed/time
In our case,
g = ((0+V)/2)/5
9.8*5 = V/2
=> V = 2*9.8*5
V = 98 m/s
<span>The velocity would be 54.2 m/s
We would use the equation 1/2mv^2top+mghtop = 1/2mv^2bottom+mghbottom where m is the mass of the bobsled(which can be ignored), vtop/bottom is the velocity of the bobsled at the top or bottom, g is gravity, and htop/bottom is the height of the bobsled at the top or bottom of the hill. Since the velocity of the bobsled at the top of the hill and height at the bottom of the hill are zero, 1/2mv^2top and mghbottom will equal zero. The equation will be mghtop=1/2mv^2bottom. Thus we would solve for v.</span>
Answer:
<em>The velocity with which the student goes down the bottom of glide is 12.48m/s.</em>
Explanation:
The Non conservative force is defined as a force which do not store energy or get he energy dissipate the energy from the system as the system progress with the motion.
Given are
<em> mass of the student 73 kg</em>
<em> height of water glide 11.8 m</em>
<em> work done as -5.5*10³ J</em>
Have to find speed at which the student goes down the glide.
According to<em> Law of Conservation of energy</em>,
K.E =P.E+Work Done
mv²/2=mgh +W
Rearranging the above eqn for v
v = √2(gh+W/m)
Substituting values,
V = 12.48 m/s.
<em>The velocity with which the student goes down the bottom of glide is 12.48m/s.</em>
