The correct graph is <u>D</u>.
The graph <em>A</em> is a straight line sloping downwards and it shows that the speed of the body is decreasing at a constant rate. Therefore, this s a graph of a body that is under a constant deceleration.
The graph B is a straight line which slopes upwards. Hence the graph shows that the speed of the body increases at a constant rate. Therefore, this is a graph of a body that is accelerating at a constant rate.
The graph C is curved line, which curves upwards. The slope of the curve increases with time. This is therefore, a graph of a body which is under increasing acceleration.
The graph D, however is a straight line parallel to the time axis. The speed of the body has the same value at all times. Therefore, Graph D is the graph which shows the motion of a body with constant speed.
Answer:
60.025m.
Explanation:
S= ut + at^2/2 (2nd equation of motion)
S= 0 + (9.8)(3.5)^2 /2 (free fall case, initial velocity = 0)
S = 4.9 x 12.25
S= 60.025 m.
Disclaimer: did math in my head, so you better double check with a calculator.
False.
The mass of a softball is approximately 200 g (0.2 kg), while the knees are located approximately at 30 cm (0.3 m) from the ground. It means that the gravitational potential energy of the ball when it is dropped is

This corresponds to the total mechanical energy of the ball at the moment it is dropped, because there is no kinetic energy (the ball starts from rest). Then the ball is dropped, and just before it hits the ground, all this energy is converted into kinetic energy: but energy cannot be created, so its final kinetic energy cannot be greater than 0.6 J.
Answer:
1m/s [E]
Explanation:
So for displacement, its saying she went 3000m east and 2000m west. Since these two are in different directions, you would change the sign of one of them to change the direction;
3000m [E] + 2000m [W] = d
3000m [E] - 2000m [E] = d
1000m [E] = d
And its saying the time is 1000 seconds, so using velocity formula:
v = d/t
v=1000m[E] / 1000s
v = 1 m/s [E]
With 30degree because incident angle = reflected angle