This is the equation for elastic potential energy, where U is potential energy, x is the displacement of the end of the spring, and k is the spring constant.
<span> U = (1/2)kx^2
</span><span> U = (1/2)(5.3)(3.62-2.60)^2
</span> U = <span>
<span>2.75706 </span></span>J
Rise over run at 1 second
It’s the same slope from 0 to 2 seconds
10/2=5mps
As a note all time points between 0and 2 will have this instantaneous velocity
Instantaneous velocity at time 2 is 0
Answer: 117.6N
Explanation:
By the second Newton's law, we know that:
F = m*a
F = force
m = mass
a = acceleration
We know that in the surface of the Earth, the gravitational acceleration is g = 9.8m/s^2.
Then we just can input that acceleration in the above equation, and also replace m by 12kg, and find that the force due the gravity is:
F = 12kg*9.8m/s^2 = 117.6N
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.
