Answer:
The graph should look like this.
Explanation:
G weight of mass m=0.55kg.
G=m×g=k×dx, whre dx=32cm is the spring stretch, computed as the difference between the actual length and original length of the spring.
dx=32cm=0.32m
g=9.81m/s²
Thus we have:
m×g=k×dx
k=m×g/dx=0.55×9.81/0.32=16.86N/m
Answer: k=16.86N/
Answer:
7.98 Newton Seconds
Explanation:
J = F * (Δt)
J = 1450 (5.5 * 10^-3)
J = 7.975
We only use three significant digits, so our answer becomes 9.98
The ball's return velocity is (-20 m/s) before reaching its starting location.
According to the 3rd equation of motion-
- When an item is propelled higher by gravity, it progressively slows down until it achieves its maximum height.
- The earth's gravitational pull causes it to turn around after reaching its highest peak and fall freely back to the ground.
An object that is falling freely has a starting velocity of "0"
The object accelerates at a rate of "g."
Let 'h' represent the distance the item travels while in free fall.
We can now get the ultimate velocity of the object shortly before it touches the earth with this equation,
v²=u²+2as
- Assume that the ball returns to its originating place at "v". Although the ball's orientation changes as it returns to its original place, its speed stays constant.
- A positive direction should point upward, while a negative direction should point below.
Calculation-
For upward motion,
providing that,
The ball's initial speed is u = +20 m/s.
So, v²= u² + 2as
⇒0= (20)² - 2gH [ where 'H' = maximum height reached]
⇒H= 400/2g
For downward motion,
v²=u²+2as
⇒v²= 0²+2g*400/2g
⇒v²= 400 m/s
⇒v= 20 m/s
Therefore, the ball's return velocity is (-20 m/s) before reaching its starting location.
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Answer:
Explanation:
Using the Gauss Law, we obtain the electric Field for a uniform large line of charge:
We calculate the potential difference from the electric field: