Answer:
Explanation:
Since the block is at rest in an elevated position, we can assume that it only has potential energy.
U=mgh is the formula for potential energy where U=potential energy, m= mass, g=acceleration due to gravity, and h=height.
Plug in known variables....
U=4kg*9.8m/s^2*20m
U=784 joules of potential energy or letter A.
For balancing the lever, force on both the sides shall be equal. so,
Force on 3 m end = m × a = 3 × 98.1 = 294.3
Now, on 6 m end, it would be: = 294.3/6 = 49.05
After rounding-off to the nearest hundredth value, it would be: 49 N
Finally, Option A would be your correct answer.
Hope this helps!
Answer:
The ball would have landed 3.31m farther if the downward angle were 6.0° instead.
Explanation:
In order to solve this problem we must first start by doing a drawing that will represent the situation. (See picture attached).
We can see in the picture that the least the angle the farther the ball will go. So we need to find the A and B position to determine how farther the second shot would go. Let's start with point A.
So, first we need to determine the components of the velocity of the ball, like this:






we pick the positive one, so it takes 0.317s for the ball to hit on point A.
so now we can find the distance from the net to point A with this time. We can find it like this:



Once we found the distance between the net and point A, we can similarly find the distance between the net and point B:







t= -0.9159s or t=0.468s
we pick the positive one, so it takes 0.468s for the ball to hit on point B.
so now we can find the distance from the net to point B with this time. We can find it like this:



So once we got the two distances we can now find the difference between them:

so the ball would have landed 3.31m farther if the downward angle were 6.0° instead.
Answer:
For every action, there is an equal and opposite reaction.
Explanation:
Physics helps alot lol
Graph B represents the velocity of the sphere changes over time when falling with constant acceleration.
- Acceleration is the measure of how quickly a body's velocity varies with regard to time, and constant acceleration occurs when a body's velocity changes proportionately over a period of time, or at a constant rate. It measures in m/s2.
- It is claimed that a body has continual positive acceleration when it begins to move with an initial velocity of zero and gradually increases to a positive value over time.
- Constant positive acceleration is demonstrated by a ball falling freely in a vertical direction.
To know more about constant acceleration. visit : brainly.com/question/9754169
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