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:
Explanation:bjbhfcvvjkkknbnnnm
Mass of gold m₁ = 47 g
Initial temperature of gold T₁ = 99 C
Specific heat of gold C₁ = 0.129 J/gC
final temperature T₂ = 38 C
Heat needed by the gold to cool down
Q =m₁ * C₁* ( T₁ - T₂)
Q = (47)(0.129)(99-38)
Q = 369.843 J
This heat will be given by the water
we need to find out mass of water m₂
and initial temperature of water is T₃ = 25 C
Specific heat of water C₂ = 4.184 J/gC
Q = m₂*C₂*(T₂ - T₃)
369.843 = m₂(4.184)(38-25)
m₂ = 6.8 g
Answer:
0.312 m/s
Explanation:
Elastic collisions conserve momentum and kinetic energy
The velocity of the center of mass will not change. It continues at
0.00521(443) / 14.80521 = 0.155893... ≈ 0.156 m/s
To conserve kinetic energy we can think of the center of mass (CoM) as an ideal spring returning to each mass that strikes it an identical speed of collision in the opposite direction.
The CoM sees the target approach at - 0.156 and will see it depart at 0.156 m/s
A ground based observer sees the target depart at the velocity of the CoM plus the relative velocity .
v = 0.156 + 0.156 = 0.312 m/s