Answer:
Energy is force times distance. For your problem, no matter how long you push, the wall still goes nowhere, so there is no obvious energy transfer. so in conclusion, you actually didn't do anything :(
Explanation:
Let denote the position vector of the ball hit by player A. Then this vector has components
where is the magnitude of the acceleration due to gravity. Use the vertical component to find the time at which ball A reaches the ground:
The horizontal position of the ball after 0.49 seconds is
So player B wants to apply a velocity such that the ball travels a distance of about 12 meters from where it is hit. The position vector of the ball hit by player B has
Again, we solve for the time it takes the ball to reach the ground:
After this time, we expect a horizontal displacement of 12 meters, so that satisfies
Answer: hello the image related to your question is missing attached below is the missing image
answer :
1.47°
Explanation:
First we have to determine the displacement
first : apply law of similar triangles
Scd / 1 = Sef / 3
Fcd ( displacement ) = 3Fef - 80
= 3(30) - 80 = 10 kip
angle of tilt ( θ ) = cos^-1 ( 1 / 10 )
= cos^-1 ( 0.1 ) = 1.47°
attached below is the detailed solution on the displacement value
Answer: 68.4 s
Explanation:
The plot points seem to indicate a 5 m/s increase in velocity in the 6 seconds between 2 and 8 second marks. This makes the acceleration
a = 5 / 6 m/s/s
At time zero, the velocity is 13 m/s, so we need to increase by
70 - 13 = 57 m/s
t = v/a = 57 / (5/6) = 68.4 s
Answer: a system
Explanation: just did the test