Since acceleration is measured in whole seconds, you need to know how many times 0.157seconds goes into 1 second... (just divide 1 by 0.157) = 6.369
So if in 0.157seconds the baseball went from zero to 38m/s, then to find out how fast it would be traveling after one whole second just multiply 38m/s by 6.369
38m/s x 6.369 = 242.038 m/s^2
The answer is 45 degrees. I am not doing a field experiment for you that involves a cannon and a day's work, for 5 points.
At the highest point of the trajectory the vertical component will have its zero velocity, and the descent caused by the force of gravity will begin.
Since the ball is thrown with a certain speed, the vertical component reaches its highest point (upwards), until returning to the receiver who will receive the ball with the same vertical component but in the opposite direction (downwards).
Therefore the vertical component will have its highest value at launch.
Answer:
Hearing, Vision and Metabolism.
Answer:
the location of 160 N from end E is 25mm
Explanation:
given:
A uniform plank of weight 100 N is 2000 mm long
and rests on a support that is 400 mm from end E
find:
At what distance from E must a 160 N weight be placed in order to balance the plank?
160 N
|---x--->|
100N \|/
======================||======================E
Δ---- 400mm---->|
|<---------------------------- 2000mm --------------------------->|
to determine the location of 160N load in order to balance the plank,
you have to take moment at support Δ.
note:
the 100 N is located at the center of a plank.
the distance of 100N from support = (2000 / 2) - 400 = 600mm
∑moment at Δ = 0
0 = -100 N (600mm) + 160 N (400-x)
0 = -60000 N.mm + 64000 N.mm - 160x N.mm
0 = 4000 - 160x
160x = 4000
x = 4000/160
x = 25 mm
therefore,
the location of 160 N from end E is 25mm
