Answer: have a subject and a verb
Explanation:
Of the forces listed I think the force of him diving and sliding across the infield acted on the player.
I think so because the slowing down was a result of an action, and I don’t think that should count as An action when it is the result of an action. However, the act of diving head-first into second base and sliding across the infield are independent actions and will cause friction, which will act upon the player.
Answer:
Behaves as a wave.
Explanation:
The dual nature of light means that light behaves as a wave. If the light consisted of small particles, the alternating light and dark bands would not have occurred.
Sometimes it behaves like a particle (called a photon), which explains how light travels in straight lines.
Vi = As * h = 1000 * 30 = 30,000 cm^3 = Vol. of the ice.
Vb = (Di/Dw) * Vi = (0.9/1.0) * 30,000 = 27,000 cm^3 = Vol. below surface - Vol. of water displaced.
27,000cm^3 * 1g/cm^3 = 27,000 grams = 27 kg = Mass of water displaced.
Answer:
Part a)

Part b)

Part c)

Part d)

Part e)

Part f)

Part g)

Explanation:
Initial speed of the launch is given as
initial speed = 
angle =
degree
Now the two components of the velocity

similarly we have

Part a)
Now we know that horizontal range is given as

maximum height is given as

so we have

time of flight is given as



Part b)
Now the speed of the ball in x direction is always constant
so at the peak of its path the speed of the ball is given as



Part c)
Initial vertical velocity is given as


Part d)
Initial speed is given as

so we will have


Part e)
Angle of projection is given as



Part f)
If we throw at same speed so that it reach maximum height
then the height will be given as


Part g)
For maximum range the angle should be 45 degree
so maximum range is

