D=s(t) so it would be d=10(.19) d=.19 FOR BITH SNDWERS
Answer:
The angle of launch of the rubber band affects the initial velocity. The more the rubber band is stretched the more force it applies to return to equilibrium and the more kinetic energy that results in.
Answer:
I = 1.875 A
Explanation:
For this exercise we use Ampere's law
∫ B . ds = μ₀ I
We use a circular path around the wire whereby B and ds are parallel, whereby the dot product is reduced to the algebraic product
ds = 2π dr
B (2πr) = μ₀ I
I = B 2π R /μ₀
r= 7.5 cm = 0.075 m
calculate
I = (50 μ₀ /π) 2π 0.075 /μ₀
I = 1.875 A
The
balloon’s potential energy was transformed to kinetic energy as it drops.
Before
the balloon was drop it possessed potential energy but at the moment the
balloon was dropped the potential energy it possessed was transformed into
kinetic energy.
<span>Hope
this answer will be a good h<span>elp for you.</span></span>
Answer:
The velocity of the ball when its hit the ground will be 54.22 m/sec
Explanation:
We have given height from which ball is dropped h = 150 m
Acceleration due to gravity 
As the ball is dropped so initial velocity will be zero so u = 0 m/sec
According to third equation of motion we know that 


So the velocity of the ball when its hit the ground will be 54.22 m/sec