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
The correct answer is (A). The speed of light would increase to a speed larger than the maximum speed of light in vacuum.
The index of refraction is the ratio of speed of light in vacuum to the speed of light in a medium.
n=C/V
here, n is the index of refraction, c the speed of light in vacuum, v is speed of light in any medium.
Now if the value of index of refraction is less than one, than the value of speed of light would be greater than the speed of light in the vacuum.
We have by the first law of thermodynamics tha energy is preserved, hence we cannot have over 840kJ per cycle. We have by the laws of thermodynamics (the 2nd one in specific) that the entropy of a system cannot increase. We cannot have an output of 840 kJ per cycle from a heat engine because then that would mean that the entropy would stay the same, while any heat engine increases it. Hence, any value
is acceptable.
80 because if you add 50 +30 =80 so yea that why I pick 80
Answer:
v = 2.82 m/s
Explanation:
For this exercise we can use the conservation of energy relations.
We place our reference system at the point where block 1 of m₁ = 4 kg
starting point. With the spring compressed
Em₀ = K_e + U₂ = ½ k x² + m₂ g y₂
final point. When block 1 has descended y = - 0.400 m
Em_f = K₂ + U₂ + U₁ = ½ m₂ v² + m₂ g y₂ + m₁ g y
as there is no friction, the energy is conserved
Em₀ = Em_f
½ k x² + m₂ g y₂ = ½ m₂ v² + m₂ g y₂ + m₁ g y
½ k x² - m₁ g y = ½ m₂ v²
v² =
let's calculate
v² =
v² = 2.7 + 5.23
v = √7.927
v = 2,815 m / s
using of significant figures
v = 2.82 m/s