Answer:
F. 25.82 s
Explanation:
Given:
Δy = 90 m
v₀ = 0 m/s
a = 0.27 m/s²
Find: t
Δy = v₀ t + ½ at²
90 m = (0 m/s) t + ½ (0.27 m/s²) t²
t = 25.82 s
By Hubble theory in which universe is expanding,
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 answer is 4.0 kg since the flywheel comes to rest the
kinetic energy of the wheel in motion is spent doing the work. Using the
formula KE = (1/2) I w².
Given the following:
I = the moment of inertia about the
axis passing through the center of the wheel; w = angular velocity ; for the
solid disk as I = mr² / 2 so KE = (1/4) mr²w². Now initially, the wheel is spinning
at 500 rpm so w = 500 * (2*pi / 60) rad / sec = 52.36 rad / sec.
The radius = 1.2 m and KE = 3900 J
3900 J = (1/4) m (1.2)² (52.36)²
m = 3900 J / (0.25) (1.2)² (52.36)²
m = 3.95151 ≈ 4.00 kg