Answer:
See below
Explanation:
Vertical position = 45 + 20 sin (30) t - 4.9 t^2
when it hits ground this = 0
0 = -4.9t^2 + 20 sin (30 ) t + 45
0 = -4.9t^2 + 10 t +45 = 0 solve for t =4.22 sec
max height is at t= - b/2a = 10/9.8 =1.02
use this value of 't' in the equation to calculate max height = 50.1 m
it has 4.22 - 1.02 to free fall = 3.2 seconds free fall
v = at = 9.81 * 3.2 = 31.39 m/s VERTICAL
it will <u>also</u> still have horizontal velocity = 20 cos 30 = 17.32 m/s
total velocity will be sqrt ( 31.39^2 + 17.32^2) = 35.85 m/s
Horizontal range = 20 cos 30 * t = 20 * cos 30 * 4.22 = 73.1 m
Answer:
Explanation:
The motion of the vehicule on a highway curve can be modelled by the following equation of equilibrium:
The maximum speed is:
B. I think is the correct answer
Answer:
frequency = 5.52 * 10² Hz
Explanation:
the equation that relates velocity, frequency and wavelength is:
velocity = frequency * wavelength
We are given that:
velocity = 331 m/sec
wavelength = 0.6 m
Substitute with the givens in the equation to get the frequency as follows:
velocity = frequency * wavelength
331 = frequency * 0.6
frequency = 331 / 0.6
frequency = 5.52 * 10² Hz
Hope this helps :)
Answer:
The size of the force that pushes the wall is 12,250 N.
Explanation:
Given;
mass of the wrecking ball, m = 1500 kg
speed of the wrecking ball, v = 3.5 m/s
distance the ball moved the wall, d = 75 cm = 0.75 m
Apply the principle of work-energy theorem;
Kinetic energy of the wrecking ball = work done by the ball on the wall
¹/₂mv² = F x d
where;
F is the size of the force that pushes the wall
¹/₂mv² = F x d
¹/₂ x 1500 x 3.5² = F x 0.75
9187.5 = 0.75F
F = 9187.5 / 0.75
F = 12,250 N
Therefore, the size of the force that pushes the wall is 12,250 N.
