Answer:
49.5 Hz.
Explanation:
From the question given above, the following data were obtained:
Period (T) = 0.0202 s
Frequency (f) =?
The frequency and period of a wave are related according to the following equation:
Frequency (f) = 1 / period (T)
f = 1/T
With the above formula, we can obtain the frequency of the wave as follow:
Period (T) = 0.0202 s
Frequency (f) =?
f = 1/T
f = 1/0.0202
f = 49.5 Hz
Therefore the frequency of the wave is 49.5 Hz.
The shot putter should get out of the way before the ball returns to the launch position.
Assume that the launch height is the reference height of zero.
u = 11.0 m/s, upward launch velocity.
g = 9.8 m/s², acceleration due to gravity.
The time when the ball is at the reference position (of zero) is given by
ut - (1/2)gt² = 0
11t - 0.5*9.8t² = 0
t(11 - 4.9t) = 0
t = 0 or t = 4.9/11 = 0.45 s
t = 0 corresponds to when the ball is launched.
t = 0.45 corresponds to when the ball returns to the launch position.
Answer: 0.45 s
You use the equation Velocity = Acceleration X Time. 4x4=16m/s.
The car travels 18m in 3 seconds.
Answer:
Explanation:
Let the velocity of projectile be v and angle of throw be θ.
The projectile takes 5 s to touch the ground during which period it falls vertically by 100 m
considering its vertical displacement
h = - ut +1/2 g t²
100 = - vsinθ x 5 + .5 x 9.8 x 5²
5vsinθ = 222.5
vsinθ = 44.5
It covers 160 horizontally in 5 s
vcosθ x 5 = 160
v cosθ = 32
squaring and adding
v²sin²θ +v² cos²θ = 44.4² + 32²
v² = 1971.36 + 1024
v = 54.73 m /s