Answer:
Explanation:
a ) V = 3 cos(0.5t)
differentiating with respect to t
dv /dt = -3 x .5 sin0.5t
= -1.5 sin0.5t.
acceleration = - 1.5 sin 0.5t
when t = 3 s
acceleration = - 1.5 sin 1.5
= - 1.496 ms⁻²
v = 3 cos.5t
b ) dx/dt = 3 cos 0.5 t
dx = 3 cos 0.5 t dt
integrating on both sides
x = 3 sin .5t / .5
x = 6 sin0.5t
At t = 2 s
x = 6 sin 1
x = 5.05 m
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Time = (displacement) / (magnitude of average velocity) .
Answer:
993.52 Hz
Explanation:
The frequency of sound emitted by the stationery train is 1057 Hz.
The car travels away from the train at 20.6 m/s.
The frequency the observer hears is given by the formula:
where v = velocity of sound = 343 m/s
vo = velocity of observer
f = frequency from source
This phenomenon is known as Doppler's effect.
Therefore:
The frequency heard by the observer is 993.52 Hz.
Answer:
(A) 10052.2 m/s²
(B) 0.00678 seconds
Explanation:
From the question,
(A) Applying
V² = U²+2as..................... Equation 1
Where V = Final velocity, U = Initial velocity, a = acceleration due to gravity, s = distance.
make a the subject of the equation
a = (V²-U²)/2s........................ Equation 2
Given: U = 0 m/s( from rest), V = 68 m/s, s = 0.230 m
Substitute these values into equation 2
a = (68²-0²)/(2×0.230)
a = 10052.2 m/s²
(B) Using,
a = (V-U)/t......................... Equation 3
Where t= time.
make t the subject of the equation
t = (V-U)/a......................... Equation 4
Given: V = 68 m/s, U = 0 m/s, a = 10052.2
Substitute into equation 4
t = (68-0)/10052.2
t = 0.00678 seconds
Ware them down, its like rubbing two pieces of chalk together.
