What frequency is received by a stationary mouse just before being dispatched by a hawk flying at it at 24.7 m/s and emitting a
1 answer:
Answer:
f' = 3665.51 Hz
Explanation:
given,
speed of the hawk = 24.7 m/s
frequency of screech emitted by the hawk = 3400 Hz
speed of sound = 331 m/s
By Doppler's effect
f' is the frequency received by the mouse
v is the speed of the sound
v_s is the speed of the hawk
now,
f' = 1.078 x 3400
f' = 3665.51 Hz
The frequency received by the stationary mouse is equal to 3665.51 Hz
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Answer:
a) x = v₀² sin 2θ / g
b) t_total = 2 v₀ sin θ / g
c) x = 16.7 m
Explanation:
This is a projectile launching exercise, let's use trigonometry to find the components of the initial velocity
sin θ = / vo
cos θ = v₀ₓ / vo
v_{oy} = v_{o} sin θ
v₀ₓ = v₀ cos θ
v_{oy} = 13.5 sin 32 = 7.15 m / s
v₀ₓ = 13.5 cos 32 = 11.45 m / s
a) In the x axis there is no acceleration so the velocity is constant
v₀ₓ = x / t
x = v₀ₓ t
the time the ball is in the air is twice the time to reach the maximum height, where the vertical speed is zero
v_{y} = v_{oy} - gt
0 = v₀ sin θ - gt
t = v_{o} sin θ / g
we substitute
x = v₀ cos θ (2 v_{o} sin θ / g)
x = v₀² /g 2 cos θ sin θ
x = v₀² sin 2θ / g
at the point where the receiver receives the ball is at the same height, so this coincides with the range of the projectile launch,
b) The acceleration to which the ball is subjected is equal in the rise and fall, therefore it takes the same time for both parties, let's find the rise time
at the highest point the vertical speed is zero
v_{y} = v_{oy} - gt
v_{y} = 0
t = v_{oy} / g
t = v₀ sin θ / g
as the time to get on and off is the same the total time or flight time is
t_total = 2 t
t_total = 2 v₀ sin θ / g
c) we calculate
x = 13.5 2 sin (2 32) / 9.8
x = 16.7 m
Answer:
Rate of change of area will be
Explanation:
We have given rate of change of radius
Radius of the circular plate r = 52 cm
Area is given by
So
Puting the value of r and
So rate of change of area will be