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disa [49]
3 years ago
6

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

screech of frequency 3400 Hz? Assume room temperature air.
Physics
1 answer:
Montano1993 [528]3 years ago
6 0

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' = (\dfrac{v}{v-v_s}})f

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' = \dfrac{341}{341-24.7}}\times 3400

   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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the earth has more rotational kinetic energy now than did the cloud of gas and dust from which it formed. where did this energy
goldfiish [28.3K]

Explanation:

Earth or any planet are actually born from huge clouds of gas and dust. Their stellar mass are fairly distributed at a radius from the axis of rotation. Gravitational force cause the cloud to come together. Now the whole gathered in smaller area. Now, individual particles come close to the roational axis. Thus, decreasing the moment of inertia of the planet.

As

I=mr^2

reducing r reduces I. However, the angular moment of the system remains always conserved. So, to conserve the angular momentum the angular velocity of the planet increases and so did the  otational kinetic energy

3 0
3 years ago
Two electrons in a vacuum exert force of F = 3.8E-09 N on each other. They are then moved such that they are separated by x = 8.
iren [92.7K]

Answer:

F_n = 5.65E-11 N

d =  1.20682E-31 m

Explanation:

F = 3.8E-09 N

where

m = Mass of electron = 9.109E−31 kilograms

G = Gravitational constant = 6.67E-11 m³/kgs²

x = Distance between them

F=G\frac{m^2}{x^2}\\\Rightarrow 3.8E-09=G\frac{m^2}{x^2}

For F_n

F_n=G\frac{m^2}{x^2}\\\Rightarrow F_n=G\frac{m^2}{(8.2x)^2}\\\Rightarrow F_n=G\frac{m^2}{67.24x^2}

Dividing the above equations we get

\frac{F}{F_n}=\frac{G\frac{m^2}{x^2}}{G\frac{m^2}{67.24x^2}}\\\Rightarrow \frac{F}{F_n}=67.24\\\Rightarrow F_n=\frac{F}{67.24}\\\Rightarrow F_n=\frac{3.8E-09}{67.24}\\\Rightarrow F_n=5.65E-11\ N

F_n = 5.65E-11 N

F=G\frac{m^2}{x^2}\\\Rightarrow x=\sqrt{\frac{Gm^2}{F}}\\\Rightarrow x=\sqrt{\frac{G}{F}}m\\\Rightarrow x=\sqrt{\frac{6.67E-11}{3.8E-09}}9.109E-31\\\Rightarrow x=1.20682E-31\ m

d =  1.20682E-31 m

8 0
4 years ago
This chart shows characteristics of three different waves all with the same wavelength of 10 m but moving at different frequenci
kap26 [50]

Answer:

the answer is B

Explanation:

wave x has the highest hertz making it the answer

6 0
3 years ago
You are trying to find the amount of heat transferred between two substances. In order to do this, you plan
Zarrin [17]

Answer:

do you have kids ?

Explanation:

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7 0
2 years ago
Consider massive gliders that slide friction-free along a horizontal air track. Glider A has a mass of 1 kg, a speed of 1 m/s, a
mamaluj [8]

Answer:

0.167m/s

Explanation:

According to law of conservation of momentum which States that the sum of momentum of bodies before collision is equal to the sum of the bodies after collision. The bodies move with a common velocity after collision.

Given momentum = Maas × velocity.

Momentum of glider A = 1kg×1m/s

Momentum of glider = 1kgm/s

Momentum of glider B = 5kg × 0m/s

The initial velocity of glider B is zero since it is at rest.

Momentum of glider B = 0kgm/s

Momentum of the bodies after collision = (mA+mB)v where;

mA and mB are the masses of the gliders

v is their common velocity after collision.

Momentum = (1+5)v

Momentum after collision = 6v

According to the law of conservation of momentum;

1kgm/s + 0kgm/s = 6v

1 =6v

V =1/6m/s

Their speed after collision will be 0.167m/s

6 0
3 years ago
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