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3 years ago

9

A police car is rapidly approaching you at a stop light. Explain what happens to the frequency of the sound as the police car dr

aws closer. What do you hear?

1 answer:

viva [34]3 years ago

7 0

When the police car is moving towards you, the sound waves are more compressed together, leading to a higher pitch sound. This is according to the Doppler Effect, in which sound waves ahead of the object is more compressed, leading to a higher pitch, and more spread out behind, leading to a lower pitch.

hope this helps

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Explain how electromagnetic waves are produced. I

madreJ [45]

Answer:

Electromagnetic are produced by gases and air and so much many things

Explanation:

Science teacher said I did awesome

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2 years ago

A ball is launched vertically with an initial speed of y˙0= 50 m/s, and its acceleration is governed by y¨=-g-cDy˙2, where the a

stira [4]

Answer:

Explanation:

Given

acceleration is given by

$a=-g-c_Dv^2$

where $\ddot{y}=a$

$\dot{y}=v$

Also acceleration is given by

$a=v\frac{\mathrm{d} v}{\mathrm{d} s}$

$ds=\frac{v}{a}dv$

$\int ds=\int \frac{v}{-g-0.001v^2}dv$

$\Rightarrow Let -g-0.001v^2=t$

$-0.001\times 2vdv=dt$

$vdv=-\frac{dt}{0.002}$

$at\ v_0=50\ m/s,\ t=-g-0.001(50)^2$

$t=-g-2.5$

at $v=0,\ t=-g$

$\int_{0}^{s}ds=\int_{-g}^{-g-2.5}\frac{-dt}{0.002t}$

$\int_{0}^{s}ds=\int^{-g}_{-g-2.5}\frac{dt}{0.002t}$

$s=\frac{1}{0.002}lnt|_{-g}^{-g-2.5}$

$s=\frac{1}{0.002}\ln (\frac{g+2.5}{g})$

$s=113.608\ m$

when air drag is neglected maximum height reached is

$h=\frac{v_0^2}{2g}$

$h=\frac{50^2}{2\times 9.8}$

$h=127.55\ m$

3 0

3 years ago

Unpolarized light with an intensity of 655 W / m2 is incident on a polarizer with an unknown axis. The light then passes through

Norma-Jean [14]

Answer:

1. $\theta=29.84^{0}$

2. $\theta=60.15^{0}$

Explanation:

Polarizes axis can create two possible angles with the vertical.

first we have to find the intensity of first polarizer

which is given as

$I=\frac{I_{0} }{2}$

$I= \frac{655\frac{W}{M^{2} } }{2}$

$I=327.5\frac{W}{m^{2} }$

For a smaller angle for the first polarizer:

According to Malus Law

$I_{2} =I_{1} Cos^{2}(90^{0} - \theta)$

$I_{2} =I_{1} sin^{2}\theta$

$\frac{I_{2} }{I_{1} }=Sin^{2}\theta$

taking square root on both sides

$\sqrt{\frac{163}{327.5} } = sin\theta$

$\theta=Sin^{-1}(0.4977)$

$\theta=29.84^{0}$

For a larger angle for the first polarizer:

According to Malus Law

$I_{2} =I_{1} cos^{2}\theta$

$\frac{I_{2} }{I_{1} }=Cos^{2}\theta$

taking square root on both sides

$\sqrt{\frac{163}{327.5} } = cos\theta$

$\theta=Cos^{-1}(0.4977)$

$\theta=60.15^{0}$

7 0

2 years ago

If izzy mass is 0.3kg he applide 657.9n force what will be the accelration

Sholpan [36]

Answer:

The acceleration of the body, a = 2193 m/s²

Explanation:

Given,

The mass of the body, m = 0.3 kg

The force acting on the body, F = 657.9 N

The force acting on an object is proportional to the product of mass and acceleration of the body.

F = m x a

Therefore, the acceleration of the body is

a = F / m

= 657.9 N / 0.3 kg

= 2193 m/s²

Hence, the acceleration of the body, a = 2193 m/s²

4 0

3 years ago

Is it using energy when someone stands against a locker that doesn’t move

soldi70 [24.7K]

When someone stands against a locker and is does not moving at all, then there will be no displacement and since displacement = 0

Work done also becomes equal to zero.

Work done is usually defined as change in energy. Since the work done is zero there has been no energy used.

5 0

2 years ago