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IgorC [24]
3 years ago
11

A police car is driving down the street with it's siren on. You are standing still on the sidewalk beside the street. If the fre

quency of the siren is 1500 Hz and the car is moving towards you at 15 m/s, what is the frequency you would hear from the siren? (v of sound in air = 343 m/s) (1 point)
1568.60 Hz
1321.43 Hz
2019.11 Hz
1676.92 Hz


If the same police car in question 1 has passed you and is now moving away from you at the same speed, what is the frequency you will hear from the siren? (1 point)
1198.67 Hz
1390.41 Hz
1437.15 Hz
1781.55 Hz
Physics
1 answer:
AleksandrR [38]3 years ago
3 0

Answer:

A) 1568.60 Hz

B) 1437.15 Hz

Explanation:

This change is frequency happens due to doppler effect

The Doppler effect is the change in frequency of a wave in relation to an observer who is moving relative to the wave source

f_(observed)=\frac{(c+-V_r)}{(C+-V_s)} *f_(emmited)\\

where

C = the propagation speed of waves in the medium;

Vr= is the speed of the receiver relative to the medium,(added to C, if the receiver is moving towards the source, subtracted if the receiver is moving away from the source;

Vs= the speed of the source relative to the medium, added to C, if the source is moving away from the receiver, subtracted if the source is moving towards the receiver.

A) Here the Source is moving towards the receiver(C-Vs)

and the receiver is standing still (Vr=0) therefore the observed frequency should get higher

f_(observed)=\frac{C}{C-V_s} *f_(emmited)\\=\frac{343}{343-15}*1500\\ =1568.60 Hz

B)Here the Source is moving away the receiver(C+Vs)

and the receiver is still not moving (Vr=0) therefore the observed frequency should be lesser

f_(observed)=\frac{C}{C+V_s} *f_(emmited)\\=\frac{343}{343+15}*1500\\ =1437.15 Hz

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3 years ago
A sealed beaker of what you are told is aqueous nickel sulfide was given to you by the local chemist and you put it on the windo
Masteriza [31]

Answer: D. It is a SUSPENSION

Explanation:

SUSPENSION

This is a combination of two or more single substances. The properties of the components involved are not however changed or lost as is the case with compounds.

For this reason this mixture can be separated due to sedimentation or filtering.

After a few days, this occurs in the aqueous nickel sulfide because the solid nickel sulfide is separating from the water.

7 0
3 years ago
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nlexa [21]

The answer is c, but I'm not sure, just so you know

4 0
3 years ago
Elect the correct answer. Two identical projectiles, A and B, are launched with the same initial velocity, but the angle of laun
allsm [11]

Answer:

A. The range of A and B are equal.

Explanation:

Let u be the initial speed of both the projectile.

The gravitational force acting in the projectile is in the downward direction, sp the speed of the projectile in the horizontal direction remains constant and equals to the initial horizontal speed.

For projectile, a projectile having initial velocity u at an angle \theta with the horizontal direction,

The speed in the horizontal direction = u\cos\theta

and the speed in the vertical direction is = u\sin\theta upward.

For A:

The speed in the horizontal direction = u\cos75^{\circ}

and the speed in the vertical direction is = u\sin75^{\circ} upward.

For B:

The speed in the horizontal direction = u\cos15^{\circ}

and the speed in the vertical direction is = u\sin15^{\circ} upward.

Let t_A and t_B are the time of flight for projectile A and B respectively.

As the range is the horizontal distance traveled by the projectile, so

The range for the projectile A = u\cos75^{\circ}\times t_A\cdots(i)

The range for the projectile B = u\cos15^{\circ}\times t_B\cdots(ii)

At the highest point, the vertical velocity is 0.

Bu using the equation of motion v^2=u^2 +2a s.

Here, the final velocity v=0, the initial velocity u = u \sin \theta , h= vertical distance up to the highest point, and a= -g (as per sign convention).

So, s= \frac{u^2\sin^2 \theta}{2g}

For projectile A: The maximum height attained.

s_A= \frac{u^2\sin^2 75^{\circ}}{2g}

For projectile B: The maximum height attained.

s_B= \frac{u^2\sin^2 15^{\circ}}{2g}

As \sin^2 75^{\circ} > \sin^2 15^{\circ}, the height of A is attained by A is more than the heigHt attained by B.

Now, the times required to reach the highest point from the ground and again form the highest point to the ground are the same.

So, the total time of flight = 2 x (Time to reach the highest point)

In a similar way, by using the equation of motion v=u+at,

The time to reach the highest point =\frac {u\sin\theta}{g}

where g is the acceleration due to gravity.

So, the total time of flight

= 2 \times \frac {u\sin\theta}{g}

The total time of flight for A

=2 \times \frac {u\sin75^{\circ}}{g}

The total time of flight for A

=2 \times \frac {u\sin15^{\circ}}{g}

Now, from equations (i) and (ii),

The range for the projectile A =

u\cos75^{\circ}\times  \frac {2u\sin75^{\circ}}{g}=\frac {u^2 \sin 150^{\circ}}{g}= \frac {u^2 \sin 30^{\circ}}{g}

The range for the projectile B =

u\cos15^{\circ}\times  \frac {2u\sin15^{\circ}}{g}=\frac {u^2 \sin 30^{\circ}}{g}.

Both the projectile have the same range.

Hence, option (A) is correct.

7 0
3 years ago
The motion of a particle is defined by the relation x 5 t 3 2 9t 2 1 24t 2 8, where x and t are expressed in inches and seconds,
shutvik [7]

Answer:

a)  t = 2.0 s,  b)  x_f = - 24.56 m,  Δx = 16.56 m

Explanation:

This is an exercise in kinematics, the relationship of position and time is indicated

          x = 5 t³ - 9t² -24 t - 8

a) ask when the velocity is zero

speed is defined by

         v = \frac{dx}{dt}

let's perform the derivative

        v = 15 t² - 18t - 24

        0 = 15 t² - 18t - 24

let's solve the quadratic equation

      t = \frac{18 \pm \sqrt{18^2 + 4 \ 15 \ 24}  }{2 \ 15}

       t = \frac{18 \pm 42}{30}

       t1 = -0.8 s

      t2 = 2.0 s

the time has to be positive therefore the correct answer is t = 2.0 s

b) the position and distance traveled for a = 0

acceleration is defined by

       a = dv / dt

       a = 30 t - 18

       a = 0

       30 t = 18

       t = 18/30

       t = 0.6 s

we substitute this time in the expression of the position

       

       x = 5 0.6³ - 9 0.6² - 24 0.6 - 8

       x = 1.08 - 3.24 - 14.4 - 8

       x = -24.56 m

we see that all the movement is in one dimension so the distance traveled is the change in position between t = 0 and t = 0.6 s

the position for t = 0

       x₀ = -8 m

the position for t = 0.6 s

      x_f = - 24.56 m

the distance

     ΔX = x_f - x₀

     Δx = | -24.56 -(-8) |

     Δx = 16.56 m

5 0
3 years ago
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