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chubhunter [2.5K]
4 years ago
15

A stationary police car emits a sound of frequency 1200 hz that bounces off a car on the highway and returns with a frequency of

1250 hz. the police car is right next to the highway, so the moving car is traveling directly toward or away from it. (a) how fast was the moving car going?
Physics
1 answer:
Whitepunk [10]4 years ago
4 0

First, since the echo frequency is greater, the car must be roaming to the police car. 
This is for the reason that the car is "running over" its own wave fronts, creating the wavelengths tinier, and henceforth increasing the frequency. 

The trick here is to understand that the relative speed between the cars is half of the speed of the echo for the reason that the echo is returning to the police car at the similar speed as the sound wave trips to the moving car. 

It would also appear that the speed of sound has been expected to be 337 m/s 

So if v is the moving car's velocity, the correct formula to use is: 

2v / 337 = (1250/1200) - 1 
2v = 14.04 m/s 
v = 7.02 m/s

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Neil pogo sticks to his science class, but stops to pick up his backpack on his way. He travels 8 m east, then 4 m west. what di
Likurg_2 [28]

Answer:

8.9 m NorthEast

Explanation:

a² + b² = c²

8² + 4² = c²

64 + 16 = c²

80 = c²

c = √80

c ≈ 8.9 m NorthEast

8 0
3 years ago
A 0.46-kg cord is stretched between two supports, 7.2 m apart. When one support is struck by a hammer, a transverse wave travels
Kryger [21]

Answer:

      T = 6.0 N

Explanation:

given,

mass of the cord = 0.46 Kg

length of the supports = 7.2 m

time taken to travel = 0.74 s

tension in the chord = ?

using formula for tension calculation

T = \dfrac{v^2.m}{l}

v = \dfrac{l}{s}

v = \dfrac{7.2}{0.74}

v = 9.73 m/s

now, calculation of tension

T = \dfrac{9.73^2\times 0.46}{7.2}

      T = 6.0 N

The tension in the cord is equal to 6.0 N.

7 0
3 years ago
Electromagnetic waves differ from other types of waves because they are
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I think it is D because u think of space and Electromagnetic Waves Travel Through Space Moving All Kinds of Particles From The Sun With It Which Creates The Northern and Southern Lights In The Poles :-)
8 0
3 years ago
A 50 kg runner runs up a flight of stairs. The runner starts out covering 3 steps every second. At the end the runner stops. Thi
nadezda [96]

To solve the problem it is necessary to take into account the concepts of kinematic equations of motion and the work done by a body.

In the case of work, we know that it is defined by,

W = F * d

Where,

F= Force

d = Distance

The distance in this case is a composition between number of steps and the height. Then,

d=h*N, for h as the height of each step and N number of steps.

On the other hand we have the speed changes, depending on the displacement and acceleration (omitting time)

V_f^2-V_i^2 = 2a\Delta X

Where,

V_f = Final velocity

V_i = Initial Velocity

a = Acceleration

\Delta X = Displacement

PART A) For the particular case of work we know then that,

W = F*d

W = m*g*(h*N)

W = 50*9.8*(0.3*30)

W = 4.41kJ

Therefore the Work to do that activity is 4.41kJ

PART B) To find the acceleration (from which we can later find the time) we start from the previously given equation,

V_f^2-V_i^2 = 2a\Delta X

Here,

V_i = \frac{0.3*3}{1} = 0.90m/s\rightarrow3 steps in one second

v_f = 0

Replacing,

V_f^2-V_i^2 = 2a\Delta X

0-0.9^2=2a(30*0.3)

Re-arrange for a,

a = -\frac{0.9^2}{2*30*0.3}

a = -45*10^{-3}m/s^2

At this point we can calculate the time, which is,

t = \frac{\Delta V}{a}

t = \frac{0-0.9}{-45*10^{-3}}

t = 20s

With time and work we can finally calculate the power

P = \frac{W}{t} = \frac{4.41}{20}

P = 0.2205kW

6 0
4 years ago
Calculate the potential energy of a 5 kg object sitting at the top of a 2 meter ramp.
trapecia [35]
The formula used to find potential energy is <em>P.E. = M * G * H</em> (P.E. is potential energy, M is mass, G is gravitational pull, and H is height). So the answer to your question is <em>5 * 9.8 * 2</em>, which equals 98.
5 0
3 years ago
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