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chubhunter [2.5K]
3 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]3 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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A student pushes on a 8-kg box with a force of 35 N forward. The force of sliding friction is 10 N backward. What is the acceler
Ne4ueva [31]

Answer:

(35 N - 10 N)/8kg = 3.125 m/s^2

Explanation:

The formula for Force is:

Force = Mass*Acceleration

(Force is equal to Mass times Acceleration)

Since we're told to find the acceleration of the box. We make acceleration the subject of the equation:

Acceleration = Force/Mass

(Acceleration equal to Force divided by Mass)

We know that the force are 35 N forward and 10 N backward, and the weight of the box is 8kg.

= (35 N - 10 N)/8kg

The reason that 35 N minus 10 N is because the 10 N is pushing the box backward.

= 25 N/8kg

= 3.125 m/s^2

Hope it helps :DD

3 0
3 years ago
Why is the answer B?
djyliett [7]

Answer:

Explanation:

The center of mass lies on a line that joins position 4 of one start with position 4 of the other star.  The shortest distance between these two points will produce the largest velocity. You are using F = m v^2/R

Small R = large force.

Large Force = increased speed.

The masses don't have any effect on the outcome: they remain constant.

7 0
3 years ago
The tow spring on a car has a spring constant of 3,086 N / m and is initially stretched 18.00 cm by a 100.0 kg college student o
sdas [7]

Answer:

The velocity of the skateboard is 0.774 m/s.

Explanation:

Given that,

The spring constant of the spring, k = 3086 N/m

The spring is stretched 18 cm or 0.18 m

Mass of the student, m = 100 kg

Potential energy of the spring, P_f=20\ J

To find,

The velocity of the car.

Solution,

It is a case of conservation of energy. The total energy of the system remains conserved. So,

P_i=K_f+P_f

\dfrac{1}{2}kx^2=\dfrac{1}{2}mv^2+20

\dfrac{1}{2}\times 3086\times (0.18)^2=\dfrac{1}{2}mv^2+20

50-20=\dfrac{1}{2}mv^2

30=\dfrac{1}{2}mv^2

v=\sqrt{\dfrac{60}{100}}

v = 0.774 m/s

So, the velocity of the skateboard is 0.774 m/s.

7 0
3 years ago
Boyles law<br> squeezing a balloon is one way to burst it. Why? ...?
KiRa [710]
According to Boyle's Law,  volume is inversely proportional to pressure. It means if the volume of a gas goes up the pressure goes down and if the volume of the gas goes up the pressure goes down. When the pressure of air inside the inflated balloon is more than the atmospheric pressure outside the balloon. And also when the density inside is greater than the density outside. The molecules inside the balloon move and bang around the inner walls which produces force, which provides the pressure of an enclosed air.
6 0
3 years ago
The block in the figure below has a mass of 5.1 kg and it rests on an incline of angle . You pull on the rope with a force F = 3
viktelen [127]

42.9°

Explanation:

Let's assume that the x-axis is aligned with the incline and the positive direction is up the incline. We can then apply Newton's 2nd law as follows:

x:\;\;\;\;F - mg\sin{\theta} = 0\;\;\;\;

\Rightarrow mg\sin{\theta} = F

Note that the net force is zero because the block is moving with a constant speed when the angle of the incline is set at \theta. Solving for the angle, we get

\sin{\theta} = \dfrac{F}{mg}

or

\theta = \sin^{-1}\left(\dfrac{F}{mg}\right)

\;\;\;=  \sin^{-1}\left[\dfrac{34\:\text{N}}{(5.1\:\text{kg})(9.8\:\text{m/s}^2)}\right]

\;\;\;=42.9°

6 0
2 years ago
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