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n200080 [17]
3 years ago
16

A student hears a police siren. What would change the frequency that the student hears? Check all that apply.

Physics
2 answers:
BlackZzzverrR [31]3 years ago
9 0

Answer:

1

2

3

4

Explanation:

It's right on Edge

ser-zykov [4K]3 years ago
4 0
<span>A student hears a police siren.

The arithmetic of the Doppler Effect shows that if the distance between
the source and observer is changing, then the observer hears a different
frequency compared to the frequency actually radiating from the source. 

Thus the first four choices would cause the student to hear a different
frequency:

-- if the student walked toward the police car
-- if the student walked away from the police car
-- if the police car moved toward the student
-- if the police car moved away from the student

The last two choices wouldn't affect the frequency heard by the student,
since the perceived frequency of a sound doesn't depend on its intensity.

-- if the intensity of the siren increased
-- if the intensity of the siren decreased.</span>
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Let’s say I am in a bumper car and have a velocity of 14 m/s, driving in the positive x-direction. I and my bumped car have a ma
AlekseyPX

Answer:

160 kg

12 m/s

Explanation:

m_1 = Mass of first car = 120 kg

m_2 = Mass of second car

u_1 = Initial Velocity of first car = 14 m/s

u_2 = Initial Velocity of second car = 0 m/s

v_1 = Final Velocity of first car = -2 m/s

v_2 = Final Velocity of second car

For perfectly elastic collision

m_{1}u_{1}+m_{2}u_{2}=m_{1}v_{1}+m_{2}v_{2}\\\Rightarrow m_2v_2=m_{1}u_{1}+m_{2}u_{2}-m_{1}v_{1}\\\Rightarrow m_2v_2=120\times 14+m_2\times 0-(120\times -2)\\\Rightarrow m_2v_2=1920\\\Rightarrow m_2=\frac{1920}{v_2}

Applying in the next equation

v_2=\frac{2m_1}{m_1+m_2}u_{1}+\frac{m_2-m_1}{m_1+m_2}u_2\\\Rightarrow v_2=\frac{2\times 120}{120+\frac{1920}{v_2}}\times 14+\frac{m_2-m_1}{m_1+m_2}\times 0\\\Rightarrow \left(120+\frac{1920}{v_2}\right)v_2=3360\\\Rightarrow 120v_2+1920=3360\\\Rightarrow v_2=\frac{3360-1920}{120}\\\Rightarrow v_2=12\ m/s

m_2=\frac{1920}{v_2}\\\Rightarrow m_2=\frac{1920}{12}\\\Rightarrow m_2=160\ kg

Mass of second car = 160 kg

Velocity of second car = 12 m/s

5 0
3 years ago
24. Most cleaning solutions like soap, with a pH of 8-9 are examples of a(n)
Sphinxa [80]

24. A, natural ph scale for it not to be acidic is 7-8

25. A, not sure about this one

26. A, looked it up

27. A, because it has to be shaken up to make the mixture appear and taste more combined.

28. D, i just guessed

29. D, that answer is stupid so it is the answer because it said not a property

30. A, looked it up

8 0
3 years ago
A net torque of magnitude 600 Nm is ex-
vlabodo [156]
A net torque of magnitude is 600

5 0
3 years ago
Consider the same roller coaster. It starts at a height of 40.0 m but once released, it can only reach a height of 25.0 m above
poizon [28]

Answer:

The magnitude of the frictional force between the car and the track is 367.763 N.

Explanation:

The roller coster has an initial gravitational potential energy, which is partially dissipated by friction and final gravitational potential energy is less. According to the Principle of Energy Conservation and Work-Energy Theorem, the motion of roller coster is represented by the following expression:

U_{g,1} = U_{g,2} + W_{dis}

Where:

U_{g,1}, U_{g,2} - Initial and final gravitational potential energy, measured in joules.

W_{dis} - Dissipated work due to friction, measured in joules.

Gravitational potential energy is described by the following formula:

U = m \cdot g \cdot y

Where:

m - Mass, measured in kilograms.

g - Gravitational constant, measured in meters per square second.

y - Height with respect to reference point, measured in meters.

In addition, dissipated work due to friction is:

W_{dis} = f \cdot \Delta s

Where:

f - Friction force, measured in newtons.

\Delta s - Travelled distance, measured in meters.

Now, the energy equation is expanded and frictional force is cleared:

m \cdot g \cdot (y_{1} - y_{2}) = f\cdot \Delta s

f = \frac{m \cdot g \cdot (y_{1}-y_{2})}{\Delta s}

If m = 1000\,kg, g = 9.807\,\frac{m}{s^{2}}, y_{1} = 40\,m, y_{2} = 25\,m and \Delta s = 400\,m, then:

f = \frac{(1000\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (40\,m-25\,m)}{400\,m}

f = 367.763\,N

The magnitude of the frictional force between the car and the track is 367.763 N.

7 0
3 years ago
At the end of the Shock Wave roller coaster ride, the 6000-kg train of cars (includes passengers) is allowed from a speed of 20
Angelina_Jolie [31]

Answer:

56250 N

Explanation:

mass, m = 6000 kg

initial speed, u = 20 m/s

final speed, v = 5 m/s

distance, s = 20 m

Use third equation of motion

v^{2}=u^{2}+2as

5 x 5 = 20 x 20 + 2 a x 20

25 = 400 + 40 a

a = - 9.375 m/s^2

Braking force, F = mass x acceleration

F = 6000 x 9.375

F = 56250 N

6 0
3 years ago
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