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Sav [38]
3 years ago
15

Suppose a radar gun sends out radio waves with a frequency of 2,000,000.0 Hz. The waves bounce off a moving car and return with

a frequency of 2,000,000.2 Hz. If the speed of light is 300,000,000 m/s, what is the velocity of the car
Physics
1 answer:
Igoryamba3 years ago
6 0

Answer:

300,000,030m/s

Explanation:

Using the relationship between frequency and speed of a wave;

F ∝ V

F = kV

k = F/V

F1/V1 = F2/V2 = k

Let F1 be the frequency of the radio wave = 2,000,000.0Hz

V1 be the speed of light = 300,000,000m/s

F2 = frequency produced by the car = 2,000,000.2Hz

V2 be the velocity of the moving car

Substituting this values in the equation above;

2,000,000/300,000,000 = 2,000,000.2/V2

Cross multiplying

2,000,000V2 = 300,000,000×2,000,000.2

2V2 = 300×2,000,000.2

V2 = 150×2,000,000.2

V2 = 300,000,030m/s

The velocity of the car is 300,000,030m/s

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An FM radio station broadcasts electromagnetic radiation at a frequency of 93.5 MHz. The wavelength of this radiation is _______
Alborosie

Answer:

3.21

Explanation:

The relation between frequency and wavelength is shown below as:

c=frequency\times Wavelength


c is the speed of light having value 3\times 10^8\ m/s

Given, Frequency = 93.5 MHz = 93.5\times 10^{6}\ Hz

Thus, Wavelength is:

Wavelength=\frac{c}{Frequency}

Wavelength=\frac{3\times 10^8}{93.5\times 10^{6}}\ m

Wavelength=3.21 \ m

<u>Answer - A.</u>

7 0
3 years ago
If mechanical energy is conserved, then a pendulum that has a potential energy of 20 J at its highest point and 0.5 J at its low
liraira [26]
At the highest point: kinetic energy is 0 due to the speed  is 0

So the total mechanical energy is 20

Assume no frictions present, then the mechanical energy is conserved

So at the lowest point, kinetic energy = mechanical energy - potential energy

Answer will be 20 - 0.5 = 19.5 J
4 0
3 years ago
A football player runs down a field with a speed of 8 m/s. How long will it take him to run 20m?
Shtirlitz [24]
To find the answer for this question, you simply need to divide 20 by 8, which is the speed he is traveling. 

20 / 8 = 2.5

The football player will run 20 yards in 2 1/2 seconds.
Hope that helped! =)
5 0
2 years ago
Read 2 more answers
PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!<br><br> Joules could be used to measure
lyudmila [28]

All of the above, work is a measurement of energy transfer, in Joules.

Potential energy = Joules

Kinetic energy = Joules

The key thing here is that anything having to do with just energy or energy transfer is measured in joules.

4 0
3 years ago
Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.
SIZIF [17.4K]

The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

Answer: (a) Total mass = 2384.5kg;

               (b) m1 = 915kg;

                   m2 = 605kg;

                   m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

F_{pull}=m_{T}.a

m_{T}=\frac{F_{pull}}{a}

m_{T}=\frac{3615}{1.516}

m_{T}=2384.5

Total mass of the system of boxes is 2384.5kg.

(b) For each mass, analyse each box and make them each a free-body diagram.

<u>For </u>m_{1}<u>:</u>

The only force acting On the m_{1} box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

m_{1} = \frac{F_{12}}{a}

m_{1} = \frac{1387}{1.516}

m_{1} = 915kg

<u>For </u>m_{2}<u>:</u>

There are two forces acting on m_{2}: tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

m_{2} = \frac{F_{23}-F_{12}}{a}

m_{2} = \frac{2304-1387}{1.516}

m_{2} = 605kg

<u>For </u>m_{3}<u>:</u>

m_{3} = m_{T} - (m_{1}+m_{2})

m_{3} = 2384.5-1520.0

m_{3} = 864.5kg

8 0
3 years ago
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