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zmey [24]
3 years ago
5

camera was able to deliver 1.3 frames per second for this photo, and that the car has a length of approximately 5.3 meters. Usin

g this information and the photo itself, approximately how fast did the car drive
Physics
1 answer:
son4ous [18]3 years ago
7 0

The question is incomplete. Here is the complete question.

The image below was taken with a camera that can shoot anywhere between one and two frames per second. A continuous series of photos was combined  for this image, so the cars you see are in fact the same car, but photographed at differene times.

Let's assume that the camera was able to deliver 1.3 frames per second for this photo, and that the car has a length of approximately 5.3 meters. Using this information and the photo itself, approximately how fast did the car drive?

Answer: v = 6.5 m/s

Explanation: The question asks for velocity of the car. Velocity is given by:

v=\frac{\Delta x}{\Delta t}

The camera took 7 pictures of the car and knowing its length is 5.3, the car's displacement was:

Δx = 7(5.3)

Δx = 37.1 m

The camera delivers 1.3 frames per second and it was taken 7 photos, so time the car drove was:

1.3 frames = 1 s

7 frames = Δt

Δt = 5.4 s

Then, the car was driving:

v=\frac{37.1}{5.4}

v = 6.87 m/s

The car drove at, approximately, a velocity of 6.87 m/s

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What frequency is received by a person watching an oncoming ambulance moving at 110 km/h and emitting a steady 800-Hz sound from
Strike441 [17]

To develop this problem we will apply the concepts related to the Doppler effect. The frequency of sound perceive by observer changes from source emitting the sound. The frequency received by observer f_{obs} is more than the frequency emitted by the source. The expression to find the frequency received by the person is,

f_{obs} = f_s (\frac{v_w}{v_w-v_s})

f_s= Frequency of the source

v_w= Speed of sound

v_s= Speed of source

The velocity of the ambulance is

v_s = 119km/h (\frac{1000m}{1km})(\frac{1h}{3600s})

v_s = 30.55m/s

Replacing at the expression to frequency of observer we have,

f_{obs} = 800Hz(\frac{345m/s}{345m/s-30.55m/s})

f_{obs} = 878Hz

Therefore the frequency receive by observer is 878Hz

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3 years ago
If the mass of one of two objects is increased, the force of attraction between them will
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It will increase..........................
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A soccer ball collides with another soccer ball at rest. the total momentum of the balls
Juliette [100K]

I am pretty sure the answer is C.

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In your daily life you come across a range of motion in which acceleration is in the direction of motion
irakobra [83]
That makes no sense to me somehow
3 0
3 years ago
*please refer to photo*
just olya [345]

Based on the calculations, the average velocity is equal to 360 m/s and the percent difference is equal to 4.72%.

<h3>What is average velocity?</h3>

An average velocity can be defined as the total distance covered by a physical object divided by the total time taken.

<h3>What is an average?</h3>

An average is also referred to as mean and it can be defined as a ratio of the sum of the total number in a data set to the frequency of the data set.

<h3>How to calculate the average velocity?</h3>

Mathematically, the average velocity for this data set would be calculated by using this formula:

Average = [F(v)]/n

Vavg = [v₁ + v₂ + v₃ + v₄ + v₅)/5

Since the values of the average velocity from the table are missing, we would assume the following values for the purpose of an explanation:

  • v₁ = 100 m/s
  • v₂ = 150 m/s
  • v₃ = 200 m/s
  • v₄ = 250 m/s
  • v₅ = 300 m/s

Substituting the parameters into the formula, we have:

Vavg = [300 + 450 + 500 + 250 + 300)/5

Vavg = 1800/5

Vavg = 360 m/s.

Next, we would calculate the percent difference by using this formula:

Percent \;difference = \frac{[V_{avg}\;-\;V_{sound}]}{V_{sound}} \times 100

Percent difference = [360 - 343]/360 × 100

Percent difference = 17/360 × 100

Percent difference = 0.0472 × 100

Percent difference = 4.72%.

Read more on average here: brainly.com/question/9550536

#SPJ1

3 0
1 year ago
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