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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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A stone with a mass of 0.100kg rests on a frictionless, horizontal surface. A bullet of mass 2.50g traveling horizontally at 500
jolli1 [7]

Answer:

Explanation:

Given that:

mass of stone (M) = 0.100 kg

mass of bullet (m) = 2.50 g = 2.5  ×10 ⁻³ kg

initial velocity of stone (u_{stone}) = 0 m/s

Initial velocity of bullet (u_{bullet}) = (500 m/s)i

Speed of the bullet after collision (v_{bullet}) = (300 m/s) j

Suppose we represent (v_{stone}) to be the velocity of the stone after the truck, then:

From linear momentum, the law of conservation can be applied which is expressed as:

m*u_{bullet} + M*{u_{stone}}= mv_{bullet}+Mv_{stone}

(2.50*10^{-3} \ kg) (500)i+0 = (2.50*10^{-3} \ kg)(300 \ m/s)j + (0.100 \ kg)v_{stone}

(2.50*10^{-3} \ kg) (500)i- (2.50*10^{-3} \ kg)(300 \ m/s)j=  (0.100 \ kg)v_{stone}

v_{stone}= (1.25\  kg.m/s)i-(0.75\ kg m/s)j

v_{stone}= (12.5\  m/s)i-(7.5\ m/s)j

∴

The magnitude now is:

v_{stone}=\sqrt{ (12.5\  m/s)^2-(7.5\ m/s)^2}

\mathbf{v_{stone}= 14.6 \ m/s}

Using the tangent of an angle to determine the direction of the velocity after the struck;

Let θ represent the direction:

\theta = tan^{-1} (\dfrac{-7.5}{12.5})

\mathbf{\theta = 30.96^0 \ below \ the \ horizontal\ level}

5 0
3 years ago
Vision is blurred if the head is vibrated at 29 Hz because the vibrations are resonant with the natural frequency of the eyeball
Likurg_2 [28]

Answer:

k = 4422.35  KN/m

Explanation:

Given that

Frequency ,f= 29 Hz

m = 7.5 g

Natural frequency ω

ω = 2 π f

We also know that for spring mass system

ω ² m =k        

k=Spring constant

So we can say that

( 2 π f)² =  m k

By putting the values

(2 x π x 29)² = 7.5 x 10⁻³  k

33167.69 = 7.5 x 10⁻³  k

k=4422.35 x  10³ N/m

k = 4422.35  KN/m

Therefore spring constant will be 4422.35  KN/m

4 0
3 years ago
All of the waves in the electromagnetic spectrum are _______ waves.
user100 [1]
Transverse, I think. I may be wrong.
5 0
3 years ago
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A kite 100 ft above the ground moves horizontally at a speed of 12 ft/s. at what rate is the angle (in radians) between the stri
Viefleur [7K]
<span>Answer: So this involves right triangles. The height is always 100. Let the horizontal be x and the length of string be z. So we have x2 + 1002 = z2. Now take its derivative in terms of time to get 2x(dx/dt) = 2z(dz/dt) So at your specific moment z = 200, x = 100âš3 and dx/dt = +8 substituting, that makes dz/dt = 800âš3 / 200 or 4âš3. Part 2 sin a = 100/z = 100 z-1 . Now take the derivative in terms of t to get cos a (da./dt) = -100/ z2 (dz/dt) So we know z = 200, which makes this a 30-60-90 triangle, therefore a=30 degrees or Ď€/6 radians. Substitute to get cos (Ď€/6)(da/dt) = (-100/ 40000)(4âš3) âš3 / 2 (da/dt) = -âš3 / 100 da/dt = -1/50 radians</span>
5 0
3 years ago
A mango falls fromthe top its tree passing a window which is 2.4m tall by taking 0.4s
Natasha2012 [34]

Explanation:

There are three points in time we need to consider.  At point 0, the mango begins to fall from the tree.  At point 1, the mango reaches the top of the window.  At point 2, the mango reaches the bottom of the window.

We are given the following information:

y₁ = 3 m

y₂ = 3 m − 2.4 m = 0.6 m

t₂ − t₁ = 0.4 s

a = -9.8 m/s²

t₀ = 0 s

v₀ = 0 m/s

We need to find y₀.

Use a constant acceleration equation:

y = y₀ + v₀ t + ½ at²

Evaluated at point 1:

3 = y₀ + (0) t₁ + ½ (-9.8) t₁²

3 = y₀ − 4.9 t₁²

Evaluated at point 2:

0.6 = y₀ + (0) t₂ + ½ (-9.8) t₂²

0.6 = y₀ − 4.9 t₂²

Solve for y₀ in the first equation and substitute into the second:

y₀ = 3 + 4.9 t₁²

0.6 = (3 + 4.9 t₁²) − 4.9 t₂²

0 = 2.4 + 4.9 (t₁² − t₂²)

We know t₂ = t₁ + 0.4:

0 = 2.4 + 4.9 (t₁² − (t₁ + 0.4)²)

0 = 2.4 + 4.9 (t₁² − (t₁² + 0.8 t₁ + 0.16))

0 = 2.4 + 4.9 (t₁² − t₁² − 0.8 t₁ − 0.16)

0 = 2.4 + 4.9 (-0.8 t₁ − 0.16)

0 = 2.4 − 3.92 t₁ − 0.784

0 = 1.616 − 3.92 t₁

t₁ = 0.412

Now we can plug this into the original equation and find y₀:

3 = y₀ − 4.9 t₁²

3 = y₀ − 4.9 (0.412)²

3 = y₀ − 0.83

y₀ = 3.83

Rounded to two significant figures, the height of the tree is 3.8 meters.

6 0
3 years ago
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