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Ulleksa [173]
3 years ago
14

How many nanoseconds does it take light to travel 3.50 ft in vacuum?

Physics
1 answer:
Fiesta28 [93]3 years ago
6 0
Answer:3.56 nanosecond

In this case, you are asked the time and given the light distance(3.5ft)
To answer this question you would need to know the velocity of light. Speed of light is <span>299792458m/s. Then the calculation would be:

time= distance/speed
time= 3.5 ft / (</span>299792458m/s) x 0.3048 meter/ 1 ft=  3.56 10^{-9} second or 3.56 nanosecond
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At which point is the potential energy the greatest? At which point is kinetic energy the greatest?
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Kinetic energy is greatest at the lowest point of a roller coaster and least at the highest point.

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How do convection cells in Earth's atmosphere cause high and low pressure belts?
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Areas of rising air causes low air pressure areas and <span>Areas of sinking air are areas of high air </span>pressure
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Find the unknown mass of the block 1 needed to balance the bar. Assume that the mass of the bar is negligible. Block 1 is locate
Natasha2012 [34]

Answer:

Incomplete question: The masses of the blocks m₂ = 1.5 kg and m₃ = 2 kg

Explanation:

Given data:

L₁ = length = 0.85 m

L₂ = 0.25 m

L₃ = 0.5 m

m₂ = 1.5 kg

m₃ = 2 kg

Question: Find the unknown mass of the block 1 needed to balance the bar, m₁ = ?

The torque is zero (intermediate point of the bar)

-m_{1} gL_{1} +m_{2} gL_{2} +m_{3} gL_{3} =0

Is negative because mass 1 is to the left of the coordinate system (see the diagram)

m_{1} =\frac{m_{2} gL_{2}+m_{3} gL_{3}  }{L_{1} } =\frac{(1.5*0.25)+(2*0.5)}{0.85} =1.6176kg

4 0
4 years ago
Water is more elastic than air. Why?
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4 years ago
Two cars are facing each other. Car A is at rest while car B is moving toward car A with a constant velocity of 20 m/s. When car
lapo4ka [179]

Answer:

Let's define t = 0s (the initial time) as the moment when Car A starts moving.

Let's find the movement equations of each car.

A:

We know that Car A accelerations with a constant acceleration of 5m/s^2

Then the acceleration equation is:

A_a(t)  = 5m/s^2

To get the velocity, we integrate over time:

V_a(t) = (5m/s^2)*t + V_0

Where V₀ is the initial velocity of Car A, we know that it starts at rest, so V₀ = 0m/s, the velocity equation is then:

V_a(t) = (5m/s^2)*t

To get the position equation we integrate again over time:

P_a(t) = 0.5*(5m/s^2)*t^2 + P_0

Where P₀ is the initial position of the Car A, we can define P₀ = 0m, then the position equation is:

P_a(t) = 0.5*(5m/s^2)*t^2

Now let's find the equations for car B.

We know that Car B does not accelerate, then it has a constant velocity given by:

V_b(t) =20m/s

To get the position equation, we can integrate:

P_b(t) = (20m/s)*t + P_0

This time P₀ is the initial position of Car B, we know that it starts 100m ahead from car A, then P₀ = 100m, the position equation is:

P_b(t) = (20m/s)*t + 100m

Now we can answer this:

1) The two cars will meet when their position equations are equal, so we must have:

P_a(t) = P_b(t)

We can solve this for t.

0.5*(5m/s^2)*t^2 = (20m/s)*t + 100m\\(2.5 m/s^2)*t^2 - (20m/s)*t - 100m = 0

This is a quadratic equation, the solutions are given by the Bhaskara's formula:

t = \frac{-(-20m/s) \pm \sqrt{(-20m/s)^2 - 4*(2.5m/s^2)*(-100m)}  }{2*2.5m/s^2} = \frac{20m/s \pm 37.42 m/s}{5m/s^2}

We only care for the positive solution, which is:

t = \frac{20m/s + 37.42 m/s}{5m/s^2} = 11.48 s

Car A reaches Car B after 11.48 seconds.

2) How far does car A travel before the two cars meet?

Here we only need to evaluate the position equation for Car A in t = 11.48s:

P_a(11.48s) = 0.5*(5m/s^2)*(11.48s)^2 = 329.48 m

3) What is the velocity of car B when the two cars meet?

Car B is not accelerating, so its velocity does not change, then the velocity of Car B when the two cars meet is 20m/s

4)  What is the velocity of car A when the two cars meet?

Here we need to evaluate the velocity equation for Car A at t = 11.48s

V_a(t) = (5m/s^2)*11.48s = 57.4 m/s

7 0
3 years ago
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