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Alex
3 years ago
6

If the Sun suddenly went dark, we would not know it until its light stopped arriving on Earth. How long would that be, in second

s, given that the Sun is 1.50 × 1011 m away?
Physics
1 answer:
Gre4nikov [31]3 years ago
5 0

Answer: 500 s

Explanation:

Speed v is defined as a relation between the distance d and time t:

v=\frac{d}{t}

Where:

v=3(10)^{8}m/s is the speed of light in vacuum

d=1.5(10)^{11}m is the distance between the Earth and Sun

t is the time it takes to the light to travel the distance d

Isolating t:

t=\frac{d}{v}

t=\frac{1.5(10)^{11}m}{3(10)^{8}m/s}

Finally:

t=500 s

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What is a good scientific question on a baloon powered car? (it's for my science fair)
Salsk061 [2.6K]

<u>Answer: </u>

Balloon powered car works on the principle of Newtons III law. Escaping air from the balloon, the car accelerates forward. The reaction is the air behind the car, pushing against it, and with the same force car moves forward is the action.

<em>Some scientific questions are:</em>

1. What is the energy stored in the balloon?

Ans: Potential energy (Potential energy is stored in elastic balloon)

2. Which energy is used in balloon-powered car?

Ans: Kinetic energy. (Whenever the balloon escapes the air, potential energy is converted into Kinetic energy)

3. What is the principle involved in balloon powered car?

Ans: Newtons III law.

6 0
4 years ago
A 20-Kg child is on a swing attached to 3.0 m-long chains. The child swings back and forth, swinging out to a 60-degree angle. (
kvv77 [185]

Answer:

 v = 29.4 m / s

Explanation:

For this exercise we can use the conservation of mechanical energy

Lowest starting point.

          Em₀ = K = ½ m v²

final point. Higher

          Em_{f} = U = m g h

Let's use trigonometry to lock her up

          cos 60 = y / L

          y = L cos 60

Height is the initial length minus the length at the maximum angle

           h = L - L cos 60

           h = L (1- cos 60)

energy is conserved

         Em₀ = Em_{f}

          ½ m v² = mgL (1 - cos 60)

         v = 2g L (1- cos 60)

 

let's calculate

          v² = 2 9.8 3.0 (1- cos 60)

          v = 29.4 m / s

6 0
3 years ago
The attractive or repulsive force between objects is called
Romashka [77]
The answer to this is electric force. Hope it helps
3 0
3 years ago
The __________ energy in food is changed into mechanical energy by your muscles.
Orlov [11]
Potential energy which is the stored energy an object has waiting to be used
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3 years ago
Tap on the photo. For each diagram, explain why the light behaves in the way that it does.
dem82 [27]

Answer:

Diagram 1, 3 and 4 can be explained with the phenomenon of refraction.

Refraction occurs when a ray of light crosses the interface between two mediums with different optical density: when this occurs, the ray of light is bent and its speed changes, according to Snell's law

n_1 sin \theta_1 = n_2 sin \theta_2

where n_1,n_2 are the refractive index of the 1st and 2nd medium

\theta_1, \theta_2 are the angle that the incident ray and the refracted ray makes with the normal to the interface

In diagram, 1, the ray of light arrives perpendicularly to the interface, so it is refracted through the medium but it doesn't change its direction (only its speed).

In diagram 3, the ray of light is refracted twice: at the 1st interface and at the 2nd interface. In the 1st case, it goes from a medium with lower refractive index to a medium with higher refractive index (n_1), this means that \theta_2, so the ray bends towards the normal. Vice-versa, in the 2nd case the ray goes from a medium with higher refractive index to a medium with lower refractive index (n_1>n_2), so it bends away from the normal (\theta_2>\theta_1).

In diagram 4, the ray of light is also refracted twice. The ray of light here acts exactly the same as in diagram 3, h

However, this time the 2nd interface is the opposite direction with respect to diagram 3, so in this case the ray of light at the 2nd interface bends in the opposite direction (still away from the normal).

Diagram 2 instead is an example of reflection, that occurs when a ray of light bounces off the interface between the two mediums, withouth entering the 2nd medium.

According to the law of reflection:

- The incoming ray, the reflected ray and the normal to the boundary are all in the same plane

- The angle of incidence is equal to the angle of reflection (both are measured relative to the normal to the boundary)

Therefore in this diagram, the ray of light hits the boundary at approx. 45 degrees from the normal, and then it is reflected back approximately at 45 degrees on the other side with respect to the normal.

3 0
4 years ago
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