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padilas [110]
4 years ago
13

We see a full moon by reflected sunlight. how much earlier did the light that enters our eye leave the sun

Physics
1 answer:
galina1969 [7]4 years ago
7 0
We know that the source of light in the universe is the Sun. Hence, the light we see as moonlight travels from the Sun's surface, to the moon, then to Earth. So, before being able to solve this problem, we have to know the distance between the Sun and the moon, and the distance between the moon and Earth. In literature, these values are 3.8×10⁵ km (Sun to moon) and 384,400 km (moon to Earth). Knowing that the speed of light is 300,000 km per second, then the total time would be

Time = distance/speed
Time = (3.8×10⁵ km + 384,400 km)/300,000 km/s
Time = 2.548 seconds

Thus, it only takes 2.548 for the light from the Sun to reach to the Earth as perceived to be what we call moonlight.
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Write a hypothesis for Part II of the lab, which is about the relationship described by F = ma. In the lab, you will use a toy c
hoa [83]

Answer:

F=ma is the relationship where, F is force, m is mass and a is acceleration.

Newton's second law states that  the unbalanced force applied to the object accelerates the object which is directly proportional to the force and inversely to the mass.

If we apply force to a toy car then It will accelerate.

This is how Newton's second law of motion is verified.

5 0
3 years ago
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A temperature of 20°c is equivalent to approximately
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8 0
3 years ago
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Experimenting with free fall, Mariana observes that her baseball takes 1.5 s to travel the last 30m before hitting the ground. F
Art [367]

Answer:

37.8 m

Explanation:

At point 0, the ball is at height y₀.

At point 1, the ball is at height 30 m.

At point 2, the ball is at height 0 m.

Given:

y₁ = 30 m

y₂ = 0 m

v₀ = 0 m/s

a = -10 m/s²

t₂ − t₁ = 1.5 s

Find: y₀

Use constant acceleration equation.

y = y₀ + v₀ t + ½ at²

Evaluate at point 1.

y₁ = y₀ + v₀ t₁ + ½ at₁²

30 m = y₀ + (0 m/s) t₁ + ½ (-10 m/s²) t₁²

30 = y₀ − 5t₁²

Evaluate at point 2.

y₂ = y₀ + v₀ t₂ + ½ at₂²

0 m = y₀ + (0 m/s) t₂ + ½ (-10 m/s²) t₂²

0 = y₀ − 5t₂²

y₀ = 5t₂²

Substitute:

y₀ = 5 (1.5 + t₁)²

y₀ = 5 (2.25 + 3t₁ + t₁²)

y₀ = 11.25 + 15t₁ + 5t₁²

30 = 11.25 + 15t₁ + 5t₁² − 5t₁²

30 = 11.25 + 15t₁

t₁ = 1.25

30 = y₀ − 5t₁²

30 = y₀ − 5(1.25)²

y₀ ≈ 37.8

4 0
4 years ago
If a car is accelerating downhill under a net force 3674 N , what additional force would cause the car to have a constant veloci
Semenov [28]

Answer:

For the car to move with constant velocity the additional force required is  F__{dg} }=  -3674 \  N

Explanation:

From the question we are told that

  The net force of the car is  F_{net} = 3674 \  N

Generally the total  force acting on the  car is the net force plus the force due to gravity acting in direction of the car (Let denote it as F__{dg})

So the total force acting on the car is mathematically represented as

        F_{net} + F__{dg}} = F

Here this F representing the total force can be mathematically represented as

      F =  ma

Now for constant velocity to be attained, the acceleration of the car will be zero  

So  at constant velocity

       F =  m * 0

=>    F =   0 \  N

So  

     F_{net} + F__{dg}} = 0

=>   F__{dg} }=  -F_{net}

=>   F__{dg} }=  -3674 \  N

6 0
3 years ago
27 miles per gallon into kilometers per liter.
ikadub [295]

Answer:

11.479 kilometres per litre

Explanation:

for an approximate result, divide the fuel economy value by 2.352

5 0
3 years ago
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