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Genrish500 [490]
4 years ago
10

If the Sun suddenly turned off, we would not know it until its light stopped coming. How long would that be, given that the Sun

is 1.50x10'' m away?
Physics
1 answer:
Yuliya22 [10]4 years ago
3 0

Answer:

8.33 minutes

Explanation:

First let us find out how much time it takes for the light for the sun to reach the earth

Distance between Earth and Sun is 1.5×10¹¹ m

Speed of light = 3×10⁸ m/s

\text{Time taken}=\frac{1.5\times 10^{11}}{3\times 10^8}\\\Rightarrow \text{Time taken}=500\ seconds

Converting to minutes

500\ seconds=\frac{500}{60}=8.33\ minutes

So, it takes 8.33 minutes for the light from the sun to reach Earth.

This means that we would be receiving light for 8.33 minutes after the Sun turned off.

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An irrigation canal has a rectangular cross section. At one point whare the canal is 16.0 m wide, and the water is 3.8 m deep, t
Irina-Kira [14]

Answer:

The depth of the water at this point is 0.938 m.

Explanation:

Given that,

At one point

Wide= 16.0 m

Deep = 3.8 m

Water flow = 2.8 cm/s

At a second point downstream

Width of canal = 16.5 m

Water flow = 11.0 cm/s

We need to calculate the depth

Using Bernoulli theorem

A_{1}V_{1}=A_{2}V_{2}

Put the value into the formula

16.0\times3.8\times2.8=16.5\times x\times 11.0

x=\dfrac{16.0\times3.8\times2.8}{16.5\times11.0}

x=0.938\ m

Hence,  The depth of the water at this point is 0.938 m.

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4 years ago
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Nick parks his car at Target and walks 150 ft. north to get to the store. On his way back to the car, he walks 50 ft. before pau
mr Goodwill [35]
75 steps is what he’s missing
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3 years ago
Two billiard balls move toward each other on a table. The mass of the number three ball, m1, is 5 g with a velocity of 3 m/s. Th
marishachu [46]

Answer:

+5.7 m/s

Explanation:

According to the law of conservation of momentum is that the momentum before the collision is equal to the momentum after the collision. In an equation form it would look like this:

M₁V₁+M₂V₂ = M₁V₁'+M₂V₂'

Where:

M₁ = mass of object 1 (kg)

V₁ = velocity of object 1 before the collision (m/s)

V₁' = Final velocity of object 1 after the collision (m/s)

M₂ = mass of object 2 (kg)

V₂ = velocity of object 2 before the collision (m/s)

V₂' = Final velocity of object 2 after the collision (m/s)

According to your problem you have the following given:

M₁ = 5 g = 0.005kg

V₁ = 3 m/s

V₁' = -5m/s (It bounced off so it is going the other direction)

M₂ = 6g = 0.006kg

V₂ = -1 m/s (It is coming from the opposite direction of the 3-ball)

V₂' = ?

So we plug in what we know and solve for what we don't know.

M_1V_1+M_2V_2 = M_1V_1' + M_2V_2'\\\\(0.005kg)(3m/s)+(0.006kg)(-1m/s) = (0.005kg)(5m/s)+(0.006kg)(V_2')\\\\(0.015kg\cdot m/s)+(-0.006kg\cdot m/s)=(-0.025kg\cdot m/s)+(0.006kg)(V_2')\\\\0.009kg\cdot m/s+0.025kg\cdot m/s = (0.006kg)(V_2')\\\\\dfrac{0.034kg\cdot m/s}{0.006kg} = V_2'\\\\5.7m/s = V_2'

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3 years ago
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Answer would have to be A
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3 years ago
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