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Paladinen [302]
3 years ago
5

When does work occur

Physics
2 answers:
iren [92.7K]3 years ago
7 0
Work occurs when an applied force results in movement of an object in the same direction as the applied force.
Anna11 [10]3 years ago
7 0
Work occurs when the object has moved. No matter how much force is applied, if you do not move the object, no work is done. A force with no motion or force perpendicular to the motion does no work.
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Which of Newton's motion laws BEST explains WHY a rock falls when it is dropped from a bridge?
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It’s A because it stays in motion whenever you drop it
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How much work is done in holding a 10 newton sack of potatoes while waiting in line at the grocery store for 3 minutes.​
hodyreva [135]

Answer:

Zero

Explanation:

W = F × s

F = 10 N,

t = 3min = 180sec

s = 0( no change in postion)

W = 10 ×0

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A boy weighs 50 kg and is running with 225 J of energy, what is his velocity?
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0.22

Explanation:

50 ÷ 225 = 0.22

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What do you call the energy from earthquakes that propagate through earth as vibrations?<br>​
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3 years ago
The intensity of light from a star (its brightness) is the power it outputs divided by the surface area over which it’s spread:
kow [346]

Answer:

\frac{d_{1}}{d_{2}}=0.36

Explanation:

1. We can find the temperature of each star using the Wien's Law. This law is given by:

\lambda_{max}=\frac{b}{T}=\frac{2.9x10^{-3}[mK]}{T[K]} (1)

So, the temperature of the first and the second star will be:

T_{1}=3866.7 K

T_{2}=6444.4 K

Now the relation between the absolute luminosity and apparent brightness  is given:

L=l\cdot 4\pi r^{2} (2)

Where:

  • L is the absolute luminosity
  • l is the apparent brightness
  • r is the distance from us in light years

Now, we know that two stars have the same apparent brightness, in other words l₁ = l₂

If we use the equation (2) we have:

\frac{L_{1}}{4\pi r_{1}^2}=\frac{L_{2}}{4\pi r_{2}^2}

So the relative distance between both stars will be:

\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{L_{1}}{L_{2}} (3)

The Boltzmann Law says, L=A\sigma T^{4} (4)

  • σ is the Boltzmann constant
  • A is the area
  • T is the temperature
  • L is the absolute luminosity

Let's put (4) in (3) for each star.

\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{A_{1}\sigma T_{1}^{4}}{A_{2}\sigma T_{2}^{4}}

As we know both stars have the same size we can canceled out the areas.

\left(\frac{d_{1}}{d_{2}}\right)^{2}=\frac{T_{1}^{4}}{T_{2}^{4}}

\frac{d_{1}}{d_{2}}=\sqrt{\frac{T_{1}^{4}}{T_{2}^{4}}}

\frac{d_{1}}{d_{2}}=\sqrt{\frac{T_{1}^{4}}{T_{2}^{4}}}

\frac{d_{1}}{d_{2}}=0.36

I hope it helps!

5 0
3 years ago
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