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3 years ago

13

Will it take longer for a train or a car traveling at 100 mi/hr to stop

2 answers:

Pani-rosa [81]3 years ago

6 0

It typically take longer for a heavier object to slow down therefor, a train will take more time. <span />

Allushta [10]3 years ago

3 0

<h3><u>Answer;</u></h3>

A train will take longer time to stop compared to a car.

<h3><u>Explanation;</u></h3>

For a body in motion to stop a braking force at a given distance is required for the body to loose kinetic energy.
The the stopping distance of a car depends on the breaking force and the the mass of the moving body. The stopping distance will depend on the kinetic energy and the breaking force acting to stop the motion of a body.
<em><u>The mass of a moving body determines its kinetic energy, and also determines the resultant force according to the newton seconds law of motion . Therefore, mass determines the breaking force and the stopping distance, such that a big body in motion will require more force to stop and will take a longer stopping distance and time.</u></em>

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Two identical metal balls a and b are mounted on insulating rods. Ball a has a charge of +q/2 and ball b is initially uncharged.

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3 years ago

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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
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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)

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σ is the Boltzmann constant
A is the area
T is the temperature
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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!

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viva [34]

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