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2 years ago

A 5kg rock and a 10kg rock are dropped from a height of 10m?

1 answer:

5 0

Look out below ! You should step nimbly to one side, to avoid being hit by one or the other of those hazardous weight objects when they arrive (at the same time).

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Molecules move from areas of low concentration to high concentration through a process of

weeeeeb [17]

Active transport move molecules from low concentrations to high concentrations.

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2 years ago

<img src="https://tex.z-dn.net/?f=what%20%5C%3A%20is%20%5C%3A%20refraction%20%5C%3A%20of%20%5C%3A%20light%20%5C%3A%20%20%7B%3F%7

topjm [15]

When light passes through one transparent medium to another transperent medium it bends and that beding of light is know as refraction of light !

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2 years ago

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How do you calculate elastic potential energy

Nimfa-mama [501]

U=1/2kx2

This image sums it up

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2 years ago

Calculate the kinetic energy in joules of a 1200 kg automobile moving at 18 m/s .

vodka [1.7K]

Answer:

194,400 joules of kinetic energy.

Explanation:

Remember that to calculate the Kinetic energy you need to use the next formula:

$Ke=\frac{1}{2}Mass*Velocity^2$

We know that Mass= 1200 kg and velocity is 18m/s, so we insert those values into the formula:

$Ke=\frac{1}{2}Mass*Velocity^2\\Ke=\frac{1}{2}1200kg*(18m/s)^2\\Ke=194,400 joules$

So the kinetic energy of a car moving at 18m/s with a mass of 1200 kg would be 194,400 joules.

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3 years ago

Read 2 more answers

12)A black body is heated from 27°C to 127° C. The ratio of their energies of radiations emitted will be

Nat2105 [25]

Answer:

$81:256$ .

Explanation:

Let $T$ denote the absolute temperature of this object.

Calculate the value of $T$ before and after heating:

$T(\text{before}) = 27 + 273 = 300\; \rm K$ .

$T(\text{after}) = 127 + 273 = 400\; \rm K$ .

By the Stefan-Boltzmann Law, the energy that this object emits (over all frequencies) would be proportional to $T^4$ .

Ratio between the absolute temperature of this object before and after heating:

$\displaystyle \frac{T(\text{before})}{T(\text{after})} = \frac{3}{4}$ .

Therefore, by the Stefan-Boltzmann Law, the ratio between the energy that this object emits before and after heating would be:

$\displaystyle \left(\frac{T(\text{before})}{T(\text{after})}\right)^{4} = \left(\frac{3}{4}\right)^{4} = \frac{81}{256}$ .

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2 years ago