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3 years ago

Please help!

Physics

1 answer:

o-na [289]3 years ago

4 0

The formula PE(Potential Energy)= mgh

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3 properties that metals have in common

Gemiola [76]

Answer:

Luster: Metals are shiny when cut, scratched, or polished.

Malleability: Metals are strong but malleable, which means that they can be easily bent or shaped. ...

Conductivity: Metals are excellent conductors of electricity and heat.

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4 years ago

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Quasar spectra often show many absorption lines that all appear to be as a result of the same electron transition (such as level

Elanso [62]

Answer: Because of different redshift of cloud.

Explanation:

We are seeing absorption lines from clouds of gas that lie between us and the quasar, and therefore each cloud has a different redshift.

A quasar's spectrum is hugely redshifted. And most astronomers think this large redshift tells us about the distance to the quasar.

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3 years ago

Amoving object is in equilibrium. Which best describes the motion of the object if no forces change?

kaheart [24]

I believe the answer is C. It will maintain its state of motion

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4 years ago

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What is momentum <br>what is momentum in words

Artist 52 [7]

Answer:

It's a strength or force

Explanation:

the quantity of motion of a moving body, measured as a product of its mass and velocity

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3 years ago

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Consider a particle moving along the x-axis where x(t) is the position of the particle at time t, x'(t) is its velocity, and x''

Lady_Fox [76]

Answer:

We know that the acceleration of the particle is defined as

$a(t)=\frac{dv}{dt}$

Since it is given that

$v(t)=\frac{5}{\sqrt{t}}\\\\\therefore a(t)=\frac{d(\frac{5}{\sqrt{t}})}{dt}\\\\a(t)=5\times \frac{dt^{-\frac{1}{2}}}{dt}\\\\=\frac{-5}{2}t^{\frac{-1}{2}-1}\\\\\therefore a(t)=x''(t)=\frac{-2.5}{t^{\frac{3}{2}}}$

Now by definition of velocity we have

$v(t)=\frac{dx(t)}{dt}\\\\\Rightarrow dx(t)=v(t)dt$

Integrating on both sides we get

$v(t)=\frac{dx(t)}{dt}\\\\\int dx(t)=\int v(t)dt\\\\x(t)=\int v(t)dt$

Applying values we get

$v(t)=\frac{dx(t)}{dt}\\\\\int dx(t)=\int v(t)dt\\\\x(t)=\int \frac{5}{t^{\frac{1}{2}}}dt\\\\\therefore x(t)=\frac{5}{0.5}\int t^{-0.5}dt\\\\x(t)=10\sqrt{t}+c$

To find the constant we note that at t=1 the particle is at x=12 Thus applying values in the above equation we get

$c=12-10\\\therefore c=2\\\\x(t)=10\sqrt{t}+2$

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3 years ago