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2 years ago

3

Physics

2 answers:

vodka [1.7K]2 years ago

6 0

Answer:

3 electron hai bro of puch mujhe sab aata h

vladimir2022 [97]2 years ago

5 0

Umm sorry but I think 3 so ya

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lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A

qaws [65]

$2.6×10^6\:\text{m}$

Explanation:

The acceleration due to gravity g is defined as

$g = G\dfrac{M}{R^2}$

and solving for R, we find that

$R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)$

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration $F_c$ experienced by the satellite is equal to the gravitational force $F_G$ or

$F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)$

The orbital velocity v is the velocity of the satellite around the planet defined as

$v = \dfrac{2\pi r}{T}$

where r is the radius of the satellite's orbit in meters and T is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

$\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}$

Solving for M, we get

$M = \dfrac{4\pi^2 r^3}{GT^2}$

Putting this expression back into Eqn(1), we get

$R = \sqrt{\dfrac{G}{g}\left(\dfrac{4\pi^2 r^3}{GT^2}\right)}$

$\:\:\:\:=\dfrac{2\pi}{T}\sqrt{\dfrac{r^3}{g}}$

$\:\:\:\:=\dfrac{2\pi}{(1.44×10^4\:\text{s})}\sqrt{\dfrac{(5×10^6\:\text{m})^3}{(3.45\:\text{m/s}^2)}}$

$\:\:\:\:= 2.6×10^6\:\text{m}$

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2 years ago

Matter can undergo chemical reactions and nuclear reactions, which

asambeis [7]

Answer:

Explanation:

they involve breaking and making chemical bonds

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1 year ago

What is the velocity of a plane to travel 3000 miles from New York to California in 5.0 hours

trapecia [35]

Answer:

10 miles per minute.

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2 years ago

A 65.0 kg football player is running at 10.0 m/s downfield when he catches a 0.420 kg football thrown in the same direction but

lukranit [14]

Explanation:

Below is an attachment containing the solution.

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3 years ago

An automobile tire is rated to last for 35,000 miles. to an order of magnitude, through how many revolutions will it turn?

Furkat [3]

The tire will rotate about 10 million times. An automobile tire is slightly less than 2 and half feet in diameter. It's circumference is that times pi with is a bit over 3. So 2.5 * 3 = 7.5 ft as an estimate for how far the tire rolls per revolution. A mile is a bit over 5000 feet, so call it 700 revolutions per mile. For the 35000 miles, call it 7 times 5000 miles. Now 7 times 7 is a bit under 50, so call 7 * 700 = 5000. And 5000 times 5000 = 25000000. The nearest order of magnitude is 10 million. So as an order of magnitude estimate, a automobile tire will rotate about 10 million times during it's life.

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2 years ago