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zhenek [66]
2 years ago
15

The formula used to find force is F=m*v. true or false

Physics
1 answer:
zepelin [54]2 years ago
4 0

It's true IF ' m ' stands for mass and ' v ' stands for acceleration. Otherwise it's false.

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Which statement best describes perigee?
pantera1 [17]

Answer:

A. The closest point in the Moon's orbit to Earth

Explanation:

The perigee is defined as the closest point in the orbit of an object (such as a satellite) from the centre of the Earth. In this case, the Earth's satellite is the Moon, so the perigee is defined as the closest point in the Moon's orbit to Earth. so option A is the correct one.

Let's see instead the names of the other options:

B. The farthest point in the Moon's orbit to Earth  --> this point is called apogee

C. The closest point in Earth's orbit of the Sun  --> this point is called perihelion

D. The Sun's orbit that is closest to the Moon --> this point has no specific name

8 0
3 years ago
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A roller of radius 12.5 cm turns at 14 revolutions per second. What is the linear velocity of the roller in meters per second?
Firdavs [7]

12.5 times 14 and convert to meters its 1.75 meters per second

7 0
3 years ago
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What is the mechanical advantage of an inclined plane that has a length of 9 feet and a height of 3 feet?
Rainbow [258]

Answer:

it would be 3

Explanation:

because you have to divide the length by the height of the incline.

5 0
3 years ago
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What is the half life-life of your 100 atoms of Carbon-14?
Naddika [18.5K]

Answer:

Answer: It takes 5,730 years for half the carbon-14 to change to nitrogen; this is the half-life of carbon-14. After another 5,730 years only one-quarter of the original carbon-14 will remain

3 0
2 years ago
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 <em>v</em> is the velocity of the satellite around the planet defined as

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

where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> 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 <em>M</em>, 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}

5 0
2 years ago
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