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Salsk061 [2.6K]
3 years ago
7

What is the average speed in 4 seconds going 8 meters

Physics
2 answers:
gulaghasi [49]3 years ago
6 0

Answer:

Explanation:

speed is define as rate of change of distance or displacement

v=s/t

s=8 m

t=4s

v=8/4

v=2 m/s

Ksenya-84 [330]3 years ago
4 0
The formula for speed is distance over time or s = d/t
Distance = 8m or meters
Time: 4s or seconds
So s = 8/4
s = 2m/s or meters per second
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In the equation F = Kq1 q2/r2 solve for q2. Solve for r.
Tpy6a [65]

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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
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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}

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