Answer:
gravity is vital because as we know space curves around objects like planets, in this curvature is our atmosphere and a gravitational pull holding everything in, without this curvature we would not have any gravitational pull.
Hope it helps!!!
Explanation:
Objects appear to us to have the color that
they reflect (paint) or transmit (jello).
<span>G = gravitational constant
M = mass of the earth
R = radius of orbit of a satellite
r = radius of orbit of a second satellite
v = speed of the satellite
P = period of a satellite
p = period of a second satellite
Equate gravitational acceleration with centripetal acceleration
g = G*M/R^2 = v^2/R
Express the orbital speed in terms of the orbit circumference and period
v = 2*pi*R/P
And insert the expression for v into the first equation
G*M/R = 4*PI^2*R^2/P^2
G*M/R^3 = 4*pi^2/P^2
R^3/P^2 = 4*pi^2/(G*M) = constant = C
We can do the above since G and M are constants for all earth orbits
So we can write a second equation of the same form for another satellite and equate to get:
R^3/P^2 = r^3/p^2
r^3 = R^3*p^2/P^2
r = R*(p^2/P^2)^(1/3)
For the second satellite we have p = 8*P
r = R*(8^2)^(1/3) = R*(64)^(1/3) = 4*R</span>
<span>A
quarterback ... runs
backward for 15 meters,
=> displacement vector = -15 j
he ... runs sideways parallel to the line of
scrimmage for 12 meters.
=> displacement vector = 12 i
he throws the ball 50 meters
perpendicular to the line of scrimmage
=> displacement vector = 50j
where it is caught by the
receiver.
=> net displacement = -15j + 12i + 50j = 12i + 35j
How far is the football from its original position?
distance ^2 = (12m)^2 + (35m)^2 = 1,369 m^2
=> distance = √ (1,369m^2) = 37 m^2
Answer: 37 m^2
</span>
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