To solve this problem, we must know the gravitational force
of the planet. The equation would be,

This would calculate the force between two objects with
masses m1 and m2 and the gravitational constant, G, is 6.67 x 10^-11 m3 s-2
kg-1 and with r as the distance between the objects.
Thus,
F = (6.67 x 10^-11 m3 s-2 kg-1) * (5.68 x 10^26 kg) * (65
kg) * ((1/6.03 x 10^7 m)^2)
F = 678 kg/s^2 or 678 N
Answer is letter B.
V = 25,364.4 cm^3 Is volumer = 2.7g/cm^3 Is density
To calculate mass you use formula:m= V*rTo avoid remembering this formula you can see the type of unit on each given variable. We can see that we have g/cm^3 and cm^3. If we multiply them, we negate cm^3 and cm^3 and we are left with g which is unit for mass.
the answer is :
m = 68,486,6 g
Force = mass times acceleration
F = 21000 x 36.9 = 774900
Therefore, 774900N force is required.
Answer:
Equal to change in momentum of larger mass.
Explanation:
We are given that
Two difference masses .
Force act on both masses for the same length of time.
We have to find the change in momentum of the smaller mass.
Let M and m are two masses
M>m
We know that
Change in momentum for large mass=
Change in momentum for small mass=
Because Force and length of time are same for both masses .
Hence, the change in momentum of smaller mass is equal to change in momentum of larger mass.
Answer:
Explanation:
Let the velocity of projectile be v and angle of throw be θ.
The projectile takes 5 s to touch the ground during which period it falls vertically by 100 m
considering its vertical displacement
h = - ut +1/2 g t²
100 = - vsinθ x 5 + .5 x 9.8 x 5²
5vsinθ = 222.5
vsinθ = 44.5
It covers 160 horizontally in 5 s
vcosθ x 5 = 160
v cosθ = 32
squaring and adding
v²sin²θ +v² cos²θ = 44.4² + 32²
v² = 1971.36 + 1024
v = 54.73 m /s