First the plane turns 100 km North, and than 200 km East. Since both the directions are perpendicular to each other, therefore we can apply the Pythagoras theorem to calculate the distance between the destination and the point where plane took off
=100^{2}+200^{2}
D=223.60 km=224 km
Therefore, The destination is 224 km from where the plane took off
First do 1.6 m (how far he jumps) 9.8 m/s (what gravity is measured at) then times 2
= 31.36
Sq root = 5.6
Answer:
vi = 4.77 ft/s
Explanation:
Given:
- The radius of the surface R = 1.45 ft
- The Angle at which the the sphere leaves
- Initial velocity vi
- Final velocity vf
Find:
Determine the sphere's initial speed.
Solution:
- Newton's second law of motion in centripetal direction is given as:
m*g*cos(θ) - N = m*v^2 / R
Where, m: mass of sphere
g: Gravitational Acceleration
θ: Angle with the vertical
N: Normal contact force.
- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:
m*g*cos(θ) - 0 = m*vf^2 / R
g*cos(θ) = vf^2 / R
vf^2 = R*g*cos(θ)
vf^2 = 1.45*32.2*cos(34)
vf^2 = 38.708 ft/s
- Using conservation of energy for initial release point and point where sphere leaves cylinder:
ΔK.E = ΔP.E
0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))
( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))
vi^2 = vf^2 - 2*g*R*( 1 - cos(θ))
vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))
vi^2 = 22.744
vi = 4.77 ft/s
let the mass of Venus is M then mass of Saturn is 100 M
similarly if the radius of Venus is R then the radius of Saturn is 10 R
now the force of gravity on a man of mass "m" at the surface of Venus is given by
now similarly the gravitational force on the man if he is at the surface of Saturn
so here if we divide the two forces
so here we can say
F1 = F2
so on both planets the gravitational force will be same
