You multiply the mass by the acceleration 82*7.5=615; that's what I would do
Answer:
Time according to earth clock (T0) = 0.22 years (Approx)
Explanation:
Given:
Time taken by light = 4.5 years
Time taken by ship = 5 years
Speed of light = c
Speed of ship (v) = 90% of c = 0.9c
Find:
Time according to earth clock (T0) = ?
Computation:
Time dilation is ,
![T(Difference) = \frac{T0}{\sqrt{1-\frac{v^2}{c^2} } }\\\\ (5-4.5)= \frac{T0}{\sqrt{1-\frac{(0.9c)^2}{c^2} } }\\\\ 0.5=\frac{T0}{\sqrt{1-0.81} }\\\\T0 =0.2179](https://tex.z-dn.net/?f=T%28Difference%29%20%3D%20%5Cfrac%7BT0%7D%7B%5Csqrt%7B1-%5Cfrac%7Bv%5E2%7D%7Bc%5E2%7D%20%7D%20%7D%5C%5C%5C%5C%20%285-4.5%29%3D%20%5Cfrac%7BT0%7D%7B%5Csqrt%7B1-%5Cfrac%7B%280.9c%29%5E2%7D%7Bc%5E2%7D%20%7D%20%7D%5C%5C%5C%5C%200.5%3D%5Cfrac%7BT0%7D%7B%5Csqrt%7B1-0.81%7D%20%7D%5C%5C%5C%5CT0%20%3D0.2179)
Time according to earth clock (T0) = 0.22 years (Approx)
-- In order to achieve constant verlocity, the net force on the mass must be zero. So if there ARE any forces acting on it, they must be balanced.
-- There is already a force on the mass that can't be eliminated . . . the force of gravity.
-- That force due to gravity is (mass x gravity) = (25 kg)(9.8 m/s²) = <em><u>245N</u></em> in the <u><em>downward</em></u> direction.
-- In order to 'balance' the forces and make them add up to zero, we have to provide another force of <em>245N</em>, all in the <em>upward</em> direction.
-- Then the forces on the object will be balanced, the NET force on it will be zero, and whichever way you start it moving, it will continue to move at a cornstant verlocity.
This is true due to the reaction that happens from water evaporating and leaving the sugar crystals behind to form.
The sad ball does not rebound after it strikes the block. This means that the collision is inelastic. If two sad balls collide with each other, we can assume completely inelastic collision. Since momentum is conserved, the kinetic energy during the collision would be twice that of each of the ball's, half of the kinetic energy of each ball will be dissipated.