In another solar system is planet driff, which
has 5 times the mass of the earth and also 5
times the radius.
how does the gravitational acceleration on
the surface of driff compare to the gravita-
tional acceleration on the surface of the earth?
1.it’s 15th as much.
2.there is no gravity on driff because 5 times 4000 miles (the radius of the earth), is 2 miles, far beyond the pull of gravity.
3.it’s 125th as much.
4.it’s 25 times as great.
5.it’s 5 times as much.
6.it’s the same, 10 m/s2
.

It is (1/5)th as much.

Explanation:

If we apply the equation

F = G*m*M / r²

where

m = mass of a man

M₀ = mass of the planet Driff

M = mass of the Earth

r₀ = radius of the planet Driff

r = radius of the Earth

G = The gravitational constant

F = The gravitational force on the Earth

F₀ = The gravitational force on the planet Driff

g = the gravitational acceleration on the surface of the earth

g₀ = the gravitational acceleration on the surface of the planet Driff

we have

F₀ = G*m*M₀ / r₀² = G*m*(5*M) / (5*r)²    

⇒  F₀ = G*m*M / (5*r²) = (1/5)*F

If

F₀ = (1/5)*F

then

W₀ = (1/5)*W   ⇒  m*g₀ = (1/5)*m*g   ⇒   g₀ = (1/5)*g

It is (1/5)th as much.

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards

helium

oxygen

neon

sodium