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.

the minimum constant speed at which she can catch the train is 2.7 m/s

given:

acceleration = 0.6

time = 4.5 s

to find:

velocity = ?

formula used:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

solution:

from 1st kinematic equation we get,

v = u + at

u = initial speed of train = 0 m/s

v = final speed of the train

a = acceleration of the train

t = time

v = 0 + (0.6 × 4.5)

v = 2.7 m/s

the minimum constant speed at which she can catch the train is 2.7 m/s

because if the tool that you are using is not balanced it there would be false mass.