If the distance between two objects is decreased to 1/10 of the original

2 answers:

mihalych1998 [28]3 years ago

5 0

F = GMm/r²

The force follows inverse square law.

If the distance is reduced to (1/10) of the original value,

F = GMm/(1/10r)²

F = GMm/(1/100)r²

F = 100* GMm/r²

Hence the force is 100 times the original value.

GaryK [48]3 years ago

4 0

The force of attraction between two objects is proportional to the mass of the objects and inversely proportional to the square of teh distance between them. In this casem if the distance is decreased to 1/10 of the original, then teh force of attraction increases by 100 times (10^2).

You might be interested in

A long solenoid consists of 1700 turns and has a length of 0.75 m.The current in the wire is 0.48 A. What is the magnitude of th

Travka [436]

Answer:

1.37 ×10^-3 T

Explanation:

From;

B= μnI

μ = 4π x 10-7 N/A2

n= number of turns /length of wire = 1700/0.75 = 2266.67

I= 0.48 A

Hence;

B= 4π x 10^-7 × 2266.67 ×0.48

B= 1.37 ×10^-3 T

4 0

4 years ago

How many neutrons and electrons are in ruthenium??? PLEASE HELP IM DROWNING IN WORK!!!!!

Ierofanga [76]

57 nuetrons and 44 elctrons just incase you need it there are 44 protons too.Hope this halps

5 0

3 years ago

Read 2 more answers

A car is moving at 19 m/s along a curve on a horizontal plane with radius of curvature 49m.

JulsSmile [24]

Answer:

$\mu =0.75$

Explanation:

Frictional Force

When the car is moving along the curve, it receives a force that tries to take it from the road. It's called centripetal force and the formula to compute it is:

$F_c=m.a_c$

The centripetal acceleration a_c is computed as

$\displaystyle a_c=\frac{v^2}{r}$

Where v is the tangent speed of the car and r is the radius of curvature. Replacing the formula into the first one

$F_c=m.\frac{v^2}{r}$

For the car to keep on the track, the friction must have the exact same value of the centripetal force and balance the forces. The friction force is computed as

$F_r=\mu N$