All categories

3 years ago

What changes would result in a decrease in the gravitational force between two objects? Check all that apply.

1 answer:

4 0

<em>I'm sorry, it says check all that apply, however there are no choices given. You should edit, and add the multiple choice answers.</em>

My Answer:

Well if the masses of two objects were both decreased, it would result in a decrease in the gravitational force. So I guess the two objects masses would need to be decreased.

You might be interested in

What instrument is used to measure volume by displacement

kicyunya [14]

Volumes of liquids such as water can be readily measured in a graduated cylinder.

8 0

3 years ago

Question 5

bezimeni [28]

Answer:

ffcvghnvb vyhgyhvthbgvgybn ytvg dfvthgbhtfgybhvtgbyhnt vfyhn fgb fvb

Explanation:

4 0

2 years ago

What is the work required for a penguin to push a box 2 meters with a force of 8 newtons?

choli [55]

Work done is given by product of force and displacement due to that force

So here we will have

$Work = Force \times displacement$

here we know that

$Force = 8 N$

$displacement = 2 m$

Now work done is given as

$W = 8\times 2$

$W = 16 J$

so it will do 16 J work to move the box

3 0

3 years ago

A 90kg mountain climber hangs from a nylon rope and streches it ny 25.0cm. if the rope was originally 30.0m long and it's diamet

bixtya [17]

Answer:

Explanation:

E = σ/ε = (F/A) / (ΔL/L)

E = (mg/(πd²/4) / (ΔL/L)

E = (4mg/(πd²) / (ΔL/L)

E = 4Lmg/(πd²ΔL)

E = 4(30.0)(90)(9.8)/(π(0.01²)0.25)

E = 1.35 x 10⁹ Pa or 1.35 GPa

7 0

2 years ago

A binary star system consists of two equal mass stars that revolve in circular orbits about their center of mass. The period of

Vladimir79 [104]

Answer:

$m = 2.23 \times 10^{-32} kg$

Explanation:

Given data:

PERIOD OF MOTION IS T = 25.5 days

orbital speeds = 220 km/s

we know that

acceleration due to centripetal force is $a = \frac{F}{m} = \frac{V^2}{r}$

Gravitational force $F= \frac{Gm m}{d^2}$

we know that

$v = \frac{2\pi R}{T}$

solving for

$R = \frac{vT}{2\pi}$

$F = \frac{Gm^2}{4(\frac{vT}{2\pi})^2}$

$F = G\times \frac{\pi m}{(vT)^2}$

$a = \frac{v^2}{\frac{vT}{2\pi}}$

$a = \frac{2\pi v}{T}$

we know that

f =ma

$G\times \frac{\pi m}{(vT)^2} = a = \frac{2\pi m v}{T}$

solving for m

$m = \frac{2Tv^3}{\pi G}$

$m = \frac{2\times 25.5 \times 86400 \times 220000^3\ m/s}{\pi \times 6.67\times 10^{-11}}$

$m = 2.23 \times 10^{-32} kg$

5 0

2 years ago