Ask question

Ask question

All categories

3 years ago

12

How does tripling the distance between two objects affect the gravitational force between them?

2 answers:

Wewaii [24]3 years ago

6 0

It weakened the gravitational force between the oblects.

Daniel [21]3 years ago

6 0

This is from Newton's Laws of Gravitation. If the distance is increased by a factor of three the force is decreased by a factor of 9. (i.e. you would this equation in this situation: F = 1/d^2, "d" being distance).

There we go, lol. I hope this helps! ♥

You might be interested in

An infinite line charge of linear density λ = 0.30 µC/m lies along the z axis and a point charge q = 6.0 µC lies on the y axis a

vitfil [10]

Answer:

$E_{net} = (3.765\hat i - 1.207\hat j)kN/C$

Explanation:

Electric field due to long line charge on position of charge at x = 4 m is given as

$E = \frac{2k\lambda}{r} \hat i$

so we have

$\lambda = 0.30 \mu C/m$

now we have

$E = \frac{2(9\times 10^9)(0.30 \mu C/m)}{4}$

$E = 1350 N/C$

Now electric field due to the charge present at y = 2.0 m

$E = \frac{kq}{r^2} \hat r$

$E = \frac{(9\times 10^9)(6 \times 10^{-6})}{2^2 + 4^2}\times \frac{4\hat i - 2\hat j}{\sqrt{4^2 + 2^2}}$

$E = 603.7 ( 4\hat i - 2\hat j)$

$E = 2415\hat i - 1207.5 \hat j$

Now total electric field is given as

$E_{net} = (1350\hat i) + (2415\hat i -1207.5\hat j)$

$E_{net} = 3765\hat i - 1207.5 \hat j$

$E_{net} = (3.765\hat i - 1.207\hat j)kN/C$

6 0

2 years ago

I have a question to whoever can tell me (10 points)

Step2247 [10]

you may just select the word answer and then without cutting it u may start to type ur answer then it will be in bold letter

:) :) :) :) :) :)

4 0

2 years ago

an incandescent lightbulb has an efficiency of 2.1% and a power of 60 w. how much light energy does the lighbulb produce in 1 se

AveGali [126]

Explanation:

Power output of the bulb:

0.021 × 60 W = 1.26 W

Energy produced by the bulb in 1 second:

E = Pt

E = (1.26 W) (1 s)

E = 1.26 J

Round as needed.

5 0

3 years ago

A car is traveling at 96km/hr. what is the acceleration of a car traveling a distance of 100m and come to rest?

8090 [49]

Answer:

Explanation:

v² = u² + 2as

v = 0

u = 96 / 3.6 = 26.7 m/s

0² = 26.7² + 2a100

a = -3.5555555... ≈ -3.6 m/s²

the negative sign indicated the acceleration vector opposes the (assumed positive) initial velocity vector direction.

6 0

2 years ago

WHAT HAPPENS TO THE VELOCITY AND ACCELERATION OF A BODY FALLING FREELY IN A VACUUM?

mixas84 [53]

both remains constant

..........

7 0

3 years ago