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2 years ago

What happens to the gravitational force between two masses when the distance between the masses is doubled?

Physics

1 answer:

vovangra [49]2 years ago

6 0

Answer:

It decreases by a factor of 4

Explanation:

The gravitational force between two masses is given by

$F=\frac{Gm_1 m_2}{r^2}$

where

G is the gravitational constant

m1, m2 are the two masses

r is the separation between the two masses

In this problem, the distance between the masses is doubled, so

r' = 2r

Substituting into the equation, we see that the new force is

$F'=\frac{Gm_1 m_2}{(2r)^2}=\frac{1}{4}\frac{Gm_1m_2}{r^2})\frac{F}{4}$

So, the force decreases by a factor of 4.

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Two girls stand 500m away from a tall wall

Mama L [17]

From the calculations, the speed of sound in this case is 16.9 m/s.

The term echo has to do with the reflection of sound waves. Sound is a mechanical wave.

we know that the speed of sound is obtained from;

V = 2x/t

x = distance covered

t = time taken

V = 2(500)/59

v = 16.9 m/s

The error in the experiment could come from;

Lack of precise time measurement
Error can also arise from the environment of the experiment

Learn more about echo:brainly.com/question/9527413

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4 0

1 year ago

car travels 60 km/h will skid 30 m when its brakes are locked. if the same car is traveling 180 km/h, what will be the skidding

Yanka [14]

Answer:

270 m

Explanation:

When the driver hits the brakes, the kinetic energy of the car is converted by friction into heat. The kinetic energy K is given by:

(1) $K=\frac{1}{2} mv^{2}$

The work W done by friction is:

(2) $W = F_fd$

where the force of friction is:

$F_f = \mu mg$

d: distance sliding

μ: friction coefficient

m: mass

g: gravitational constant

setting equation 1 and 2 equal:

(3) $\frac{1}{2}mv^{2}=\mu mgd$

simplifying:

(4) $d = \frac{v^{2} }{2\mu g}$

Use equation 4 to find the ratio between the two cases gives:

(5) $\frac{d_1}{d_2} = \frac{v_1^{2} }{v_2^{2} }$

plugging in:

$\frac{30}{d_2}=\frac{60^{2} }{180^{2}}$

$d_2=270$

8 0

3 years ago

A 980-kg sports car travelling with a speed of 21 m/s collides into the rear end of a 2300-kg SUV that is stationary at a red li

stiv31 [10]

Answer:

Explanation:

After the collision , both the car will have common velocity , which according to conservation of momentum will be as follows

v = 980 x 21 / (980 + 2300)

= 20580 / 3280

= 6.274 m /s

The kinetic energy of both the cars after collision

= 1/2 x (980+2300) x 6.274²

= 64555.44 J .

frictional force = μ mg where μ is coefficient of friction , mg is weight of both the cars

= .8 x (980+2300) x 9.8

= 25715.2 N

work done by friction will be equal to kinetic energy of car

25715.2 x d = 64555.44 ; where d is displacement of both the cars

d = 2.5 m

4 0

3 years ago

A car is traveling at 20.0 m/s on tires with a diameter of 70.0 cm. The car slows down to a rest after traveling 300.0 m. If the

cupoosta [38]

Answer: deceleration of $1.904\ rad/s^2$

Explanation:

Given

Car is traveling at a speed of u=20 m/s

The diameter of the car is d=70 cm

It slows down to rest in 300 m

If the car rolls without slipping, then it must be experiencing pure rolling i.e. $a=\alpha \cdot r$

Using the equation of motion

$v^2-u^2=2as\\$

Insert $v=0,u=20,s=300$

$0-(20)^2=2\times a\times 300\\\\a=\dfrac{-400}{600}\\\\a=-\dfrac{2}{3}\ m/s^2$

Write acceleration as $a=\alpha \cdot r$

$-\dfrac{2}{3}=\alpha \times 0.35\\\\\alpha =-\dfrac{2}{1.05}\\\\\alpha =-1.904\ rad/s^2$

So, the car must be experiencing the deceleration of $1.904\ rad/s^2$ .

4 0

3 years ago

A candle burns 1.2 m away from a lens with a focal length of 0.4 m. how far away from the lens will an image form?

rusak2 [61]

Answer:

0.6 m

Explanation:

We can solve the problem by using the lens equation:

$\frac{1}{f}=\frac{1}{p}+\frac{1}{q}$

where

f is the focal length

p is the distance between the lens and the candle

q is the distance between the image and the lens

In this problem, we have

p = 1.2 m is the distance of the candle from the lens

f = 0.4 m is the focal length

Solving the equation for q, we find

$\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{0.4 m}-\frac{1}{1.2 m}=1.67 m^{-1}\\q=\frac{1}{1.67 m^{-1}}=0.6 m$

6 0

2 years ago