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Alenkinab [10]
3 years ago
10

Two asteroids exert a gravitational force, F, on each other. Some time later, the asteroids are now three times as far from each

other as before. Which of the following represents the gravitational force at this distance? Select one:
a. F/9
b. F/6
c. F/3
d. F/2
Physics
1 answer:
dolphi86 [110]3 years ago
7 0

Answer:

a.F/9

Explanation:

Hello, I can help you wit this

the gravitational Force is described by the next equation

F=G*\frac{m_{1}*m_{2} }{r^{2} } \\

Where, G  is   the universal gravitational constant, M1 and M2 are the masses of the objects, and r is the distance between then.

Step 1

time 1

Let

m1= mass of the asteroid 1

m2= mass of the asteroid 2

distance 1=r1

put the values into the equation to find the force

F_{1}=G*\frac{m_{1}*m_{2} }{r^{2} }\\F_{1}=G*\frac{m_{1}*m_{2} }{r_{1}^{2} }

Step 2

time 2

r2=3r1

F_{2} =G*\frac{m_{1}*m_{2} }{(3*r_{1} )^{2} }\\

Step 3

compare the forces

F_{1}=G*\frac{m_{1}*m_{2} }{r^{2} }\\F_{2} =G*\frac{m_{1}*m_{2} }{9(r_{1}^{2}) }\\F_{2}=\frac{F_{1}}{9}

a.F/9

here it is observed that the force is inversely proportional to the square of the distance, at double distance, it is divided into 4, triple in 9, etc.

Have a great day.

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Am arrow of mass 1000kg is shot into a wooden block of mass 5000lg lying at rest in a smooth surface.If the arrow travels 15m/s
Ber [7]

Answer:

Vf=3

Explanation:

you must first write your data

data before impact

M1=1000 M2=5000

V1=0 m/s V2 =0m/s

data after impact

M1=1000 M2=5000

V1=15m/s V2=?

M1V1 +M2V2=M1V1 +M2V2f

(1000)(0)+(5000)(0)=(1000)(15)+(5000)Vf

0=15000+5000Vf

- 15000÷5000=5000Vf÷5000

Vf= -3

Vf =3

6 0
2 years ago
How loudness of a sound wave depends on the intensity of the sound wave?
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Hope this helps!
3 0
3 years ago
he membrane that surrounds a certain type of living cell has a surface area of 6.0 x 10-9 m2 and a thickness of 1.6 x 10-8 m. As
k0ka [10]

Answer:

1.54481175\times 10^{-12}\ C

Explanation:

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

A = Area = 6\times 10^{-9}\ m^2

d = Thickness = 1.6\times 10^{-8}\ m

k = Dielectric constant = 5.4

V = Voltage = 86.2 mV

Charge is given by

Q=CV\\\Rightarrow Q=k\epsilon\dfrac{A}{d}V\\\Rightarrow Q=5.4\times 8.85\times 10^{-12}\times \dfrac{6\times 10^{-9}}{1.6\times 10^{-8}}\times 86.2\times 10^{-3}\\\Rightarrow Q=1.54481175\times 10^{-12}\ C

The charge on the outer surface is 1.54481175\times 10^{-12}\ C

4 0
2 years ago
A block pushed along the floor with velocity V0 slides a distance d after the pushing force is removed. a) if the mass of the bl
KengaRu [80]

Answer:

a) \ d_2=d_1\\b) \ d_2=4d_1

Explanation:

Assume that the distance travelled initially is d.

In order to stop the block you need some external force which is friction.

If we use the law of energy conservation:

E_i=E_f\\\frac{mv^2}{2}= E_{Friction}\\E_{Friction}=F_{Friction}*d\\F_{Friction}= \mu_kmg\\\frac{mv^2}{2}= \mu_kmgd\\ d=\frac{v^2}{2\mu_kg}

a)

Looking at the formula you can see that the mass doesn't affect the distance travelled, as lng as the initial velocity is constant (Which indicates that the force must be higher to push the block to the same speed) therefore the distance is the same.

b) If the velocity is doubled, then the distance travelled is multiplied by 4, because the distance deppends on the square of the velocity.

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