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3 years ago

When the moon slowly moves away from the earth.How is it affecting the gravitational force between the earth and the moon?

1 answer:

3 0

Well, think about how the tides will be affected when the moon moves farther away. If the moon first started off very close the earth, we would have more tsunamis. (Scientists have found that the moon has possibly been closer to earth long ago.) While it moves away, soon there will no longer be many tides.

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Weight is proportional to but not equal to mass. In which of the following situations would a person show an increase in weight

meriva

Answer: c living in a camber in an under water habitat

Explanation:

4 0

2 years ago

Two small spheres with mass 10 gm and charge q, are suspended from a point by threads of length L=0.22m. What is the charge on e

Nataliya [291]

Answer:

$q=1.95*10^{-7}C$

Explanation:

According to the free-body diagram of the system, we have:

$\sum F_y: Tcos(15^\circ)-mg=0(1)\\\sum F_x: Tsin(15^\circ)-F_e=0(2)$

So, we can solve for T from (1):

$T=\frac{mg}{cos(15^\circ)}(3)$

Replacing (3) in (2):

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=F_e$

The electric force ( $F_e$ ) is given by the Coulomb's law. Recall that the charge q is the same in both spheres:

$(\frac{mg}{cos(15^\circ)})sin(15^\circ)=\frac{kq^2}{r^2}(4)$

According to pythagoras theorem, the distance of separation (r) of the spheres are given by:

$sin(15^\circ)=\frac{\frac{r}{2}}{L}\\r=2Lsin(15^\circ)(5)$

Finally, we replace (5) in (4) and solving for q:

$mgtan(15^\circ)=\frac{kq^2}{(2Lsin(15^\circ))^2}\\q=\sqrt{\frac{mgtan(15^\circ)(2Lsin(15^\circ))^2}{k}}\\q=\sqrt{\frac{10*10^{-3}kg(9.8\frac{m}{s^2})tan(15^\circ)(2(0.22m)sin(15^\circ))^2}{8.98*10^{9}\frac{N\cdot m^2}{C^2}}}\\q=1.95*10^{-7}C$

5 0

3 years ago

Will a substance that contains an o-h group dissolve in water?

Stella [2.4K]

Usually, it increases the solubility in water.

4 0

3 years ago

If the pitch of the sound coming out of a speaker increases, which statement is true about the sound wave?

S_A_V [24]

Answer:

frequency and amplitude increases

3 0

2 years ago

Question 2 (1 point)

Tju [1.3M]

Answer:

I know someone anwsered but it would be 400M

Explanation:

i initial velocity (u)=10m/s

acceleration (a)=0

time taken (t) =40s

then distance (s)=u t +1/2 a t^2

s= u t +0 (as a is 0)

s= 10 x 40

s= 400M

7 0

3 years ago