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2 years ago

What is gravity’s role in tides?

2 answers:

8 0

The moon and sun and their distances from each other do play a critical role in affecting the oceans tides. Gravity is one of the main forces that creates the tides.Tidal generating forces are based on the gravitational attractive force.

Hope this helps! :)

kotegsom [21]2 years ago

6 0

The relationship between the masses of the Earth, moon and sun and their distances to each other play critical roles in affecting tides

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Um comentarista de futebol certa vez comentou:"A bola bateu na trave e voltou duas vezes mais forte". Sabendo que quando a bola

ryzh [129]

Answer:

Por ela ter batido na trave, não tem como voltar 2x mais forte, por que toda ação correspondente a uma reação de igual intensidade, mas que atua no sentido oposto

Explanation:

7 0

2 years ago

Little Timmy wants to measure how tall his house is. He doesn’t have a tape measure but does have a stopwatch. He recruits Big B

Komok [63]

The distance covered by an object accelerating from rest is

D = (1/2) · (acceleration) · (time)² .

In this particular case, 'acceleration' is 9.8 m/s² ... due to gravity.

D = (1/2) · (9.8 m/s²) · (1.67 s)²

D = (4.9 m/s²) · (2.789 s²)

D = 13.67 meters

6 0

3 years ago

You and a friend each carry a 15 kg suitcase up two flights of stairs, walking at a constant speed. Take each suitcase to be the

AlekseyPX

Answer:

Both of you did the same work but you expended more power.

Explanation:

<em>Work done</em> by an object is calculated by force applied multiplied by the distance.

W=F*d

From the figure given below let us calculate force applied bith you and yopur friend.

Let us take the stairs in positive x direction,

Work done by you W₁ ,

The force applied Fₓ = F - mgsinθ =maₓ

here aₓ = 0, because both of you move with constant speed

F - mgsinθ = 0

F= mgsinθ

The work done by you on the suitcase is

W = F L cos0° , where L is he length of the staircase.

W = FL = mgsinθL , by substituting value of F

Work done by you is W₁ = mgLsinθ

Similarly work done by your friend is W₂ = mgLsinθ.

Because both of you carry suitcase of same weight and in staircase is in same angle the force applied is same .

Therefore <em>work done by both of you is same</em> . Both of you did equal work.

The power , is defined as amount of energy converted or transfered per second or rate at which work is done .

P = $\frac{W}{t}$ = $\frac{FL}{t}$

Power spend by you P₁ = mgLsinθ/t

P₁ = 15*9.8*Lsinθ/30

P₁ = 4.9L sinθ eqn 1

Power spend by your friend is P₂ = mgLsinθ/t

P₂ =15*9.8*Lsinθ/60

P₂ = 2.45Lsinθ eqn 2

Dividing eqn 1 and eqn 2

P₁ = 2P₂

You have spend more power than your friend .

Hence Both of you did equal work but you spend more power.

7 0

3 years ago

Imagine you are riding on a yacht in the ocean and traveling at 20 mph. You then hit a golf ball at 100 mph from the deck of the

Whitepunk [10]

Before solving this question, first we have to understand the special theory of relative.

As per classical mechanics, the velocity of light will be different in different frame of reference. The light moves in the ether medium which exists every where in the entire universe.

Let us consider a body which moves with a velocity v. Let light is coming along the direction of the body. As per classical mechanics,the velocity of light with respect to the body will be [ c-v].

Let us consider that light is coming from opposite direction. Hence, the velocity of light with respect to the observer will be c+v.

From above we see that velocity of light is different in both the cases which is wrong.

As per Einstein's special theory of relativity, the velocity of light will be same in every frame of reference i.e c=300000 km/s.

As per the question ,the space craft is moving with a velocity 0.1 c.

We are asked to calculate the velocity of the light with respect to an observer present in Mars.

Considering Einstein's theory of relativity, the velocity of light will be c [300000 km/s] with respect to the person in Mars.

6 0

2 years ago

Read 2 more answers

A solid conducting sphere with radius R that carries positive charge Q is concentric with a very thin insulating shell of radius

chubhunter [2.5K]

Answer:

a) 0 < r < R: E = 0, R < r < 2R: E = KQ/r^2, r > 2R: E = 2KQ/r^2

b) See the picture

Explanation:

We can use Gauss's law to find the electric field in all the regions:

EA = qen/e0 where qen is the enclosed charge

Remember that the electric field everywhere outside a sphere is:

E(r) = q/(4*pi*eo*r^2) = Kq/r^2

For 0 < r < R: There is not enclosed charge because all of it remains on the outer layer of the conducting sphere, therefore E = 0 EA = 0/e0 = 0 E = 0
For R < r < 2R: Here the enclosed charge is equal Q E = Q/(4*pi*eo*r^2) = KQ/r^2
For r > 2R: Here the enclosed charge is equal 2Q E = Q/(4*pi*eo*r^2) + Q/(4*pi*eo*r^2) = 2Q/(4*pi*eo*r^2) = 2KQ/r^2

b) At the beginning there is no electric field this is why you see a line in zero, In R the electric field is maximum and then it starts to decrease exponentially with the distance and finally in 2R the field increase a little due to the second sphere to then continue decreasing exponentially with the distance

7 0

3 years ago