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OLga [1]
3 years ago
15

If the moon did not rotate at the same rate that it revolved, what would happen to the gravity of the earth?

Physics
1 answer:
dem82 [27]3 years ago
5 0

Answer:

The gravity, which is an acceleration to the center of the earth, will be the same.

Explanation:

The gravity on earth depends only on the masses and distance, between two objects. We can see it in the gravitational force equation.  

F=G\frac{m\cdot M}{r^{2}}  

Now if we put a man, with mass m, on the surface of the earth, with mass M, the distance from the center of mass and the man will be R (earth radius). Knowing that F = m*a, we can find the accelerations due to this mass M and this value will be 9.81 m/s².  

On the other hand, the moon has a gravity value and is less than the earth, because its mass, and affects the water sea due to the gravitational force between earth and moon. If the moon changes the rate of its rotate it changes probably the distance between them, let's recall they must conserve angular momentum, but the gravity won't be affected.

Therefore, the gravity, which is an acceleration to the center of the earth, will be the same.

   

I hope it helps you!

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What affect does doubling the net force have on the acceleration of the object (when
Triss [41]
<h3>Answer: The acceleration doubles</h3>

===========================================================

Explanation:

Consider a mass of 10 kg, so m = 10

Let's say we apply a net force of 20 newtons, so F = 20

The acceleration 'a' is...

F = ma

20 = 10a

20/10 = a

2 = a

a = 2

The acceleration is 2 m/s^2. Every second, the velocity increases by 10 m/s.

---------------

Now let's double the net force on the object

F = 20 goes to F = 40

m = 10 stays the same

F = ma

40 = 10a

10a = 40

a = 40/10

a = 4

The acceleration has also doubled since earlier it was a = 2, but now it's a = 4.

---------------

In summary, if you double the net force applied to the object, then the acceleration doubles as well.

8 0
3 years ago
Read 2 more answers
If the mass of the ladder is 12.0 kgkg, the mass of the painter is 55.0 kgkg, and the ladder begins to slip at its base when her
Marysya12 [62]

Answer:

 μ = 0.336

Explanation:

We will work on this exercise with the expressions of transactional and rotational equilibrium.

Let's start with rotational balance, for this we set a reference system at the top of the ladder, where it touches the wall and we will assign as positive the anti-clockwise direction of rotation

          fr L sin θ - W L / 2 cos θ - W_painter 0.3 L cos θ  = 0

          fr sin θ  - cos θ  (W / 2 + 0,3 W_painter) = 0

          fr = cotan θ  (W / 2 + 0,3 W_painter)

Now let's write the equilibrium translation equation

     

X axis

        F1 - fr = 0

        F1 = fr

the friction force has the expression

       fr = μ N

Y Axis

       N - W - W_painter = 0

       N = W + W_painter

       

we substitute

      fr = μ (W + W_painter)

we substitute in the endowment equilibrium equation

     μ (W + W_painter) = cotan θ  (W / 2 + 0,3 W_painter)

      μ = cotan θ (W / 2 + 0,3 W_painter) / (W + W_painter)

we substitute the values ​​they give

      μ = cotan θ  (12/2 + 0.3 55) / (12 + 55)

      μ = cotan θ  (22.5 / 67)

      μ = cotan tea (0.336)

To finish the problem, we must indicate the angle of the staircase or catcher data to find the angle, if we assume that the angle is tea = 45

       cotan 45 = 1 / tan 45 = 1

the result is

    μ = 0.336

5 0
4 years ago
Kathy 82 kg performer standing on a diving board at the carnival dive straight down into a small pool of water. Just before stri
mixas84 [53]

Solution :

Given weight of Kathy = 82 kg

Her speed before striking the water, $V_o $ = 5.50 m/s

Her speed after entering the water, $V_f$= 1.1 m/s

Time = 1.65 s

Using equation of impulse,

$dP = F \times  dT$

Here, F =  the force ,

       dT =  time interval over which the force is applied for

            = 1.65 s

       dP  = change in momentum

dP = m x dV

    $= m \times [V_f - V_o] $

    = 82 x (1.1 - 5.5)

    = -360 kg

∴ the net force acting will be

$F=\frac{dP}{dT}$

$F=\frac{-360}{1.65}$

  = 218 N

8 0
3 years ago
!!! 100 POINTS !!! PLEASE HELP !!!
hram777 [196]

Q1. Option 2: basketball


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<span>Q3. A basketball sitting on the floor stays there and a basketball rolling on court keeps on rolling.</span>

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<span>Q4 Second law says acceleration is dependent upon net force and mass of the object.</span>


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<span>Q6. Third law says f<span>or every action, there is an equal and opposite reaction.</span></span>

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<span><span>Q7. As a player dribbles, the force the basketball hits the floor with is the same as the force from the floor on the ball. That is why the ball bounces back up in air.</span></span>

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8 0
3 years ago
Read 2 more answers
What is true of all particles of gas enclosed within a container?
Andre45 [30]
Theyre all in constant motion

6 0
4 years ago
Read 2 more answers
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