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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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Heat is conducted in the axial direction through a cylinder by heating one end. If the diameter of the cylinder is doubled (but
CaHeK987 [17]

Answer:

hello your question is incomplete attached below is the complete question with the options

answer : Doubling the Diameter increases the heat transfer rate by a factor of 4  ( B )

Explanation:

The heat transfer in the axial direction of the cylinder can be calculated/determined as attached below

\frac{q_{2} }{q_{1} } = \frac{(2D)^2}{D^2}

q_{2} = 4q_{1}  ( This shows that Doubling the diameter increases the heat transfer rate by a factor of 4 )

4 0
3 years ago
a 1150 kg car is on a 8.70 hill. using x-y axis tilted down the plane, what is the x-component of the normal force(unit=N)
rodikova [14]

The x-component of the normal force is equal to <u>1706.45 N.</u>

Why?

To solve the problem, and since there is no additional information, we can safely assume that the x-axis is parallalel to the hill surface and the y-axis is perpendicular to the x-axis. Knowing that, we can calculate the components of the normal force (or weight for this case), using the following formulas:

N_{x}=W*Sin(\alpha)=mg*Sin(\alpha)\\\\N_{y}=W*Cos(\alpha)=mg*Cos(\alpha)

Now, using the given information, we have:

mass=m=1150Kg\\\alpha=8.70\°\\g=9.81\frac{m}{s^{2}}

Calculating, we have:

N_{x}=mg*Sin(\alpha)

N_{x}=1150Kg*9.81\frac{m}{s^{2}}*Sin(8.70\°)\\\\N_{x}=11281.5\frac{Kg.m}{s^{2} }*Sin(8.70\°)=1706.45\frac{Kg.m}{s^{2} }=1706.45.23N

Hence, we have that the x-component of the normal force is equal to  <u>1706.45 N.</u>

Have a nice day!

3 0
4 years ago
Steam enters an adiabatic turbine at 5 MPa and 4500C and leaves at a pressure of 1.4 MPa. Determine the work output of the turbi
REY [17]

Answer:350.92 KJ/kg

Explanation:

Given the process is reversible adiabatic i.e it is isentropic

P_1=5 MPa

T_1=4500 MPa

P_2=1.4 MPa

From steam table

h_1=3317.03KJ/kg

For isentropic process s_1=s_2

at P_2=1.4 MPa

s_2=6.82 KJ/kg-k

h_2=2966.11 KJ/kg

Therefore Work output of the turbine per unit mass of steam is =h_1-h_2

=3317.03-2966.11

=350.92 KJ/kg

8 0
3 years ago
A pump can throw a stream of water up to 29.6m. What is the initial speed of the stream when it leaves a hose nozzle.
zhannawk [14.2K]
Hello
We know that the water's kinetic energy upon leaving will be converted into potential energy as it is projected straight up. The maximum height attained by the water is 29.6m.
We know Kinetic Energy is
1/2 * m * v^2
And Potential energy is
m*g*h
Equating,
(m*v^2)/2 = m*g*h
The mass cancels out from each side, making an object's mass irrelevant when its Kinetic and Potential energies are interchanging.
v = sprt(2gh)
v = sqrt(2*9.81*29.6)
v = 24.1 m/s
7 0
4 years ago
A 50 kg laboratory worker is exposed to 30 mJ of neutron radiation with an RBE of 10. Part A What is the dose in mSv
lys-0071 [83]

Answer:

The dose is 6 mSV

Explanation:

The absorbed dose (in gray - Gy) is the amount of energy that ionizing radiation deposits per unit mass of tissue. That is,

Absorbed dose = Energy deposited / Mass

while Dose equivalent (DE) (in Seivert -Sv) is given by

DE = Absorbed dose × RBE (Relative biological effectiveness)

First, we will determine the Absorbed dose

From the question, Energy deposited = 30mJ and Mass = 50kg

From,

Absorbed dose = Energy deposited / Mass

Absorbed dose = 30mJ/50kg

Absorbed dose = 0.6 mGy

Now, for the Dose equivalent (DE)

DE = Absorbed dose × RBE

From the question, RBE = 10

Hence,

DE = 0.6mGy × 10

DE = 6 mSv

5 0
3 years ago
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