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blondinia [14]
2 years ago
10

5. Gravity is an attractive force betweenA. all massive objects.B. Earth and objects on Earth.C. Earth and Moon, and objects on

Earth.D. all objects everywhere.
Background image
Physics
1 answer:
qwelly [4]2 years ago
7 0

The correct answer is option D gravity is an attractive force between any objects in the universe i.e. gravitational force.

Is gravity a force that draws two items together?

A mass attracts a mass; the amount of the gravitational force is directly proportional to the masses of the two items and inversely proportional to the square of the distance between the two objects. Gravitational force is an attractive force that exists between all objects with mass.

We can see that the gravitational force is inversely proportional to the distance between masses and directly proportional to the product of the masses of two objects. As a result, any item in the cosmos can be attracted to another by the force of gravity.

To learn more about the gravitational force from the given link

brainly.com/question/72250

#SPJ1

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3 years ago
The tonga trench in the pacific ocean is 36,000 feet deep. assuming that sea water has an average density of 1.04 g/cm3, calcula
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Let's start by calculating how many cm deep is 36,000 feet. 

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 1097280 cm * 1.04 g/cm^3 = 1141171.2 g/cm^2   

 We now have a number using g/cm^2 as it's unit and we desire a unit of Pascals ( kg/(m*s^2) ).  

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3 years ago
Compute the resistance in ohms of a silver block 10 cm long and 0.10 cm2 in cross-sectional area. ( = 1.63 x 10-6 ohm-cm)
Vedmedyk [2.9K]
The resistance of the silver block is given by
R= \frac{\rho L}{A}
where
\rho=1.63 \cdot 10^{-6} \Omega \cdot cm is the silver resistivity
L=10 cm is the length of the block
A=0.10 cm^2 is the cross-sectional area of the block

If we plug the data into the equation, we find the resistance of the silver block:
R= \frac{(1.63 \cdot 10^{-6} \Omega \cdot cm)(10 cm)}{0.10 cm^2}=1.63 \cdot 10^{-4} \Omega
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