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2 years ago

Explain the relationships between gravity, mass, and distance.

Physics

2 answers:

Mazyrski [523]2 years ago

7 0

Answer:

As the mass of an object increases, its gravitational force increases.

As an object's distance to other objects increases, its gravitational force on those objects increases.

Explanation:

The gravitational force of one object on another is calculated with the equation

F = (G*m1*m2)/(r²),

where G is the gravitational constant,

M1 and M2 are the masses of the two objects, and

r is the distance between them

We can see that the force has a direct relationship with both of the mass values, and an inverse square relationship with the distance between them.

Hope this helped!

dexar [7]2 years ago

7 0

Simpler way to put it:

Gravity has a direct relationship with mass. An increase in mass results in an increase in gravitational force. Gravity also has an inverse relationship with distance. An increase in distance results in a decrease in gravitational force.

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Define SI unit of measurement.

ivanzaharov [21]

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2 years ago

the amount of surface area of the block contact with the surface is 2.03*10^-2*m2 what is the average pressure exerted on the su

CaHeK987 [17]

Complete question:

A block of solid lead sits on a flat, level surface. Lead has a density of 1.13 x 104 kg/m3. The mass of the block is 20.0 kg. The amount of surface area of the block in contact with the surface is 2.03*10^-2*m2, What is the average pressure (in Pa) exerted on the surface by the block? Pa

Answer:

The average pressure exerted on the surface by the block is 9655.17 Pa

Explanation:

Given;

density of the lead, ρ = 1.13 x 10⁴ kg/m³

mass of the lead block, m = 20 kg

surface area of the area of the block, A = 2.03 x 10⁻² m²

Determine the force exerted on the surface by the block due to its weight;

F = mg

F = 20 x 9.8

F = 196 N

Determine the pressure exerted on the surface by the block

P = F / A

where;

P is the pressure

P = 196 / (2.03 x 10⁻²)

P = 9655.17 N/m²

P = 9655.17 Pa

Therefore, the average pressure exerted on the surface by the block is 9655.17 Pa

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2 years ago

Consider the power dissipated in a series R–L–C circuit with R=280Ω, L=100mH, C=0.800μF, V=50V, and ω=10500rad/s. The current an

ki77a [65]

Answer:

0.28802

2.57162 W

14.28 W

53.55 W

6.07142 W

Explanation:

R = 280Ω

L = 100 mH

C = 0.800 μF

V = 50 V

ω = 10500rad/s

For RLC circuit impedance is given by

$Z=\sqrt{R^2+(X_L-X_C)^2}\\\Rightarrow Z=\sqrt{R^2+(\omega L-\dfrac{1}{\omega C})^2}\\\Rightarrow Z=\sqrt{280^2+(10500\times 100\times 10^{-3}-\dfrac{1}{10500\times 0.8\times 10^{-6}})^2}\\\Rightarrow Z=972.1483\ \Omega$