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2 years ago

Planets A and B have the same mass, but planet A is half the size of planet B.

Physics

2 answers:

pychu [463]2 years ago

7 0

Answer:

c. You would weigh less on planet A because the distance between

you and the planet's center of gravity would be smaller.

Explanation:

The statement that best describes your weight on each planet is that you would weigh less on planet A because the distance between you and the planet's center of gravity would be smaller.

This is based on Newton's law of universal gravitation which states that "the force of gravity between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distances between them".

Since weight is dependent on the force of gravity and mass, the planet with more gravitational pull will have masses on them weighing more.

Since the distance between the person and the center of the planet is smaller, therefore, the weight will be lesser.

Whitepunk [10]2 years ago

7 0

Answer:it’s D for ape x

Explanation:

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3 years ago

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serg [7]

Answer:

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Explanation:

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2 years ago

How is the independent variable affected by the dependent variable

stiv31 [10]

Answer:

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2 years ago

What is x-rays ? and how it's formed ?

tia_tia [17]

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2 years ago

Using 0.500 g of nichrome, you are asked to fabricate a wire with uniform cross-section. The resistance of the wire is 0.673 Ω.

mojhsa [17]

Explanation:

Given that,

Mass of Nichrome, m = 0.5 g

The resistance of the wire, R = 0.673 ohms

Resistivity of the nichrome wire, $\rho=10^{-6}\ \Omega -m$

Density, $d=8.31\times 10^3\ kg/m^3$

(A) The length of the wire is given by using the definition of resistance as :

Volume,

$V=A\times l\\\\A=\dfrac{V}{l}\\\\Since, V=\dfrac{m}{d}\\\\V=\dfrac{m}{d}\\\\V=\dfrac{0.5\times 10^{-3}}{8.31\times 10^3}\\\\V=6.01\times 10^{-8}\ m^3$

Area,

$A=\dfrac{V}{l}\\\\A=\dfrac{6.01\times 10^{-8}}{l}$ ....(1)

$R=\rho \dfrac{l}{A}\\\\l=\dfrac{RA}{\rho}\\\\l=\dfrac{0.673\times 6.01\times 10^{-8}}{l\times 10^{-6}}\\\\l=0.201\ m$

(b) Equation (1) becomes :

$A=\dfrac{6.01\times 10^{-8}}{l}\\\\A=\dfrac{6.01\times 10^{-8}}{0.201}\\\\\pi r^2=3\times 10^{-7}\\\\r=\sqrt{\dfrac{3\times 10^{-7}}{\pi}} \\\\r=3.09\times 10^{-4}\ m$

Hence, this is the required solution.

5 0

3 years ago