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3 years ago

What would the person’s mass be on the earth? Part B pls

Physics

1 answer:

Dafna1 [17]3 years ago

4 0

Answer:

I hopes it helps

thank you

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#2 A car accelerates from rest at 4 m/s^2. What is the velocity of the car after 4 second?

solmaris [256]

You use the equation Velocity = Acceleration X Time. 4x4=16m/s.

The car travels 18m in 3 seconds.

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4 years ago

Is the process of changing from one phase of matter to another a physical change or a chemical change? Explain why. please asap

Andreas93 [3]

The process of changing from one phase of matter to another is a physical matter.

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3 years ago

Represent 7468 N with SI units having an appropriate prefix. Express your answer to four significant figures and include the app

Musya8 [376]

Answer:

7.468 kN

Explanation:

Here the force is given in Newton

Some of the prefixes of the SI units are

kilo = 10³

Mega = 10⁶

Giga = 10⁹

Tera = 10¹²

The number is 7468.0

Here, the only solution where the number of significant figures is kilo. If any other prefix is chosen then the significant figures will increase.

1 kilonewton = 1000 Newton

$1\ Newton=\frac{1}{1000}\ kilonewton$

$\\\Rightarrow 7468\ Newton=\frac{7468}{1000}\ kilonewton\\ =7.468\ kilonewton$

So, 7468 N = 7.468 kN

7 0

3 years ago

Two planets A and B, where B has twice the mass of A, orbit the Sun in elliptical orbits. The semi-major axis of the elliptical

lozanna [386]

Answer:

2.83

Explanation:

Kepler's discovered that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit, that is called Kepler's third law of planet motion and can be expressed as:

$T=\frac{2\pi a^{\frac{3}{2}}}{\sqrt{GM}}$ (1)

with T the orbital period, M the mass of the sun, G the Cavendish constant and a the semi major axis of the elliptical orbit of the planet. By (1) we can see that orbital period is independent of the mass of the planet and depends of the semi major axis, rearranging (1):

$\frac{T}{a^{\frac{3}{2}}}=\frac{2\pi}{\sqrt{GM}}$

$\frac{T^{2}}{a^{3}}=(\frac{2\pi }{\sqrt{GM}})^2$ (2)

Because in the right side of the equation (2) we have only constant quantities, that implies the ratio $\frac{T^{2}}{a^{3}}$ is constant for all the planets orbiting the same sun, so we can said that: