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crimeas [40]
3 years ago
9

Calculate how much an astronaut with mass of 90 kg would weigh while standing on the surface of Mars. The acceleration due to gr

avity on Mars is approximately 3.7 m/s^2. Express your answer in Newtons. Show your work
Physics
1 answer:
Mars2501 [29]3 years ago
6 0

The weight of anything is

(the thing's mass) times (gravity in the place where the thing is located).

Gravity on Earth is 9.8 m/s
² .
90 kg of mass weighs  (90 kg) · (9.8 m/s²)  =  882 Newtons on Earth.

Gravity on the Moon is 1.62 m/s² .
The same 90 kg weighs (90 kg)·(1.62 m/s²) = 146 Newtons on the Moon.

Gravity on Mars is 3.7 m/s² .
The same 90 kg of mass weighs (90 kg)·(3.7 m/s²) = 333 Newtons on Mars.
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Because Any object or substance that is less dense than a fluid will float in that fluid, so hot water rises (floats) in colder water. When fluids are cooled, they contract and therefore become more dense. Any object or substance that is more dense than a fluid will sink in that fluid, so cold water sinks in warmer water
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3 years ago
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Two masses traveling at the same speed v₀ collide head on. Nothing is known about the collision other than that it is not comple
alex41 [277]

Answer:

false

Explanation:

Elastic collision: the collision is said to be elastic if the kinetic energy is conserved during the collision.

If the kinetic energy is not conserved, then the collision is not elastic.

So, here no information is given so we cant say about the type of collision.

So, it is False.

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3 years ago
Which of the following is a<br> fossil fuel?<br> A. Coal<br> B. Wind energy<br> C. Biomass energy
almond37 [142]

A. Coal

is the correct answer

5 0
3 years ago
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lozanna [386]

Answer:

D) a Battery

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6 0
3 years ago
The loop is in a magnetic field 0.20 T whose direction is perpendicular to the plane of the loop. At t = 0, the loop has area A
love history [14]

Answer:

Part a)

EMF = 14 \times 10^{-3} V

Part b)

EMF = 15.67 \times 10^{-3} V

Explanation:

As we know that magnetic flux through the loop is given as

\phi = B.A

now we have

\phi = B\pi r^2

now rate of change in flux is given as

\frac{d\phi}{dt} = B(2\pi r)\frac{dr}{dt}

now we know that

A = \pi r^2

0.285 = \pi r^2

r = 0.30 m

Now plug in all data

EMF = (0.20)\times 2\pi\times (0.30) \times (0.037)

EMF = 14 \times 10^{-3} V

Part b)

Now the radius of the loop after t = 1 s

r_1 = r_0 + \frac{dr}{dt}

r_1 = 0.30 + 0.037

r_1 = 0.337 m

Now plug in data in above equation

EMF = (0.20)\times 2\pi\times (0.337) \times (0.037)

EMF = 15.67 \times 10^{-3} V

5 0
3 years ago
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