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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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Bill Nye help.. I didn’t get link to vid, hoping y’all have seen it. 20 POINTS!
Lyrx [107]

Answer:

Potential, Kinetic and Chemical energy.

Explanation:

btw, congratulations for turning into an expert.

6 0
3 years ago
A satellite of mass M = 270kg is in circular orbit around the Earth at an altitude equal to the earth's mean radius (6370 km). A
zubka84 [21]

To solve this problem we will apply the concepts related to Orbital Speed as a function of the universal gravitational constant, the mass of the planet and the orbital distance of the satellite. From finding the velocity it will be possible to calculate the period of the body and finally the gravitational force acting on the satellite.

PART A)

V_{orbital} = \sqrt{\frac{GM_E}{R}}

Here,

M = Mass of Earth

R = Distance from center to the satellite

Replacing with our values we have,

V_{orbital} = \sqrt{\frac{(6.67*10^{-11})(5.972*10^{24})}{(6370*10^3)+(6370*10^3)}}

V_{orbital} = 5591.62m/s

V_{orbital} = 5.591*10^3m/s

PART B) The period of satellite is given as,

T = 2\pi \sqrt{\frac{r^3}{Gm_E}}

T = \frac{2\pi r}{V_{orbital}}

T = \frac{2\pi (2*6370*10^3)}{5.591*10^3}

T = 238.61min

PART C) The gravitational force on the satellite is given by,

F = ma

F = \frac{1}{4} mg

F = \frac{270*9.8}{4}

F = 661.5N

5 0
3 years ago
Sugar dissolved in water is an example of?
lbvjy [14]

Answer:

D. Solution

Explanation:

Sugar dissolved in water is an example of solution.

A solution is a homogenous mixture of solutes and solvents.

In a solution the solute particles ae distributed uniformly in the solvents. The solute is the substance and it is the sugar here that is dissolved to make a solution.

The solvent is the water in this instance that helps to dissolve the solute.

5 0
3 years ago
Read 2 more answers
A light bulb does 100 joules of work in 2.5 seconds. how much work power dos it have
Lesechka [4]
100/2.5 because power=energy/time
5 0
3 years ago
A 0.75μF capacitor is charged to 70 V . It is then connected in series with a 55Ω resistor and a 140 Ω resistor and allowed to d
Ipatiy [6.2K]

Answer:

Explanation:

Capacitor of 0.75μF, charged to 70V and connect in series with 55Ω and 140 Ω to discharge.

Energy dissipates in 55Ω resistor is given by V²/R

Since the 55ohms and 140ohms l discharge the capacitor fully, the voltage will be zero volts and this voltage will be shared by the resistor in ratio.

So for 55ohms, using voltage divider rule

V=R1/(R1+R2) ×Vt

V=55/(55+140) ×70

V=19.74Volts is across the 55ohms resistor.

Then, energy loss will be

E=V²/R

E=19.74²/55

E=7.09J

7.09J of heat is dissipated by the 55ohms resistor

6 0
4 years ago
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