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vivado [14]
3 years ago
7

The planet Mars is host to five functioning spacecraft, three in orbit about the planet and two on the surface of the planet. Th

anks to those spacecraft, we know that the planet Mars has a mass that is 0.11 times that of Earth and a radius that is 0.53 times that of Earth. The acceleration of an object in free-fall near the surface of Mars is most nearly what in terms of the local value of g on Earth
Physics
1 answer:
Naily [24]3 years ago
3 0

Answer:

0.392

Explanation:

Mm = 0.11Me

Rm = 0.53Re

g = GM / r^2

G = 6.67 * 10^-11

gmars = (G * 0.11Me) / (0.53Re)^2

Recall:

gearth = GMe /Re^2

Hence, gmars in terms of gearth equals

gmars = gearth * (0.11 / 0.53^2)

gmars = gearth * 0.3915984

gmars = 0.392gearth

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1.15) What distance will 650 joules of work move a box weighing 50 newtons?​
son4ous [18]

Answer:

The answer is 13 however make sure if they ask for a certain measurement like meter answer it by saying 13 meters.

Explanation:

This basically turns into basic algebra if you know the formula for work. The formula for work is W=F*d  

Here are the variables that you know 650J=50N*d so you need d.

All you do is divide 650J by 50N and you get a total of 13 (meters since I don't know what they want you to put it in).

6 0
3 years ago
Describe how you can determine:<br>a) Volume of an irregular body<br>b) Density of a liquid​
vlada-n [284]

Density of liquid try thank you so much

7 0
3 years ago
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I need help with question 17
Alexxx [7]
It is b i had that qustion 

6 0
3 years ago
Two tiny conducting spheres are identical and carry charges of -19.8μC and +40.7μC. They are separated by a distance of 3.59 cm.
romanna [79]

Answer:

(a): \rm -5.627\times 10^3\ N.

(b):  \rm 7.626\times 10^2\ N.

Explanation:

<u>Given:</u>

  • Charge on one sphere, \rm q_1 = -19.8\ \mu C = -19.8\times 10^{-6}\ C.
  • Charge on second sphere, \rm q_2 = +40.7\ \mu C = +40.7\times 10^{-6}\ C.
  • Separation between the spheres, \rm r=3.59\ cm = 3.59\times 10^{-2}\ m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two static point charges is given by

\rm F=k\cdot\dfrac{q_1q_2}{r^2}

where,

k is called the Coulomb's constant, whose value is \rm 9\times 10^9\ Nm^2/C^2.

From Newton's third law of motion, both the spheres experience same force.

Therefore, the magnitude of the force that each sphere experiences is given by

\rm F=k\cdot\dfrac{q_1q_2}{r^2}\\=9\times 10^9\times \dfrac{(-19.8\times 10^{-6})\times (+40.7\times 10^{-6})}{(3.59\times 10^{-2})^2}\\=-5.627\times 10^3\ N.

The negative sign shows that the force is attractive in nature.

Part (b):

The spheres are identical in size. When the spheres are brought in contact with each other then the charge on both the spheres redistributes in such a way that the net charge on both the spheres distributed equally on both.

Total charge on both the spheres, \rm Q=q_1+q_2=-19.8\ \mu C+40.7\ \mu C = 20.9\ \mu C.

The new charges on both the spheres are equal and given by

\rm q_1'=q_2'=\dfrac Q2 = \dfrac{20.9}{2}\ \mu C=10.45\ \mu C = 10.45\times 10^{-6}\ C.

The magnitude of the force that each sphere now experiences is given by

\rm F'=k\cdot \dfrac{q_1'q_2'}{r^2}'\\=9\times 10^9\times \dfrac{10.45\times 10^{-6}\times 10.45\times 10^{-6}}{(3.59\times 10^{-2})^2}\\=7.626\times 10^2\ N.

7 0
3 years ago
A cyclist acceleration from 0m/s to 8m/s in 3second. what is his acceleration?​
umka2103 [35]

Answer:

2.66 m/s² .

Explanation:

Initial velocity , u = 0 m/s

Final Velocity , v = 8 m/s

Time Taken , t = 3 s

So , Acceleration = (v-u)/t = (8 m/s - 0 m/s) /3 sec . = 8/3 m/s² = 2.66 m/s²

4 0
3 years ago
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