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posledela
3 years ago
12

A NASA spacecraft measures the rate of at which atmospheric pressure on Mars decreases with altitude. The result at a certain al

titude is: Convert to .
Physics
1 answer:
Vlada [557]3 years ago
5 0

Answer:

= 4.21\times 10^{-10}J/cm^4

Explanation:

According to the given situation, the computation is shown below:-

1kPa=10^3Pa

So,

1Km = 10^5cm

1kPa=0.01Pa/Cm

1kPa=10^{-8}J/cm^4

Now as per the question it is given that

r = 0.0421KPa/Km

= 0.0421KPa/Km \times \frac{10^{-8}J/cm^4}{1KPa/Km}

after solving the above equation we will get

= 0.0421\times 10^{-8}J/cm^4

or we can write the above equation in this manner, that is

= 4.21\times 10^{-10}J/cm^4

Therefore we have solved the equation to reach the right answer.

You might be interested in
A geosynchronous Earth satellite is one that has an orbital period of precisely 1 day. Such orbits are useful for communication
koban [17]

Answer:

r = 4.24x10⁴ km.  

     

Explanation:

To find the radius of such an orbit we need to use Kepler's third law:

\frac{T_{1}^{2}}{T_{2}^{2}} = \frac{r_{1}^{3}}{r_{2}^{3}}

<em>where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km.                           </em>                              

From equation (1), r₁ is:

r_{1} = r_{2} \sqrt[3] {(\frac{T_{1}}{T_{2}})^{2}}                            

r_{1} = 3.84\cdot 10^{5} km \sqrt[3] {(\frac{1 d}{0.07481 y \cdot \frac{365 d}{1 y}})^{2}}      

r_{1} = 4.24 \cdot 10^{4} km      

Therefore, the radius of such an orbit is 4.24x10⁴ km.

I hope it helps you!

3 0
3 years ago
A spring balance attempts to drag an object across a table, but the object does not move because of the force of friction acting
Luda [366]
Moving
.................................................
4 0
2 years ago
A 45 kg bear slides, from rest, 11 m down a lodgepole pine tree, moving with a speed of 5.8 m/s just before hitting the ground.
Viefleur [7K]

Answer:

Explanation:

a )

change  in the gravitational potential energy of the bear-Earth system during the slide  = mgh

= 45 x 9.8 x 11

= 4851 J

b )

kinetic energy of bear just before hitting the ground

= 1/2 m v²

= .5 x 45 x 5.8²

= 756.9 J

c ) If  the average frictional force that acts on the sliding bear be F

negative work done by friction

= F x 11 J

then ,

4851 J -  F x 11 =  756.9 J

F x 11 = 4851 J -   756.9 J

= 4094.1 J

F = 4094.1 / 11

= 372.2 N  

4 0
3 years ago
Is joules law and Ohm's law same thing or different? Please tell:​
Zielflug [23.3K]

different because joules law talks about heat produce in an electric whiles ohm' law talks about potential difference

4 0
2 years ago
The sunspot cycle is a pattern of solar activity where the average number of sunspots gradually _______________________________
zzz [600]

Answer:

option D

Explanation:

Sunspots are the spot that appears on the sun, this spot appears darker than the surrounding surface of the sun.

Sun magnetic field goes through a cycle and this cycle is called the Sunspot cycle. Every 11 years the magnetic field of the sun completely flips. This sunspot cycle affects activity on the surface of the sun.

Sunspot cycle is the pattern of solar activity where an average number of sunspot gradually increase and decrease.

Hence, the correct answer is option D

8 0
3 years ago
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