Ask question

Ask question

All categories

3 years ago

7

Why is Jupiter's rotation dangerous for human survivability?

1 answer:

krok68 [10]3 years ago

7 0

Answer:

1. Why is Jupiter's rotation dangerous for human survivability?

<h2>=> </h2>

<em><u>Jupiter is the fastest rotating planet in our solar system. One day lasts about 9.5 Earth hours. This creates powerful winds that can whip around the planet at more than 300 mph. About 75 miles below the clouds, you reach the limit of human exploration.</u></em>

2 .Why is Jupiter's planet axis tilt an issue for human survivability?

<h2>=></h2>

<em><u>Jupiter, like Venus, has an axial tilt of only 3 degrees, so there is literally no difference between the seasons. ... The length of each season is roughly three years. Jupiter is the fastest spinning planet in our Solar System, which causes the planet to flatten at the poles and bulge at the </u></em><em><u>equator.</u></em>

3.Why is the diameter of Jupiter an issue for human survivability?

<h2>=></h2>

<em><u>Since </u></em><em><u>,</u></em><em><u>The </u></em><em><u>Jupiter </u></em><em><u>is </u></em><em><u>so </u></em><em><u>huge </u></em><em><u>in </u></em><em><u>mass</u></em><em><u> </u></em><em><u>,</u></em><em><u>The </u></em><em><u>central</u></em><em><u> </u></em><em><u>force</u></em><em><u> </u></em><em><u>toward</u></em><em><u> </u></em><em><u>the </u></em><em><u>centre </u></em><em><u>will </u></em><em><u>be </u></em><em><u>high</u></em><em><u> </u></em><em><u>and</u></em><em><u> </u></em><em><u>we'll</u></em><em><u> </u></em><em><u>be </u></em><em><u>forced</u></em><em><u> </u></em><em><u>toward</u></em><em><u> </u></em><em><u>it </u></em><em><u>causing</u></em><em><u> </u></em><em><u>Several</u></em><em><u> </u></em><em><u>problems</u></em><em><u>.</u></em>

You might be interested in

A hydrogen ion of mass 'm' and charge 'q' travels with a speed 'v' in a circle of radius 'r' in a magnetic field of intensity 'B

Allisa [31]

Answer:

$r=\frac{mv}{qB}$ , $T=\frac{2\pi m}{qB}$

Explanation:

The force experienced by the ion due to the magnetic field is given by:

$F=qvB$

where

q is the charge

v is the speed

B is the intensity of the magnetic field

The cetripetal force is

$F=\frac{mv^2}{r}$

where

m is the mass

r is the radius of the circle

Since the magnetic force acts as centripetal force, we can equate the two expressions:

$qvB=m\frac{v^2}{r}$

Re-arranging it, we find the radius:

$r=\frac{mv}{qB}$

Now, if we want to find the time it takes for the ion to make one complete circle (=the period), we just need to divide the length of one circumference by the speed:

$T=\frac{2\pi r}{v}$

And susbstituting the expression we found before for r, we find

$T=\frac{2\pi mv}{qBv}=\frac{2\pi m}{qB}$

8 0

3 years ago

Which of the following describe a neutron?

Leokris [45]

Neutral charge, because neutron means neutral

5 0

4 years ago

Which expression should appear in the blue box to compute percent error? Accepted value – Experimental value Accepted value × Ex

Alexxx [7]

The last choice ... Experimental value -- <span>Accepted value ... is the place
to start, but it doesn't quite get you there.

To find the FRACTIONAL error, you'd have to divide that difference
by the Accepted Value.

And then . . .

To find the PERCENTAGE error, you'd have to multiply
the fractional error by 100 .
</span>

5 0

3 years ago

A space station shaped like a giant wheel has a radius of a radius of 153 m and a moment of inertia of 4.16 × 10⁸ kg·m² (when it

Naya [18.7K]

Answer:

a = 1.709g

Explanation:

Given the absence of external forces being applied in the space station, it is possibly to use the Principle of Angular Momentum Conservation, which states that:

$I_{o} \cdot \omega_{o} = I_{f} \cdot \omega_{f}$

The required initial angular speed is obtained herein:

$g= \omega_{o}^{2}\cdot R_{ss}$

$\omega_{o}=\sqrt{\frac{g}{R_{ss}} }$

$\omega_{o}= \sqrt{\frac{9.807\,\frac{m}{s^{2}} }{153\,m} }$

$\omega_{o} \approx 0.253\,\frac{rad}{s}$

The initial moment of inertia is:

$I_{o} =I_{ss}+n\cdot m_{person}\cdot R_{ss}^{2}$

$I_{o} = 4.16\times 10^{8}\,kg\cdot m^{2}+(150)\cdot (65\,kg)\cdot (153\,m)^{2}$

$I_{o} = 6.442\times 10^{8}\,kg\cdot m^{2}$

The final moment of inertia is:

$I_{f} =I_{ss}+n\cdot m_{person}\cdot R_{ss}^{2}$

$I_{f} = 4.16\times 10^{8}\,kg\cdot m^{2}+(50)\cdot (65\,kg)\cdot (153\,m)^{2}$

$I_{f} = 4.921\times 10^{8}\,kg\cdot m^{2}$

Now, the final angular speed is obtained:

$\omega_{f} = \frac{I_{o}}{I_{f}}\cdot \omega_{o}$

$\omega_{f} = \frac{6.442\times 10^{8}\,{kg\cdot m^{2}}}{4.921\times 10^{8}\,kg\cdot m^{2}} \cdot (0.253\,\frac{rad}{s} )$

$\omega_{f} = 0.331\,\frac{rad}{s^}$

The apparent acceleration is:

$a_{f} = \omega_{f}^{2}\cdot R_{ss}$

$a_{f} = (0.331\,\frac{rad}{s} )^{2}\cdot (153\,m)$

$a_{f} = 16.763\,\frac{m}{s^{2}}$

This is approximately 1.709g.

4 0

4 years ago

What happens to electron flow with a conductor of the voltage source is removed?

9966 [12]

The electrons stop flowing

4 0

3 years ago