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3 years ago

Why is saturn still giving off extra energy?

1 answer:

8 0

There is not a "for sure" answer yet, but scientist thinks that it might be due to Saturn's cloud cover, if the cloud cover fluctuated, then Saturn would scatter infrared light differently, giving off more energy than it needs to.

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Earth has a mass of 5.97 × 1024 kg and a radius of 6.38 × 106 m, while Saturn has a mass of 5.68 × 1026 kg and a radius of 6.03

hammer [34]

Answer: mass on earth = 625.792 N, mass on saturn = 666.75 N

Explanation: weight of an object = mg

Where g = acceleration due to gravity =GM/r²

Where G = gravitational constant, M = mass of planet and r = radius of planet.

Let us start with earth.

G =gravitational constant of the earth =6.67×10^-11 m³ /kgs²

Me = mass of earth = 5.97×10^24 kg

re = radius of earth = 6.38×10^6 m

Acceleration due to gravity on earth = (6.67×10^-11×5.97×10^24) /(6.38×10^6)²

Acceleration due to gravity on earth = 3.98×10^14 /4.07×10^13

Acceleration due gravity on earth = 9.778 m/s²

The mass of the person is 64 kg, hence his weight on earth is given as

W = mass × Acceleration due gravity on earth = 64 ×9.778 = 625.792 N

For saturn.

G =gravitational constant of the earth =6.67×10^-11 m³ /kgs²

Ms = mass of saturn = 5.68×10^26kg

rs= radius of saturn= 6.03×10^7 m

Acceleration due to gravity on saturn= (6.67×10^-11×5.68×10^26 /(6.03×10^7)²

Acceleration due to gravity on saturn= 3.788×10^16/3.636×10^15

Acceleration due gravity on saturn= 10.417 m/s²

The mass of the person is 64 kg, hence his weight on saturn is given as

W = mass × Acceleration due gravity on saturn = 64 ×10.417 = 666.75 N

3 0

3 years ago

An unmarked police car traveling a constant 38.6 m/s is passed by a speeder traveling 53.4 m/s. Precisely 2.2 seconds after the

Juliette [100K]

Answer:

The unmarked police car needs approximately 71.082 seconds to overtake the speeder.

Explanation:

Let suppose that speeder travels at constant velocity, whereas the unmarked police car accelerates at constant rate. In this case, we need to determine the instant when the police car overtakes the speeder. First, we construct a system of equations:

Unmarked police car

$s = s_{o}+v_{o,P}\cdot (t-t') + \frac{1}{2}\cdot a\cdot (t-t')^{2}$ (1)

Speeder

$s = s_{o} + v_{o,S}\cdot t$ (2)

Where:

$s_{o}$ - Initial position, measured in meters.

$s$ - Final position, measured in meters.

$v_{o,P}$ , $v_{o,S}$ - Initial velocities of the unmarked police car and the speeder, measured in meters per second.

$a$ - Acceleration of the unmarked police car, measured in meters per square second.

$t$ - Time, measured in seconds.

$t'$ - Initial instant for the unmarked police car, measured in seconds.

By equalizing (1) and (2), we expand and simplify the resulting expression:

$v_{o,P}\cdot (t-t')+\frac{1}{2}\cdot a\cdot (t-t')^{2} = v_{o,S}\cdot t$

$v_{o,P}\cdot t -v_{o,P}\cdot t' +\frac{1}{2}\cdot a\cdot t^{2}-a\cdot t'\cdot t+\frac{1}{2}\cdot a\cdot t'^{2} = v_{o,S}\cdot t$

$\frac{1}{2}\cdot a\cdot t^{2}+[(v_{o,P}-v_{o,S})-a\cdot t']\cdot t -\left(v_{o,P}\cdot t'-\frac{1}{2}\cdot a\cdot t'^{2}\right) = 0$

If we know that $a = 1.6\,\frac{m}{s^{2}}$ , $v_{o,P} = 0\,\frac{m}{s}$ , $v_{o,S} = 53.4\,\frac{m}{s}$ and $t' = 2.2\,s$ , then we solve the resulting second order polynomial: