The springs of a 1800 kg car compress 5.4 mm when its 66 kg driver gets into the driver's seat. If the car goes over a bump, wha
t will be the frequency of oscillations
1 answer:
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Answer:
a ) 11.1 *10^3 m/s = 39.96 Km/h
b) T_{o2} =1.58*10^5 K
Explanation:
a)= 11.1 km/s =11.1 *10^3 m/s = 39.96 Km/h
b)
M_O2 = 32.00 g/mol =32.0*10^{-3} kg/mol
gas constant R = 8.31 j/mol.K
So,
multiply each side by M_{o2}, so we have
solving for temperature T_{o2}
In the question given,
T_{o2} =1.58*10^5 K
Answer:
Explanation: please see attached file I attached the answer to your question.
Answer:
R=3818Km
Explanation:
Take a look at the picture. Point A is when you start the stopwatch. Then you stand, the planet rotates an angle α and you are standing at point B.
Since you travel 2π radians in 24H, the angle can be calculated as:
t being expressed in hours.
From the triangle formed by A,B and the center of the planet, we know that:
Solving for r, we get:
