Answer:
Eastward, at 11 m/s^2
Explanation:
64N-31N=unbalanced force of 33N
F=ma
33N=(3kg)a
a=11m/s^2 to the East
Answer:
Initial pressure = 6 atm. Work = 0.144 J
Explanation:
You need to know the equation P1*V1=P2*V2, where P1 is the initial pressure, V1 is the initial volume, and P2 and V2 are the final pressure and volume respectively. So you can rearrange the terms and find that (1.2*0.05)/(0.01) = initial pressure = 6 atm. The work done by the system can be obtained calculating the are under the curve, so it is 0.144J
To solve this problem it is necessary to apply the concepts related to the Third Law of Kepler.
Kepler's third law tells us that the period is defined as

The given data are given with respect to known constants, for example the mass of the sun is

The radius between the earth and the sun is given by

From the mentioned star it is known that this is 8.2 time mass of sun and it is 6.2 times the distance between earth and the sun
Therefore:


Substituting in Kepler's third law:






Therefore the period of this star is 3.8years
Because mass and distance determine gravity, so the more mass you have, the more gravity.