we have to use newtons law of gravitation which is
F=GMm/r^2
G=6.67 x 10^<span>-11N kg^2/m^2
</span>M=<span>(15kg)
</span>m=15 kg
r=(3.0m)^2<span>
</span>putting values we have
<span>=(6.67 x 10^-11N kg^2/m^2)(15kg)(15kg)/(3.0m)^2 </span>
=1.67 x 10^-9N
Answer:
Explanation:
To solve this problem we use the Hooke's Law:
(1)
F is the Force needed to expand or compress the spring by a distance Δx.
The spring stretches 0.2cm per Newton, in other words:
1N=k*0.2cm ⇒ k=1N/0.2cm=5N/cm
The force applied is due to the weight

We replace in (1):
We solve the equation for m:
Answer:
Explanation:
Given
mass of saturated liquid water 
at
specific volume is
(From Table A-4,Saturated water Temperature table)



Final Volume 


Specific volume at this stage



Now we see the value and find the temperature it corresponds to specific volume at vapor stage in the table.


