Answer:
0.00417 kW/K or 4.17 W/K
Second law is satisfied.
Explanation:
Parameters given:
Rate of heat transfer, Q = 2kW
Temperature of hot reservoir, Th = 800K
Temperature of cold reservoir, Tc = 300K
The rate of entropy change is given as:
ΔS = Q * [(1/Tc) - (1/Th)]
ΔS = 2 * (1/300 - 1/800)
ΔS = 2 * 0.002085
ΔS = 0.00417 kW/K or 4.17 W/K
Since ΔS is greater than 0, te the second law of thermodynamics is satisfied.
A neutron star has more mass than a bowling ball,
and is about the same size as Chicago.
Answer:
2697.75N/m
Explanation:
Step one
This problem bothers on energy stored in a spring.
Step two
Given data
Compression x= 2cm
To meter = 2/100= 0.02m
Mass m= 0.01kg
Height h= 5.5m
K=?
Let us assume g= 9.81m/s²
Step three
According to the principle of conservation of energy
We know that the the energy stored in a spring is
E= 1/2kx²
1/2kx²= mgh
Making k subject of formula we have
kx²= 2mgh
k= 2mgh/x²
k= (2*0.01*9.81*5.5)/0.02²
k= 1.0791/0.0004
k= 2697.75N/m
Hence the spring constant k is 2697.75N/m
