Answer:
The work done on the system is -616 kJ
Explanation:
Given;
Quantity of heat absorbed by the system, Q = 767 kJ
change in the internal energy of the system, ΔU = +151 kJ
Apply the first law of thermodynamics;
ΔU = W + Q
Where;
ΔU is the change in internal energy
W is the work done
Q is the heat gained
W = ΔU - Q
W = 151 - 767
W = -616 kJ (The negative sign indicates that the work is done on the system)
Therefore, the work done on the system is -616 kJ
surface wave is a wave that travels along the surface of a medium. The medium is the matter through which the wave travels. Ocean waves are the best-known examples of surface waves. They travel on the surface of the water between the ocean and the air.
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Answer:
Also, as stream depth increases, the hydraulic radius increases thereby making the stream more free flowing. Both of these factors lead to an increase in stream velocity. The increased velocity and the increased cross-sectional area mean that discharge increases.
Below is the solution:
Heat soda=heat melon
<span>m1*cp1*(t-t1)=m2*cp2*(t2-t); cp2=cpwater </span>
<span>12*0.35*3800*(t-5)=6.5*4200*(27-t) </span>
<span>15960(t-5)=27300(27-t) </span>
<span>15960t-136500=737100-27300t </span>
<span>43260t=873600 </span>
<span>t=873600/43260 </span>
<span>t=20.19 deg celcius</span>
Answer:
a)15 N
b)12.6 N
Explanation:
Given that
Weight of block (wt)= 21 N
μs = 0.80 and μk = 0.60
We know that
Maximum value of static friction given as
Frs = μs m g = μs .wt
by putting the values
Frs= 0.8 x 21 = 16.8 N
Value of kinetic friction
Frk= μk m g = μk .wt
By putting the values
Frk= 0.6 x 21 = 12.6 N
a)
When T = 15 N
Static friction Frs= 16.8 N
Here the value of static friction is more than tension T .It means that block will not move and the value of friction force will be equal to the tension force.
Friction force = 15 N
b)
When T= 35 N
Here value of tension force is more than maximum value of static friction that is why block will move .We know that when body is in motion then kinetic friction will act on the body.so the value of friction force in this case will be 12.6 N
Friction force = 12.6 N
