Answer:
-58.876 kJ
Explanation:
m = mass of air = 1 kg
T₁ = Initial temperature = 15°C
T₂ = Final temperature = 97°C
Cp = Specific heat at constant pressure = 1.005 kJ/kgk
Cv = Specific heat at constant volume = 0.718 kJ/kgk
W = Work done
Q = Heat = 0 (since it is not mentioned we are considering adiabatic condition)
ΔU = Change in internal energy
Q = W+ΔU
⇒Q = W+mCvΔT
⇒0 = W+mCvΔT
⇒W = -mCvΔT
⇒Q = -1×0.718×(97-15)
⇒Q = -58.716 kJ
Electrostatic repulsion is the force between two charges having the same sign, that tends to separate them further. The force is proportional to the product of the charges, and inversely proportional to the square of the distance between them.
Answer:10842.33m/s
Explanation:
F=qvBsine
V=f/(qBsine)
V=(3.5×10^-2)÷(8.4×10^-4×6.7×10^-3×sin35)
V=10842.33m/s
Answer:
D.The striking foot pushes backward against the ground. The friction with the ground provides an equal and opposite force forward
