Answer:
Explanation:
Constant pressure molar heat capacity Cp = 29.125 J /K.mol
If Cv be constant volume molar heat capacity
Cp - Cv = R
Cv = Cp - R
= 29.125 - 8.314 J
= 20.811 J
change in internal energy = n x Cv x Δ T
n is number of moles , Cv is molar heat capacity at constant volume , Δ T is change in temperature
Putting the values
= 20 x 20.811 x 15
= 6243.3 J.
Answer:
E = k Q 1 / (x₀-x₂) (x₀-x₁)
Explanation:
The electric field is given by
dE = k dq / r²
In this case as we have a continuous load distribution we can use the concept of linear density
λ= Q / x = dq / dx
dq = λ dx
We substitute in the equation
∫ dE = k ∫ λ dx / x²
We integrate
E = k λ (-1 / x)
We evaluate between the lower limits x = x₀- x₂ and higher x = x₀-x₁
E = k λ (-1 / x₀-x₁ + 1 / x₀-x₂)
E = k λ (x₂ -x₁) / (x₀-x₂) (x₀-x₁)
We replace the density
E = k (Q / (x₂-x₁)) [(x₂-x₁) / (x₀-x₂) (x₀-x₁)]
E = k Q 1 / (x₀-x₂) (x₀-x₁)
Answer:
Pressure is equal to the ratio of thrust to the area in contact. Upthrust is a force exerted by the fluids on an object placed in the fluid . Upthrust acts in upward direction.
k = spring constant of the spring = 85 N/m
m = mass of the box sliding towards the spring = 3.5 kg
v = speed of box just before colliding with the spring = ?
x = compression the spring = 6.5 cm = 6.5 cm (1 m /100 cm) = 0.065 m
the kinetic energy of box just before colliding with the spring converts into the spring energy of the spring when it is fully compressed.
Using conservation of energy
Kinetic energy of spring before collision = spring energy of spring after compression
(0.5) m v² = (0.5) k x²
m v² = k x²
inserting the values
(3.5 kg) v² = (85 N/m) (0.065 m)²
v = 0.32 m/s
