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:
t₁ = 0.95 s
Explanation:
In this chaos we must use the definition of Newton's second law
F = m a = m dv / dt
dv = F dt / m
Let's replace and integrate, let's take the upward direction of the plane as positive, the force is positive
dv = ∫ (3 + 2t) dt / m
v = (3 t + 2 t²/ 2) /m
Let's evaluate between the lower limit t = 0 v = -6 ft / s (going down) to the upper limit t = t and v = 0
0 - (-6) = (3 (t- 0) + (t² -0)) / m
t² + 3t -6m = 0
Let's look for the mass
W = mg
m = W / g
m = 20/32
m = 0.625 slug
Let's solve the second degree equation
t² + 3t -3.75 = 0
t = (-3 ± √ (32 + 4 1 3.75)) / 2
t = (-3 ± 4,899) / 2
t₁ = 0.95 s
t₂ = -3.95 s
We take the positive time
Answer:
(a) Steel rod: 
Copper rod: 
(b) Steel rod: 
Copper rod: 
Explanation:
Length of each rod = 0.75 m
Diameter of each rod = 1.50 cm = 0.015 m
Tensile force exerted = 4000 N
(a) Strain is given as the ratio of change in length to the original length of a body. Mathematically, it is given as
Strain = 
where Y = Young modulus
F = Fore applied
A = Cross sectional area
For the steel rod:
Y = 200 000 000 000 
F = 4000N
A = 
(r = d/2 = 0.015/2 = 0.0075 m)
=> A = 
=> A = 0.000177 
∴ 
For the copper rod:
Y = 120 000 000 000 N/m²
F = 4000N
A = (r = d/2 = 0.015/2 = 0.0075 m)
=> A =
=> A = 0.000177
(b) We can find the elongation by multiplying the Strain by the original length of the rods:
Elongation = Strain * Length
For the steel rod:
Elongation = 
For the copper rod:
Elongation = 
Answer:
b ≈ 64 Kg/s
Explanation:
Given
Fd = −bv
m = 2.5 kg
y = 6.0 cm = 0.06 m
g = 9.81 m/s²
The object in the pan comes to rest in the minimum time without overshoot. this means that damping is critical (b² = 4*k*m).
m is given and we find k from the equilibrium extension of 6.0 cm (0.06 m):
∑Fy = 0 (↑)
k*y - W = 0 ⇒ k*y - m*g = 0 ⇒ k = m*g / y
⇒ k = (2.5 kg)*(9.81 m/s²) / (0.06 m)
⇒ k = 408.75 N/m
Hence, if
b² = 4*k*m ⇒ b = √(4*k*m) = 2*√(k*m)
⇒ b = 2*√(k*m) = 2*√(408.75 N/m*2.5 kg)
⇒ b = 63.9335 Kg/s ≈ 64 Kg/s
