Answer:
a) Q₂ = 150 cm³/s
b) Q₂ = 33.33 cm³/s
c) Q₂ =25.0 cm³/s
d) Q₂ = 0.01 cm³/s.
e) Q₂ = 0.03 cm³/s
Explanation:
a) Original flow rate, Q₁ = 100 cm³/s
Based on the Poiseulle's equation
ΔP = (128цLQ/πD⁴)
ΔP₂/ΔP₁ = 1.50
ΔP ∝ Q
ΔP₂/ΔP₁ = Q₂/Q₁
1.5 = Q₂/Q₁
Q₂ = 1.5 Q₁
Q₂ = 1.5 * 100
Q₂ = 150 cm³/s
b)ΔP = (128цLQ/πD⁴)
Q ∝1/ц
Q = k/ц
Q₁ = k/ц₁
Q₂ = k/ц₂
Q₂/Q₁ = ц₁/ц₂
ц₂/ц₁ = 3
ц₁/ц₂ =1/3
Q₂/100 = 1/3
Q₂ = 100/3
Q₂ = 33.33 cm³/s
c)Q ∝1/L
Q = k/L
Q₁ = k/L₁
Q₂ = k/L₂
Q₂/Q₁ = L₁/L₂
L₂/L₁ = 4
L₁/L₂ =1/4
Q₂/100 = 1/4
Q₂ = 100/4
Q₂ =25.0 cm³/s
d) Q ∝ D⁴
Q₂/Q₁ = (D₂/D₁)⁴
2R₂ = D₂
2R₁ = D₁
D₂/D₁ = R₂/R₁ = 0.1
Q₂/Q₁ = 0.1⁴
Q₂/Q₁ = 0.0001
Q₂ = 0.0001Q₁
Q₂ = 0.0001 * 100
Q₂ = 0.01 cm³/s.
e) Q ∝ D⁴∝ΔP∝1/L
Q = k D⁴ΔP/L
Q₂/Q₁ = (D₂/D₁)⁴(ΔP₂/ΔP₁)(L₁/L₂)
D₂/D₁ = R₂/R₁ = 0.1
L₂/L₁ = 1/2
L₁/L₂ = 2
ΔP₂/ΔP₁ = 1.5
Q₂/Q₁ = (0.1)⁴*(1.5)*(2)
Q₂/Q₁ = 0.0003
Q₂ = 0.0003 * 100
Q₂ = 0.03 cm³/s
Answer:
they're divided in two ways extensive/intensive or physical/chemical
intensive properties do not depend on the amount of matter this includes (boiling point and color)
extensive properties depend on the amount of matter that is being measured (mass, and volume)
Answer:
0.0031 m
Explanation:
y = Length of pixel = 281 μm
L = Distance to screen = 1.3 m
= Wavelength = 550 nm
d = Pupil diameter
= Angle
We have the expression
We have the expression
The pupil diameter is 0.0031 m
Answer:
Answered
Explanation:
v= 1 m/s
A= 1 m^2
m= 100 kg
y= 1 mm
μ = ?
ζ= viscosity of SAE 20 crankcase oil of 15° C= 0.3075 N sec/m^2
forces acting on the block are
F_s ← ↓ →F_f
mg
N= mg
F_s= shear force = ζAv/y F_f= friction force = μN
now in x- direction F_s= F_f
ζAv/y = μN
0.3075×1×1×1/1×10^{-3} = μ×100
⇒μ=0.313 (coefficient of sliding friction for the block)
Now, as the velocity is increased shear force also increases and due to this frictional force also increases.
Now, to compensate this frictional force friction coefficient must increase
as v∝μ