Answer:W = 1.23×10^-6BTU
Explanation: Work = Surface tension × (A1 - A2)
W= Surface tension × 3.142 ×(D1^2 - D2^2)
Where A1= Initial surface area
A2= final surface area
Given:
D1=0.5 inches , D2= 3 inches
D1= 0.5 × (1ft/12inches)
D1= 0.0417 ft
D2= 3 ×(1ft/12inches)
D2= 0.25ft
Surface tension = 0.005lb ft^-1
W = [(0.25)^2 - (0.0417)^2]
W = 954 ×10^6lbf ft × ( 1BTU/778lbf ft)
W = 1.23×10^-6BTU
Answer:
Explanation:
Let the equilibrium position of third charge be x distance from q₁.
Force on third charge due to q₁
= 9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x²
Force on third charge due to q₂
= 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²
Both the force will act in opposite direction and for balancing , they should be equal.
9 x 10⁹ x 5 x 10⁻⁹ x 15 x 10⁺⁹ / x² = 9 x 10⁹ x 2 x 10⁻⁹ x 15 x 10⁺⁹ /( .40-x)²
5 / x² = 2 / ( .4 - x )²
Taking square root on both sides
2.236 / x = 1.414 / .4 - x
2.236 ( .4 - x ) = 1.414 x
.8944 - 2.236 x = 1.414 x
.8944 = 3.65 x
x = .245 m
24.5 cm
So the third charge should be at a distance of 24.5 cm from q₁ .
Answer:
Explanation:
The formula to determine the size of a capillary tube is
h = 2•T•Cos θ / r•ρ•g
Where
h = height of liquid level
T = surface tension
r = radius of capillary tube
ρ = density of liquid
θ = angle of contact = 0°
g =acceleration due to gravity=9.81m/s²
The liquid is water then,
ρ = 1000 kg / m³
Given that,
T = 0.0735 N/m
h = 0.25mm = 0.25 × 10^-3m
Then,
r = 2•T•Cos θ / h•ρ•g
r = 2 × 0.0735 × Cos0 / 2.5 × 10^-3 × 1000 × 9.81
r = 5.99 × 10^-3m
Then, r ≈ 6mm
The radius of the capillary tube is 6mm
So, the minimum size is
Volume = πr²h
Volume = π × 6² × 0.25
V = 2.83 mm³
The minimum size of the capillary tube is 2.83mm³
Answer:
3rd picture straight line going up right
Explanation:
3rd picture
