Answer:
A. 5.4 * 10^(-4) m
B. 500V
Explanation:
A. Electric potential, V is given as:
V = kq/r
This means that radius, r is
r = kq/V
r = (9 * 10^9 * 30 * 10^(-12))/500
r = (270 * 10^(-3))/500
r = 5.4 * 10^(-4) m
B. Now the radius is doubled and the charge is doubled,
V = (9 * 10^9 * 2 * 30 * 10^(-12))/(2 * 5.4 * 10^(-4) * 2)
V = 500V
Answer:
18 times
Explanation:
According to the security purposes which is set under the rules and regulation OSHA, which describes all the rights to the worker.
In the boom hoist receiving system all the sheaves which are used should have a pitch diameter of rope not less than 18 times the diameter of the nominal rope which is used.
Answer:
X= 700 Joules
Explanation:
The question asked about the efficiency of the work done.
The formula for efficiency is: Efficiency = (Useful output / input work) * 100%
The useful output given in the question is 140J, the question asked for input work. Let X be the input work. It is also given that the efficiency is 20%.
Using the formula of efficiency,
20 = (140/X) * 100
So, we simply solve the above equation.
X= 140*100/20
X= 700 Joules
Answer:
R₂ / R₁ = D / L
Explanation:
The resistance of a metal is
R = ρ L / A
Where ρ is the resistivity of aluminum, L is the length of the resistance and A its cross section
We apply this formal to both configurations
Small face measurements (W W)
The length is
L = W
Area
A = W W = W²
R₁ = ρ W / W² = ρ / W
Large face measurements (D L)
Length L = D= 2W
Area A = W L
R₂ = ρ D / WL = ρ 2W / W L = 2 ρ/L
The relationship is
R₂ / R₁ = 2W²/L
