Answer:
a) 23.551 hp
b) 516.89 hp
Explanation:
<u>given:</u>
<u>required:</u>
the power in hp
<u>solution:</u>
.............(1)
by substituting in the equation (1)
=353.27 lbf
..........(2)
by substituting in the equation (2)
= 2769.29 lbf
power is defined by
P=F.V
353.27*36.67
=12954.411 lbf.ft/s
=12954.411*.001818
=23.551 hp
2769.29*102.67
= 284323 lbf.ft/s
= 284323*.001818
= 516.89 hp
Answer:
a) V =10¹¹*(1.5q₁ + 3q₂)
b) U = 1.34*10¹¹q₁q₂
Explanation:
Given
x₁ = 6 cm
y₁ = 0 cm
x₂ = 0 cm
y₂ = 3 cm
q₁ = unknown value in Coulomb
q₂ = unknown value in Coulomb
A) V₁ = Kq₁/r₁
where r₁ = √((6-0)²+(0-0)²)cm = 6 cm = 0.06 m
V₁ = 9*10⁹q₁/(0.06) = 1.5*10¹¹q₁
V₂ = Kq₂/r₂
where r₂ = √((0-0)²+(3-0)²)cm = 3 cm = 0.03 m
V₂ = 9*10⁹q₂/(0.03) = 3*10¹¹q₂
The electric potential due to the two charges at the origin is
V = ∑Vi = V₁ + V₂ = 1.5*10¹¹q₁ + 3*10¹¹q₂ = 10¹¹*(1.5q₁ + 3q₂)
B) The electric potential energy associated with the system, relative to their infinite initial positions, can be obtained as follows
U = Kq₁q₂/r₁₂
where
r₁₂ = √((0-6)²+(3-0)²)cm = √45 cm = 3√5 cm = (3√5/100) m
then
U = 9*10⁹q₁q₂/(3√5/100)
⇒ U = 1.34*10¹¹q₁q₂
Answer: OHMMETER & MEGOHMMETER:
Explanation: The ohmmeter measures circuit resistance; the megohmmeter measures the high resistance of insulation. A meter used to measure electric current. It is connected as part of a circuit.
Answer:
A. smallest wire is No. 12
