Answer:
Explanation:
Let the resistance of resistor be R .
Power of resistor V² / R , where V is potential applied .
V² / R = 100
120² / R = 100
R = 120² / 100
= 144 ohm .
Now potential diff applied = 220 V
current = potential diff / resistance
= 220 / 144
= 1.53 A approx .
Answer:
1.) 11 km/s
2.) 9.03 × 10^-5 metres
Explanation:
Given that an electron enters a region of uniform electric field with an initial velocity of 64 km/s in the same direction as the electric field, which has magnitude E = 48 N/C.
Electron q = 1.6×10^-19 C
Electron mass = 9.11×10^-31 Kg
(a) What is the speed of the electron 1.3 ns after entering this region?
E = F/q
F = Eq
Ma = Eq
M × V/t = Eq
Substitute all the parameters into the formula
9.11×10^-31 × V/1.3×10^-9 = 48 × 1.6×10^-19
V = 7.68×10^-18 /7.0×10^-22
V = 10971.43 m/s
V = 11 Km/s approximately
(b) How far does the electron travel during the 1.3 ns interval?
The initial velocity U = 64 km/s
S = ut + 1/2at^2
S = 64000×1.3×10^-6 + 1/2 × 8.4×10^12 × ( 1.3×10^-9)^2
S =8.32×10^-5 + 7.13×10^-6
S = 9.03 × 10^-5 metres
Answer:
V = 6.36 m³
Explanation:
For this exercise we will use fluid mechanics relations, starting with the continuity equation.
Let's write the flow equation
Q = v₁ A₁
The area of a circle is
A = π r²
Radius is half the diameter
A = π/4 d²
Q = v₁ π/4 d₁²
Q = π/ 4 15 0.03 2
Q = 0.0106 m3 / s
The volume of water in t = 10 min = 10 60 = 600 s
Q = V / t
V = Q t
V = 0.0106 600
V = 6.36 m³
