From Ohm's law . . . Resistance = (voltage) / (current)
Resistance = (120 volts) / (7.6 Amperes)
<em>Resistance = 15.8 Ω</em>
Answer:
Explanation:
V1 = 31 V
V2 = 168 V
charge on proton, q = e
(a)
The change in potential energy is given by
ΔU = q ( V2 - V1)
ΔU = e (v2 - V1)
(b)
ΔU = e (168 - 31)
ΔU = 137 eV
(c)
Work done = change in potential energy
W = 137 eV
W = 137 x 1.6 x 10^-19
W = 2.19 x 10^-17 J
A particle has centripetal acceleration whenever it's a making a turn of radius R. If the particle is moving at a constant tangential speed v throughout the turn, then the magnitude of centripetal acceleration is
v²/R
If the particle is following a uniformly circular path, then it moves in a circle of radius R and travels a distance equal to its circumference, 2πR. Let T be the time it takes to complete one such loop. Then the entire circle is traversed with speed v = 2πR/T, so that the centripetal acceleration is also given by
v²/R = (2πR/T)²/R = 4π²R/T²
<span>The distance between the sources of a two point surface
interference determines the number of lines that can be created for the nodal
and anti-nodal lines. When the sources are moved closer together, the number of
lines becomes minimal. Thus, the spaces between the lines become bigger. The
answer to this item is letter A. The spacing increases. </span>
Answer:
0.15A
Explanation:
The parameters given are;
R=20.0 Ω
C= 2.50 μF
V= 3.00 V
f= 2.48×10^-3 Hz
Xc= 1/2πFc
Xc= 1/2×3.142 × 2.48×10^-3 × 2.5 ×10^-6
Xc= 25666824.1
Z= 1/√(1/R)^2 +(1/Xc)^2
Z= 1/√[(1/20)^2 +(1/25666824.1)^2]
Z= 1/√(2.5×10^-3) + (1.5×10^-15)
Z= 20 Ω
But
V=IZ
Where;
V= voltage
I= current
Z= impedance
I= V/Z
I= 3.00/20
I= 0.15A