Answer:
<u>The correct answer is 0.556 Watts</u>
Explanation:
The computer monitor uses 200 Watts of power in an hour, that is the standard measure.
If we want to know, how much energy the computer monitor uses in one second, we will have to divide both sides of the equation into 3,600.
1 hour = 60 minutes = 3,600 seconds (60 x 60)
Energy per second = 200/3600
Energy per second = 0.0556 Watts
Therefore to calculate how much energy is used in 10 seconds, we do this:
Energy per second x 10
<u>0.0556 x 10 = 0.556 Watts</u>
<u>The computer monitor uses 0.556 Watts in 10 seconds</u>
Answer:
good because piggy loves peppa pig :3
Explanation:
Answer:
The potential difference is the drop in voltage that occurs across a resistor as current flows through it in a circuit, potential difference or voltage(V) = current (I) *resistance (R), or to abbrevate V = I*R. In this case, I = 5amps and R = 10 ohms, so V = 5 * 10 = 50volts
Answer: 0.516 ft/s
Explanation:
Given
Length of ladder L=20 ft
The speed at which the ladder moving away is v=2 ft/s
after 1 sec, the ladder is 5 ft away from the wall
So, the other end of the ladder is at
Also, at any instant t
differentiate w.r.t.
Answer:
a) The current density ,J = 2.05×10^-5
b) The drift velocity Vd= 1.51×10^-15
Explanation:
The equation for the current density and drift velocity is given by:
J = i/A = (ne)×Vd
Where i= current
A = Are
Vd = drift velocity
e = charge ,q= 1.602 ×10^-19C
n = volume
Given: i = 5.8×10^-10A
Raduis,r = 3mm= 3.0×10^-3m
n = 8.49×10^28m^3
a) Current density, J =( 5.8×10^-10)/[3.142(3.0×10^-3)^2]
J = (5.8×10^-10) /(2.83×10^-5)
J = 2.05 ×10^-5
b) Drift velocity, Vd = J/ (ne)
Vd = (2.05×10^-5)/ (8.49×10^28)(1.602×10^-19)
Vd = (2.05×10^-5)/(1.36 ×10^10)
Vd = 1.51× 10^-5
