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>
(a) The work done by the applied force is 26.65 J.
(b) The work done by the normal force exerted by the table is 0.
(c) The work done by the force of gravity is 0.
(d) The work done by the net force on the block is 26.65 J.
<h3>
Work done by the applied force</h3>
W = Fdcosθ
W = 14 x 2.1 x cos25
W = 26.65 J
<h3>
Work done by the normal force</h3>
W = Fₙd
W = mg cosθ x d
W = (2.5 x 9.8) x cos(90) x 2.1
W = 0 J
<h3>Work done force of gravity</h3>
The work done by force of gravity is also zero, since the weight is at 90⁰ to the displacement.
<h3> Work done by the net force on the block</h3>
∑W = 0 + 26.65 J = 26.65 J
Thus, the work done by the applied force is 26.65 J.
The work done by the normal force exerted by the table is 0.
The work done by the force of gravity is 0.
The work done by the net force on the block is 26.65 J.
Learn more about work done here: brainly.com/question/8119756
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B.) <span>The amp is the unit for "Current"
Hope this helps!</span>
The formula for the energy in a capacitor , u in terms of q and c is q²/2c
<h3>What is the energy of a capacitor?</h3>
The energy of a capacitor u = 1/2qv where
- q = charge on capacitor and
- v = voltage across capacitor.
<h3>What is the capacitance of a capacitor?</h3>
Also, the capacitance of a capacitor c = q/v where
- q = charge on capacitor and
- v = voltage across capacitor.
So, v = q/c
<h3>
The formula for energy of the capacitor in terms of q and c</h3>
Substituting v into u, we have
u = 1/2qv
= 1/2q(q/c)
= q²/2c
So, the formula for the energy in a capacitor , u in terms of q and c is q²/2c
Learn more about energy in a capacitor here:
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Answer:
How far will the electron travel beforehitting a plate is 248.125mm
Explanation:
Applying Gauss' law:
Electric Field E = Charge density/epsilon nought
Where charge density=1.0 x 10^-6C/m2 & epsilon nought= 8.85× 10^-12
Therefore E = 1.0 x 10^-6/8.85× 10^-12
E= 1.13×10^5N/C
Force on electron F=qE
Where q=charge of electron=1.6×10^-19C
Therefore F=1.6×10^-19×1.13×10^5
F=1.808×10^-14N
Acceleration on electron a = Force/Mass
Where Mass of electron = 9.10938356 × 10^-31
Therefore a= 1.808×10^-14 /9.11 × 10-31
a= 1.985×10^16m/s^2
Time spent between plate = Distance/Speed
From the question: Distance=1cm=0.01m and speed = 2×10^6m/s^2
Therefore Time = 0.01/2×10^6
Time =5×10^-9s
How far the electron would travel S =ut+ at^2/2 where u=0
S= 1.985×10^16×(5×10^-9)^2/2
S=24.8125×10^-2m
S=248.125mm
