(a) The time for the capacitor to loose half its charge is 2.2 ms.
(b) The time for the capacitor to loose half its energy is 1.59 ms.
<h3>
Time taken to loose half of its charge</h3>
q(t) = q₀e-^(t/RC)
q(t)/q₀ = e-^(t/RC)
0.5q₀/q₀ = e-^(t/RC)
0.5 = e-^(t/RC)
1/2 = e-^(t/RC)
t/RC = ln(2)
t = RC x ln(2)
t = (12 x 10⁻⁶ x 265) x ln(2)
t = 2.2 x 10⁻³ s
t = 2.2 ms
<h3>
Time taken to loose half of its stored energy</h3>
U(t) = Ue-^(t/RC)
U = ¹/₂Q²/C
(Ue-^(t/RC))²/2C = Q₀²/2Ce
e^(2t/RC) = e
2t/RC = 1
t = RC/2
t = (265 x 12 x 10⁻⁶)/2
t = 1.59 x 10⁻³ s
t = 1.59 ms
Thus, the time for the capacitor to loose half its charge is 2.2 ms and the time for the capacitor to loose half its energy is 1.59 ms.
Learn more about energy stored in capacitor here: brainly.com/question/14811408
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Formula:
F = ma
F: force (N) m: mass (kg) a: acceleration (m/s^2)
Solution:
F = ma
F = 1200 × 1.89
= 2268N
Answer:
a) E = 1.47 × 10^5 N/C
b) south
Explanation:
The magnitude of an electric field can be defined mathematically as;
E = F/q ........1
Where,
E = magnitude of the electric field
F = electric force
q = charge on the proton
Given;
F = 2.36 × 10^-14 N
Note that charge on a proton is known as Qp = 1.602 × 10^-19 C
q = 1.602 × 10^-19 C
Substituting into equation 1, we have;
E = 2.36 × 10^-14 N/1.602 × 10^-19 C
E = 1.47 × 10^5 N/C
b) The direction of the electric field;
From equation 1
E = F/q ........1
since both electric field and electric force are vector quantity and q is a positive charge (constant), then both the electric field and electric force would be parallel to each other. Therefore the electric field is directed to the south also.
(When a vector is multiplied by a positive constant the direction remains the same)
Answer:
a) 
b) 
c) 
d) No,
No.
Explanation:
Given:
length of rope, 
weight of the rope, 
mass suspended at the lower end of the rope, 
<u>Now the mass of the rope:</u>
<u>So the linear mass density of rope:</u>
We know that the speed of wave in a tensed rope is given as:
where:
tension force in the rope
a)
At the bottom of the hanging rope we have an extra mass suspended. So the tension at the bottom of the rope:
Therefore the speed of the wave at the bottom point of the rope:
b)
Tension at a point in the middle of the rope:
Now wave speed at this point:
c)
Tension at a point in the top of the rope:
Now wave speed at this point:
d)
Tension at the middle of the rope is not the average tension of tension at the top and bottom of the rope because we have an extra mass attached at the bottom end of the rope.
Also the wave speed at the mid of the rope is not the average f the speeds at the top and the bottom of the ropes because it depends upon the tension of the rope at the concerned points.
The gravitational potential energy becomes work driving the pipe
<span>m g h = f d </span>
<span>450 * g * 1.5 = f * .45</span>
