To solve this problem, we will apply the concepts related to Faraday's law that describes the behavior of the emf induced in the loop. Remember that this can be expressed as the product between the number of loops and the variation of the magnetic flux per unit of time. At the same time the magnetic flux through a loop of cross sectional area is,
Here,
= Angle between areal vector and magnetic field direction.
According to Faraday's law, induced emf in the loop is,
At time 
, Induced emf is,
Therefore the magnitude of the induced emf is 10.9V
sorry - late reply...just stumbled across tis...hope u can still use it :)
By the mirror equation: 1/di + 1/do = 1/f
<span>
</span>
<span>where di = distance to image = +12cm (+ for real image)</span>
and do = distance to object = +8cm
Substitute and solve for f, the focal length
<span><span>
1/12 + 1/8 = 1/f
</span><span>
1/f = (8 + 12) / 12 * 8 = 20/96
</span><span>
so f = 96/20 = 4.8 cm</span>
</span>
Answer:
<h2>
a) Q = 0.759µC</h2><h2>
b) E = 39.5µJ</h2>
Explanation:
a) The charge Q on the positive charge capacitor can be gotten using the formula Q = CV
C = capacitance of the capacitor (in Farads )
V = voltage (in volts) = 100V
C = ∈A/d
∈ = permittivity of free space = 8.85 × 10^-12 F/m
A = cross sectional area = 600 cm²
d= distance between the plates = 0.7cm
C = 8.85 × 10^-12 * 600/0.7
C = 7.59*10^-9Farads
Q = 7.59*10^-9 * 100
Q = 7.59*10^-7Coulombs
Q = 0.759*10^-6C
Q = 0.759µC
b) Energy stored in a capacitor is expressed as E = 1/2CV²
E = 1/2 * 7.59*10^-9 * 100²
E = 0.0000395Joules
E = 39.5*10^-6Joules
E = 39.5µJ
Answer:
1.73 seconds
Explanation:
The velocity the ball first hits the ground with is:
v² = v₀² + 2aΔx
v² = (0 m/s)² + 2 (-10 m/s²) (-20 m)
v = -20 m/s
The velocity it rebounds with is 3/4 of that in the opposite direction, or 15 m/s.
The time it takes to return to the ground is:
Δx = v₀ t + ½ at²
0 = (15 m/s) t + ½ (-10 m/s²) t²
0 = t (15 − 5t²)
t = √3
t ≈ 1.73 seconds
