Answer:
The maximum energy that can be stored in the capacitor is 6.62 x 10⁻⁵ J
Explanation:
Given that,
dielectric constant k = 5.5
the area of each plate, A = 0.034 m²
separating distance, d = 2.0 mm = 2 x 10⁻³ m
magnitude of the electric field = 200 kN/C
Capacitance of the capacitor is calculated as follows;
Maximum potential difference:
V = E x d
V = 200000 x 2 x 10⁻³ = 400 V
Maximum energy that can be stored in the capacitor:
E = ¹/₂CV²
E = ¹/₂ x 8.275 x 10⁻¹⁰ x (400)²
E = 6.62 x 10⁻⁵ J
Therefore, the maximum energy that can be stored in the capacitor is 6.62 x 10⁻⁵ J
Answer:
When there is nearsightedness or myopia
Explanation:
As, in myopia the image is formed in front of the retina.
Which makes things looking at things near very easy, but looking at far away things very difficult.
So concave lens makes things look bigger, so therefore, it is used during myopia, to make things look bigger when they are far away
Answer:
a) v = √(v₀² + 2g h), b) Δt = 2 v₀ / g
Explanation:
For this exercise we will use the mathematical expressions, where the directional towards at is considered positive.
The velocity of each ball is
ball 1. thrown upwards vo is positive
v² = v₀² - 2 g (y-y₀)
in this case the height y is zero and the height i = h
v = √(v₀² + 2g h)
ball 2 thrown down, in this case vo is negative
v = √(v₀² + 2g h)
The times to get to the ground
ball 1
v = v₀ - g t₁
t₁ = 
ball 2
v = -v₀ - g t₂
t₂ = - \frac{v_{o} + v }{ g}
From the previous part, we saw that the speeds of the two balls are the same when reaching the ground, so the time difference is
Δt = t₂ -t₁
Δt = ![\frac{1}{g} \ [(v_{o} - v) - ( - v_{o} - v) ]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bg%7D%20%5C%20%5B%28v_%7Bo%7D%20-%20v%29%20%20-%20%28%20-%20v_%7Bo%7D%20%20-%20v%29%20%5D)
Δt = 2 v₀ / g
Answer:
The direction a wave propagates is perpendicular to the direction it oscillates for transverse waves. A wave does not move mass in the direction of propagation; it transfers energy.
Explanation:
Answer:
I_v = 2,700 W / m^2
I_m = 610 W / m^2
I_s = 16 W / m^2
Explanation:
Given:
- The Power of EM waves emitted by Sun P_s = 4.0*10^26 W
- Radius of Venus r_v = 1.08 * 10^11 m
- Radius of Mars r_m = 2.28 * 10^11 m
- Radius of Saturn r_s = 1.43 * 10^12 m
Find:
Determine the intensity of electromagnetic waves from the sun just outside the atmospheres of (a) Venus, (b) Mars, and (c) Saturn.
Solution:
- We know that Power is related to intensity and surface area of an object follows:
I = P / 4*pi*r^2
Where, A is the surface area of a sphere models the atmosphere around the planets.
a)
- The intensity at the surface of Venus is calculated as:
I_v = P_s / 4*pi*r^2_v
I_v = 4.0*10^26 / 4*pi*(1.08*10^11)^2
I_v = 2,700 W / m^2
b)
- The intensity at the surface of Mars is calculated as:
I_m = P_s / 4*pi*r^2_m
I_m = 4.0*10^26 / 4*pi*(2.28*10^11)^2
I_m = 610 W / m^2
c)
- The intensity at the surface of Saturn is calculated as:
I_s = P_s / 4*pi*r^2_s
I_s = 4.0*10^26 / 4*pi*(1.43*10^12)^2
I_s = 16 W / m^2
