Answer:
The change in the charge on the positve plate when the Teflon is inserted is +2.5 nC.
Explanation:
- It can be showed that the capacitance of a parallel-plate capacitor, can be expresssed as follows:
- Where ε, is the dielectric constant of the material that fills the space between plates.
- When this space is filled with air, ε= ε₀ = 8,85*10⁻¹² F/m.
- At the same time, the capacitance of a capacitor, by definition, is as follows:
- If we insert a Teflon slab, in such a way that fills completely the gap between the plates, all other parameters being equal, if ε = 2*ε₀, this means that C₂ = 2* C₁. = 50 pF
- As V₂=V₁ (due to the capacitor remains connected to the same battery) the charge must be the double, so Q₂ = 2* Q₁ = 5 nC.
- So, the change in the charge of the positive plate is +2.5 nC.
Velocity =displacement
Change in time
D=72km/hr
Time=20s
But the S.I unit of velocity is m/s so you woul have to change 72km/hr to m/s
Changing 72km to m
1 kilometer=1000meters
Then, 72 kilometers =?
72•1000/1
=72000m
Changing 72hours to seconds
If 1 hour = 3600 seconds
Then 72 hours=?
72•3600/1
=259200 seconds
Velocity =displacement
Change in time
V= 72,000
259,2005
=0.028m/s
Answer:
f = 1.69*10^5 Hz
Explanation:
In order to calculate the frequency of the sinusoidal voltage, you use the following formula:
(1)
V_L: voltage = 12.0V
i: current = 2.40mA = 2.40*10^-3 A
L: inductance = 4.70mH = 4.70*10^-3 H
f: frequency = ?
you solve the equation (1) for f and replace the values of the other parameters:
The frequency of the sinusoidal voltage is f
I can't make sense of this question. Julie's throwing the ball, so it's leaving her rather than arriving at her ???
Answer:
v₂ = 70 m / s
Explanation:
For this exercise let's use Bernoulli's equation
where subscript 1 is for the top of the mountain and subscript 2 is for Tuesday's level
P₁ + ½ ρ v₁² + ρ g y₁ = P₂ +1/2 ρ v₂² + ρ g y₂
indicate that the pressure in the two points is the same, y₁ = 250 m, y₂ = 0 m, the water in the upper part, because it is a reservoir, is very large for which the velocity is very small, we will approximate it to 0 (v₁ = 0), we substitute
ρ g y₁ = ½ ρ v₂²
v₂ = 
let's calculate
v₂ = √( 2 9.8 250)
v₂ = 70 m / s
