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.
Series,effective resistance =R₁+R₂+R₃...
parallel,effective resistance 1/R=1/R₁ +1/R₂ +1/R₃...
Here,effective resistance 1/R =1/7 +1/5+ 1/4+1/2
=1.092
R = 1/1.092 =0.915Ω
voltage V=9 V
current I=V/R
I=9 / 0.915
=9.83 A
Explanation:
Kepler’s third law states that for all objects orbiting a given body, the cube of the semimajor axis (A) is proportional to the square of the orbital period (P).
For each of our planets orbiting the Sun, the relationship between the orbital period and semimajor axis can be represented by the equation as:
k is constant of proportionality
It is required to solve the above equation for k
Since the stone was dropped from height, its initial velocity = 0 m/s
Using v² = u² + 2gs.
Where g ≈ 10 m/s², u = initial velocity = 0 m/s, s = height from drop = 2.5 m
v² = u² + 2gs
v² = 0² + 2*10*2.5
v² = 0 + 50
v² = 50
v = √50
v ≈ 7.07 m/s
Hence velocity just before hitting the ground is ≈ 7.07 m/s
<h2>Question:</h2>
An automobile is driving uphill. Which form of energy is not involved in this process?
<h2>Choosing:</h2>
electromagnetic
potential
kinetic
chemical
<h2>Answer:</h2>
<u>Electromagnetic</u><u> </u>
<h3>
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<u>#CARRYONLEARNING</u><u> </u></h3><h3>
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