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.
Answer:
metal
Explanation:
metal is an excellent conductor for heat and electricity therefore making it perfect to experiment with different temperatures.
The kinetic energy K = 0.5 * m * v² must be equal to the potential energy U = m * g * h.
m mass
v velocity
h height
g = 9.81m/s²
The mass m cancels out:
0.5 * v² = g * h
Solve for height h and transform to distance traveled.
(sin (4°) = height / distance)
Answer:
minimum length of a surface crack is 18.3 mm
Explanation:
Given data
plane strain fracture toughness K = 82.4 MPa m1/2
stress σ = 345 MPa
Y = 1
to find out
the minimum length of a surface crack
solution
we will calculate length by this formula
length = 1/π ( K / σ Y)²
put all value
length = 1/π ( K / σ Y)²
length = 1/π ( 82.4 
/ 345× 1)²
length = 18.3 mm
minimum length of a surface crack is 18.3 mm
Answer:
22Volts
Explanation:
The pd at the terminal is known as the emf
Since there are Ten 2.2V cells
Terminal voltage = number of cells * pd of one cell
Terminal voltage = 10 * 2.2
Terminal voltage = 22V
Hence the pd at the battery terminals is 22Volts
