components of the speed of the coin is given as
now the time taken by the coin to reach the plate is given by
now in order to find the height
so it is placed at 1.52 m height
applied forces would be push for example.
normal forces would seem to be a force such as gravity.
friction for example when you try to slide on carpet but the fabric or whatever its made of stops you.
Well, first of all, there's no such thing as "fully charged" for a capacitor.
A capacitor has a "maximum working voltage", because of mechanical
or chemical reasons, just like a car has a maximum safe speed. But
anywhere below that, cars and capacitors do their jobs just fine, without
any risk of failing.
So we have a capacitor that has some charge on it, and therefore some
voltage across it. From the list of choices above . . .
<span>-- Both plates have the same amount of charge.
Yes. And both plates have opposite TYPES of charge.
One plate is loaded with electrons and is negatively charged.
The other plate is missing electrons and is positively charged.
-- There is a potential difference between the plates.
Yes. That's the "voltage" mentioned earlier.
It's a measure of how badly the extra electrons want to jump
from the negative plate to the positive plate.
-- Electric potential energy is stored.
Yes. It's the energy that had to be put into the capacitor
to move electrons away from one plate and cram them
onto the other plate.
</span>
Answer:
A) 26V
Explanation:
(a) the potential difference between the plates
Initial capacitance can be calculated using below expresion
C1= A ε0/ d1
Where d1= distance between = 2.70 mm= 2.70× 10^-3 m
ε0= permittivity of space= 8.85× 10^-12 Fm^-1
A= area of the plate = 7.90 cm2 = 7.90 ×10^-4 m^2
If we substitute the values we
C1= A ε0/ d1
=( 7.90 ×10^-4 × 8.85× 10^-12 )/2.70× 10^-3
C1=2.589 ×10^-12 F= 2.59 pF
Initial charge can be determined using below expresion
q1= C1 × V1
V1=2.589 ×10^-12 F
V1= voltage=7.90 V
If we substitute we have
q1= 2.589 ×10^-12 × 7.90
q1= 20.45×10^-12C
20.45 pC
Final capacitance can be calculated as
C2= A ε0/ d2
d2=8.80 mm= /8.80× 10^-3
7.90 ×10^-4 × 8.85× 10^-12 )/8.80× 10^-3
C1=0.794 ×10^-12 F= 0.794 pF
Final charge= initial charge
q2=q1 (since the battery is disconnected)
q2=q1= 20.45 pC
Final potential difference
V2= q/C2
= 20.45/0.794
= 26V
Answer:
Ф = 28.9°
Explanation:
given:
radius (r) = 117m
velocity (v) = 25.1 m/s
required: angle Ф
Ф = inv tan (v² / (r * g)) we know that g = 9.8
Ф = inv tan (25.1² / (117 * 9.8))
Ф = 28.9°
