Help for physics please!
1 answer:
You might be interested in
Answer:
a) E = 4.5*10⁴ V/m
b) C= 17.7 nF
c) Q = 159. 3 nC
Explanation:
a)
- By definition, the electric field is the electrostatic force per unit charge, and since the potential difference between plates is just the work done by the field, divided by the charge, assuming a uniform electric field, if V is the potential difference between plates, and d is the separation between plates, the electric field can be expressed as follows:
b)
- For a parallel-plate capacitor, applying the definition of capacitance as the quotient between the charge on one of the plates and the potential difference between them, and assuming a uniform surface charge density σ, we get:
From (1), we know that V = E*d, but at the same time, applying Gauss'
Law at a closed surface half within the plate, half outside it , it can be
showed than E= σ/ε₀, so finally we get:
c)
- From (3) we can solve for Q as follows:
- Displacement = 10 m
- Time = 5 s
- We know,
- Therefore, the car's velocity
The car's velocity is 2 m/s.
Hope you could get an idea from here.
Doubt clarification - use comment section.
Answer:
The force applied on one wheel during braking = 6.8 lb
Explanation:
Area of the piston (A) = 0.4
Force applied on the piston(F) = 6.4 lb
Pressure on the piston (P) =
⇒ P =
⇒ P = 16
This is the pressure inside the cylinder.
Let force applied on the brake pad =
Area of the brake pad ()= 1.7
Thus the pressure on the brake pad () =
When brake is applied on the vehicle the pressure on the piston is equal to pressure on the brake pad.
⇒ P =
⇒ 16 =
⇒ = 16 ×
Put the value of we get
⇒ = 16 × 1.7
⇒ = 27.2 lb
This the total force applied during braking.
The force applied on one wheel = =
= 6.8 lb
⇒ The force applied on one wheel during braking.