) A 1000-nF capacitor with circular parallel plates with a radius of 1cm is accumulating charge at the rate of 52 mC/s at some p
1 answer:
You might be interested in
Answer:
C) 7.35*10⁶ N/C radially outward
Explanation:
- If we apply the Gauss'law, to a spherical gaussian surface with radius r=7 cm, due to the symmetry, the electric field must be normal to the surface, and equal at all points along it.
- So, we can write the following equation:
- As the electric field must be zero inside the conducting spherical shell, this means that the charge enclosed by a spherical gaussian surface of a radius between 4 and 5 cm, must be zero too.
- So, the +8 μC charge of the solid conducting sphere of radius 2cm, must be compensated by an equal and opposite charge on the inner surface of the conducting shell of total charge -4 μC.
- So, on the outer surface of the shell there must be a charge that be the difference between them:
- Replacing in (1) A = 4*π*ε₀, and Qenc = +4 μC, we can find the value of E, as follows:
- As the charge that produces this electric field is positive, and the electric field has the same direction as the one taken by a positive test charge under the influence of this field, the direction of the field is radially outward, away from the positive charge.
Newton's third law of motion sates that force is directly proportional to the rate of change of momentum produced
(a) The final speeds of the ice sleds is approximately 0.49 m/s each
(b) The impulse on the cat is 11.0715 kg·m/s
(c) The average force on the right sled is 922.625 N
The reason for arriving at the above values is as follows:
The given parameters are;
The masses of the two ice sleds, m₁ = m₂ = 22.7 kg
The initial speed of the ice, v₁ = v₂ = 0
The mass of the cat, m₃ = 3.63 kg
The initial speed of the cat, v₃ = 0
The horizontal speed of the cat, v₃ = 3.05 m/s
(a) The required parameter:
The final speed of the two sleds
For the first jump to the right, we have;
By the law of conservation of momentum
Initial momentum = Final momentum
∴ m₁ × v₁ + m₃ × v₃ = m₁ × v₁' + m₃ × v₃'
Where;
v₁' = The final velocity of the ice sled on the left
v₃' = The final velocity of the cat
Plugging in the values gives;
22.7 kg × 0 + 3.63 × 0 = 22.7 × v₁' + 3.63 × 3.05
∴ 22.7 × v₁' = -3.63 × 3.05
v₁' = -3.63 × 3.05/22.7 ≈ -0.49
The final velocity of the ice sled on the left, v₁' ≈ -0.49 m/s (opposite to the direction to the motion of the cat)
The final speed ≈ 0.49 m/s
For the second jump to the left, we have;
By conservation of momentum law, m₂ × v₂ + m₃ × v₃ = m₂ × v₂' + m₃ × v₃'
Where;
v₂' = The final velocity of the ice sled on the right
v₃' = The final velocity of the cat
Plugging in the values gives;
22.7 kg × 0 + 3.63 × 0 = 22.7 × v₂' + 3.63 × 3.05
∴ 22.7 × v₂' = -3.63 × 3.05
v₂' = -3.63 × 3.05/22.7 ≈ -0.49
The final velocity of the ice sled on the right = -0.49 m/s (opposite to the direction to the motion of the cat)
The final speed ≈ 0.49 m/s
(b) The required parameter;
The impulse of the force
The impulse on the cat = Mass of the cat × Change in velocity
The change in velocity, Δv = Initial velocity - Final velocity
Where;
The initial velocity = The velocity of the cat before it lands = 3.05 m/s
The final velocity = The velocity of the cat after coming to rest =
∴ Δv = 3.05 m/s - 0 = 3.05 m/s
The impulse on the cat = 3.63 kg × 3.05 m/s = 11.0715 kg·m/s
(c) The required information
The average velocity
Impulse = × Δt
Where;
Δt = The time of collision = The time it takes the cat to finish landing = 12 ms
12 ms = 12/1000 s = 0.012 s
We get;
∴
The average force on the right sled applied by the cat while landing, = 922.625 N
Learn more about conservation of momentum here: