A 4.0-mF capacitor initially charged to 50 V and a 6.0-mF capacitor charged to 30 V are connected to each other with the positiv
2 answers:
Answer:
The 6.0 mF capacitor's final charge will be 12 mC.
Explanation:
The charge on the 4 mF capacitor is calculated as:
4 mF × 50 V = 200 mC
The 6 mF capacitor starts with:
6 mF × 30 V = 180 mC
The total charge when the negative terminals are combined is given by:
q =
q = 200 - 180
q = 20 mC
To determine the final charge on the 6.0 mF capacitor, we need to calculate the combined voltage:
q = (4 x V) + (6 x V)
20 = 10 V
V = 2 V
Now for the final charge on the 6.0 mF:
q = CV
q = 6.0 mF × 2 V
q = 12 mC
Thus, the final charge on the 6.0 mF capacitor will be 12 mC.
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Response:
Details:
Utilizing Faraday's Newmann Lenz law enables the assessment of the induced emf within the loop:
where:
represents the change in magnetic flux
symbolizes the change over time.
#The magnetic flux linked to the coil can be represented as:
Where:
N represents the number of loops.
A denotes the area for each loop ().
B indicates the strength of the magnetic field.
represents the angle between the magnetic field direction and the normal to the loop's area.
=0.0250T/s is indicated as the rate of magnetic field increase.
#Plugging in values into the emf equation:
Thus, the induced emf is
Answer:
176.38 rpm
Explanation:
The proportion of mass for arms and legs is 13%.
For legs and trunk, it's 80% of the total mass.
Additionally, the head accounts for 7% of the total mass.
Overall, the mass of the skater is 74.0 kg.
Each arm length is 70 cm, or 0.7 m.
The skater's height is 1.8 m, and the trunk diameter measures 35 cm, or 0.35 m.
Starting angular momentum is 68 rpm.
We make the following assumptions:
- The skater is modeled as a vertical cylinder (head, trunk, and legs), with arms extending horizontally as two uniform rods.
- Friction between the skater and the ice is considered negligible.
To analyze her body, we divide it into two parts, treating the arms as spinning rods.
1. Each arm (rod) has a moment of inertia of .
The arms comprise 13% of 74 kg, calculated as 0.13 x 74 = 9.62 kg.
Each arm is then evaluated as 9.62/2 = 4.81 kg.
Let L represent the length of each arm.
Thus,
I = x 4.81 x
= 0.79 kg-m for each arm.
2. The body, treated as a cylinder, has a moment of inertia of .
For the body, the radius r is half of the trunk diameter: r = 0.35/2 = 0.175 m.
The mass of the trunk amounts to (80% + 7%) of 74 kg, which calculates to 0.87 x 74 = 64.38 kg.
Therefore, I = x 64.38 x
= 0.99 kg-m.
Two cases are considered:
case 1: Body spinning with arms extended.
Total moment of inertia equals the combined moments of inertia of both arms and the trunk.
I = (0.79 x 2) + 0.99 = 2.57 kg-m.
The angular momentum is given by Iω.
Here, ω = angular speed = 68.0 rpm = x 68 = 7.12 rad/s.
The angular momentum then becomes 2.57 x 7.12 = 18.29 kg-rad/m-s.
case 2: Arms drawn in alongside the trunk.
The moment of inertia is attributed solely to the trunk. This is 0.91 kg-m.
The angular momentum equals Iω.
= 0.99 x ω = 0.91ω.
By the principle of conservation of angular momentum, the two angular momentum quantities are equal. Therefore,
18.29 = 0.99ω.
Solving gives ω = 18.29/0.99 = 18.47 rad/s.
This leads to 18.47 ÷ = 176.38 rpm.