A particle of mass m moves in the xy plane with a velocity of v = vxî + vyĵ. Determine the angular momentum of the particle abou

In a demonstration, a 4.00 cm2 square coil with 10 000 turns enters a larger square region with a uniform 1.50 T magnetic field

serg [3582]

Explanation:

It is stated that,

Area of square coil, $A=4\ cm^2=0.0004\ m^2$

Length of one side of the square, L = 0.02 m

Number of coils, N = 10000

Consistent magnetic field, B = 1.5 T

Velocity, v = 100 m/s

An electromotive force is produced in the coil calculated as:

$E=NBLv$

$E=10000\times 1.5\times 0.02\times 100$

E = 30000 V

Breakdown voltage of air, $V=4000\ V/cm=400000\ V/m$

Let d be the distance between the ends of the coil wires that can still produce a spark. Therefore,

Electric field, $E'=\dfrac{V}{d}$

$\dfrac{30000}{d}=400000$

d = 0.075 m

Thus, this forms the final answer.

6 0

1 month ago

An advertiser who calls a basement apartment a “garden apartment” is using the rhetorical device called

serg [3582]

I would choose B or D, based on the location. If it’s situated in South Kensington, London, then D might be appropriate. Conversely, if it’s in an underprivileged area, I'd opt for B.

3 0

1 month ago

A 1.97-pF capacitor with a plate area of 5.86 cm2 and separation between the plates of 2.63 mm is connected to a 9.0-V battery a

inna [3103]

If the plate separation is modified after the battery is disconnected, the updated distance between plates is 9.21 mm

If changes are made while the battery remains connected, the new separation becomes 0.11 mm

The capacitance for an air-filled parallel plate capacitor can be expressed as:

$C=\frac{\epsilon_0A}{d}$

In this equation, $\epsilon_0$ refers to the permittivity of free space, A stands for the plate area, and D represents the separation distance.

Thus,

$C \alpha \frac{1}{d}$ .......(1)

Therefore, should the distance between the plates shift from d₁ to d₂, the capacitance ratio in both scenarios can be represented as:

$\frac{C_1}{C_2} =\frac{d_2}{d_1}$ ......(2)

Scenario (i)

When the capacitor is fully charged and then disconnected from the battery before adjusting the plate distance, the charge will remain steady while the capacitance varies.

The initial energy E₁ stored in the capacitor can be expressed as:

$E_1=\frac{Q^2}{2C_1}$ ......(3)

Once the separation changes to d₂, capacitance becomes C₂, but the charge Q remains unchanged.

Thus,

$E_2=\frac{Q^2}{2C_2}$ ......(4)

By dividing equation (4) by (3),

$\frac{E_2}{E_1} =\frac{C_1}{C_2}$

According to equation (2),

$\frac{E_2}{E_1} =\frac{C_1}{C_2}=\frac{d_2}{d_1}$

This results in a 3.5 fold increase in energy.

$\frac{E_2}{E_1} =\frac{d_2}{d_1}=3.5\\ d_2=3.5*2.63 mm\\ =9.205 mm=9.21 mm$

Scenario (2)

If the capacitor is kept connected to the power source, the voltage V across the plates will remain unchanged.

The initial energy is described as

$E_1=\frac{1}{2} C_1V^2$ ......(5)

The final energy when the plate separation transitions to d₂ can be written as:

$E_2=\frac{1}{2} C_2V^2$ .....(6)

Referencing equations (5) and (6)

$\frac{E_2}{E_1} =\frac{C_2}{C_1}$

From equation (2),

$\frac{E_2}{E_1} =\frac{C_2}{C_1}=\frac{d_1}{d_2}$

Thus, in this particular scenario,

$\frac{E_2}{E_1} =\frac{d_1}{d_2}\\d_2=\frac{d_1}{3.5} \\ =\frac{2.63 mm}{3.5} \\ =0.109 mm=0.11 mm$

Therefore,

Adjusting plate separation after battery disconnection yields 9.21 mm

If modified while connected, the new separation measures 0.11 mm

6 0

2 months ago

Four students measured the acceleration of gravity. The accepted value for their location is 9.78m/s2. Which student’s measureme

Ostrovityanka [3204]

Upon comparison, it is evident that the student with the highest percent error is A. Student 4, who measured 9.61 m/s². Four students recorded the acceleration of gravity, with the accepted local value being 9.78 m/s². Now let's find out which student's measurement exhibited the greatest percent error.

4 0

2 months ago

Most calculators operate on 6.0 V. If, instead of using batteries, you obtain 6.0 V from a transformer plugged into 110-V house

kicyunya [3294]

1/0.0545. The transformation ratio of primary coil turns to secondary coil turns is directly proportional to the voltage transformation occurring. With 6.0 V on the secondary side (output) and 110 V on the primary side (input), the voltage ratio is calculated as 6/110 = 0.0545. This means for each turn in the primary coil, there are 0.0545 turns in the secondary coil.

6 0

2 months ago