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2 years ago

Some students set up a circuit and decided to measure the voltage at different points around

Physics

1 answer:

Fiesta28 [93]2 years ago

3 0

Answer:

V₁ = 6 V , V₂ = V₃ = 3 V

Explanation:

To solve this circuit we must remember that there are two fundamental types of construction in series and parallel.

* a serial circuit there is only one path for current

in this circuit the constant current in the entire circuit and the voltage is the sum of the voltage of each term

* Parallel circuit in this there are two or more paths for the current

in this circuit the voltage is constant and the east is divided between each branch

with these principles let's analyze the proposed circuit

The DC battery is in parallel with resistor R1 and the equivalent of the other branch,

as in a parallel circuit the voltage is constant

V₁ = 6 V

in the other branch (23) it forms a series construction, where the current is constant

6 = iR₂ + iR₃

as they indicate that each resistance has the same value

6 = 2 iR

V = V₂ = V₃ = 3 V

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Answer:

1.09

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8 0

2 years ago

Example 7.3

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2 years ago

Suppose you wish to construct a motor that produces a maximum torque whose magnitude is 1.7 × 10-2 N·m. The coil of the motor ha

Len [333]

Answer:

86 turns

Explanation:

Parameters given:

Magnetic torque, τ = 1.7 * 10^(-2) Nm

Area of coil, A = 9 * 10^(-4) m²

Current in coil, I = 1.1 A

Magnetic field, B = 0.2 T

The magnetic toque is given mathematically as:

τ = N * I * A * B

Where N = number of turns

To find the number of turns, we make N subject of formula:

N = τ/(I * A * B)

Therefore:

N = (1.7 * 10^(-2)) / (1.1 * 9 * 10^(-4) * 0.2)

N = 85.85 = 86 turns (whole number)

The number of turns must be 86.

3 0

3 years ago

In a double slit experiment, the distance between the slits is 0.2 mm and the distance to the screen is 100 cm. What is the phas

Anni [7]

Answer:

The phase difference is $\Delta \phi = 180^o$

Explanation:

From the question we are told that

The distance between the slits is $d = 0.2 \ mm = \frac{0.2}{1000} = 0.2 *10^{-3} \ m$

The distance to the screen is $D = 100 cm = \frac{100}{100} = 1 \ m$

The wavelength is $\lambda = 400nm$

The distance of the wave from the central maximum is $L = 5mm = 5*10^{-3} m$

Generally the path difference of this waves is mathematically represented as

$y = d sin \theta$

Here $\theta$ is the angle between the the line connecting the mid-point of the slits with the screen and the line connecting the mid-point of the slits to the central maximum

This implies that

$tan \theta = \frac{L}{D}$

=> $\theta = tan ^{-1} \frac{L}{D}$

$\theta = tan ^{-1} [\frac{5*10^{-3}}{1}]$

$\theta =0.2865$

Substituting values into the formula for path difference

$y = 0.2 *10^{-3} sin(0.2864)$

$y = 9.997*10^{-7} \ m$

The phase difference is mathematically represented as

$\Delta \phi = \frac{2 \pi }{\lambda } * y$

Substituting values

$\Delta \phi = \frac{2 \pi }{400 *10^{-9} } \ * 9.997*10^{-7}$

$\Delta \phi =5 \pi$

Converting to degree

$\Delta \phi =5 \pi radians = 5 (180^o) = 180^o$

the solution is subtracted by 360° in order to get the actual angle

4 0

2 years ago

Your electric drill rotates initially at 5.21 rad/s. You slide the speed control and cause the drill to undergo constant angular

RUDIKE [14]

Answer:

The drill's angular displacement during that time interval is 24.17 rad.

Explanation:

Given;

initial angular velocity of the electric drill, $\omega _i$ = 5.21 rad/s

angular acceleration of the electric drill, α = 0.311 rad/s²

time of motion of the electric drill, t = 4.13 s

The angular displacement of the electric drill at the given time interval is calculated as;

$\theta = \omega _i t \ + \ \frac{1}{2}\alpha t^2\\\\\theta = (5.21 \ \times \ 4.13) \ + \ \frac{1}{2}(0.311)(4.13)^2\\\\\theta = (21.5173 ) \ + \ (2.6524)\\\\\theta =24.17 \ rad$

Therefore, the drill's angular displacement during that time interval is 24.17 rad.

6 0

2 years ago