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Consider a single turn of a coil of wire that has radius 6.00 cm and carries the current I = 1.50 A . Estimate the magnetic flux through this coil as the product of the magnetic field at the center of the coil and the area of the coil. Use this magnetic flux to estimate the self-inductance L of the coil.

Answers

Answer 1

Answer:

Answer:

$\phi = 1.78 *10^(-7) \ Weber$

$L = 1.183 *10^(-7) \ H$

Explanation:

From the question we are told that

The radius is $r = 6 \ cm = (6)/(100) = 0.06 \ m$

The current it carries is $I = 1.50 \ A$

The magnetic flux of the coil is mathematically represented as

$\phi = B * A$

Where B is the magnetic field which is mathematically represented as

$B = (\mu_o * I)/(2 * r)$

Where $\mu_o$ is the magnetic field with a constant value $\mu_o = 4\pi * 10^(-7) N/A^2$

substituting value

$B = (4\pi * 10^(-7) * 1.50 )/(2 * 0.06)$

$B = 1.571 *10^(-5) \ T$

The area A is mathematically evaluated as

$A = \pi r ^2$

substituting values

$A = 3.142 * (0.06)^2$

$A = 0.0113 m^2$

the magnetic flux is mathematically evaluated as

$\phi = 1.571 *10^(-5) * 0.0113$

$\phi = 1.78 *10^(-7) \ Weber$

The self-inductance is evaluated as

$L = (\phi )/(I)$

substituting values

$L = (1.78 *10^(-7) )/(1.50 )$

$L = 1.183 *10^(-7) \ H$

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A flat loop of wire consisting of a single turn of cross-sectional area 7.80 cm2 is perpendicular to a magnetic field that increases uniformly in magnitude from 0.500 T to 3.30 T in 1.00 s. What is the resulting induced current if the loop has a resistance of 1.20 ?

Answers

Answer:

Explanation:

Area of crossection, A = 7.80 cm²

Initial magnetic field, B = 0.5 T

Final magnetic field, B' = 3.3 T

Time, t = 1 s

resistance of the coil, R = 1.2 ohm

The induced emf is given by

$e=(d\phi)/(dt)=A(B' - B)/(t)$

where, Ф is the rate of change of magnetic flux.

e = 7.80 x 10^-4 x (3.3 - 0.5) / 1

e = 2.184 mV

i = e/R

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As the captain of the scientific team sent to Planet Physics, one of your tasks is to measure g. You have a long, thin wire labeled 1.73 g/m and a 1.30 kg weight. You have your accurate space cadet chronometer but, unfortunately, you seem to have forgotten a meter stick. Undeterred, you first find the midpoint of the wire by folding it in half. You then attach one end of the wire to the wall of your laboratory, stretch it horizontally to pass over a pulley at the midpoint of the wire, then tie the 1.30 kg weight to the end hanging over the pulley. By vibrating the wire, and measuring time with your chronometer, you find that the wire's second harmonic frequency is 200 Hz . Next, with the 1.30 kg weight still tied to one end of the wire, you attach the other end to the ceiling to make a pendulum. You find that the pendulum requires 313 s to complete 200 oscillations. Pulling out your trusty calculator, you get to work.What value of g will you report back to headquarters?

Answers

Answer:

The value of g is $g =76.2 m/s^2$

Explanation:

From the question we are told that

The mass of the weight is $m = 1.30 kg$

The spring constant $k = 1.73 g/m = 1.73 *10^(-3) \ kg/m$

The second harmonic frequency is $f = 100 \ Hz$

The number of oscillation is $N = 200$

The time taken is $t = 315 \ s$

Generally the frequency is mathematically represented as

$f = (v)/(\lambda)$

At second harmonic frequency the length of the string vibrating is equal to the wavelength of the wave generated

$l = \lambda$

Noe from the question the vibrating string is just half of the length of the main string so

Let assume the length of the main string is $L$

So $l = (L)/(2)$

The velocity of the vibrating string is mathematically represented as

$v = \sqrt{(T)/(\mu) }$

Where T is the tension on the string which can be mathematically represented as

$T = mg$

$v = \sqrt{(mg)/(k) }$

Then

$f = (v)/((L)/(2) )$

=> $v = (fL )/(2)$

=> $\sqrt{(mg)/(k) } = (fL)/(2)$

=> $g = (f^2 L^2 \mu)/(4m)$

substituting values

$g = ((100) * (1.73 *10^(-3) ))/((4 * 1.30)) L^2$

$g = 3.326 m^(-1) s^(-2) L^2$

Generally the period of oscillation is mathematically represented as

$T_p = 2 \pi \sqrt{(L)/(g) }$

=> $L = (T^2 g)/(4 \pi ^2)$

The period can be mathematically evaluated as

$T_p = (t)/(N)$

substituting values

$T_p = (315)/(200)$

$T_p = 1.575 \ s$

Therefore

$L = (1.575^2 * g )/(4 \pi ^2)$

$L = 0.0628 ^2 g$

$g = 3.326 m^(-1) s^(-2) L^2$

substituting for L

$g = 3.326 ((0.0628) g)^2$

=> $g = (1)/((3.326)* (0.0628)^2)$

$g =76.2 m/s^2$

The deepest point of the Pacific Ocean is 11,033 m, in the Mariana Trench. What is the water pressure at that point? The density of seawater is 1025 kg/m3. The deepest point of the Pacific Ocean is 11,033 m, in the Mariana Trench. What is the water pressure at that point? The density of seawater is 1025 kg/m3. 1.11 × 104 Pa 1.09 × 105 Pa 1.13 × 107 Pa 1.11 × 108 Pa 2.18 × 105 Pa

Answers

Answer: 1.11 x 10⁸ Pa

Explanation:

At any deep, the absolute pressure is the same for all points located at the same level, and can be expressed as follows:

p = p₀ + δ. g . h, where p₀ = atmospheric pressure = 101, 325 Pa

Replacing by the values, we get:

p= 101,325 Pa + 1025 Kg/m³ . 9.8 m/s². 11,033 m = 1.11 x 10⁸ Pa.

What advantage is there in using a set of helmholtz coils over just a single small magnet?

Answers

Two parallel coils separated by a distance equal to the radius of the coils are known as Helmholtz coils. They are frequently used because they generate a magnetic field that is uniform over an appreciable region about its midpoint.

In this example the motion is in a vertical circle. Passengers in a carnival ride travel in a circle with radius 5.0 m. The ride moves at a constant speed and makes one complete circle in a time T=4.0s. What is the acceleration of the passengers? If the ride increases in speed so that T=3.0s, what is arad? (This question can be answered by using proportional reasoning, without much arithmetic.)

Answers

Answer:

a. $12.3m/s^(2)$

b. $21.93m/s^(2)$

Explanation:

From the data given, the radius is 5.0m, and the time taken to complete one circle is 4.0secs

Since the motion is in a circular part, we can conclude that the total distance covered in this time is given as circumference of the circle.

which is expressed as

$Distance=2\pi R$

To determine the speed, we use the equation

$speed=(distnce)/(time)\n Speed=(2\pi R)/(time)\n speed=(2\pi*5 )/(4)\n Speed=7.85m/s$

The acceleration as required is expressed as

$a=(v^(2))/(r)\n a=(7.85^(2))/(5)\n a=12.3m/s^(2)$

if the speed increase and it takes 3secs to complete one circle, the speed is

$speed=(distnce)/(time)\n Speed=(2\pi R)/(time)\n speed=(2\pi*5 )/(3)\n Speed=10.47m/s$

and the acceleration becomes

$a=(v^(2))/(r)\n a=(10.47^(2))/(5)\n a=21.93m/s^(2)$

The acceleration of the passengers in the vertical circle carnival ride is 19.6 m/s^2. When the time taken to complete one circle is 3.0 s, the new acceleration is 26.13 m/s^2.

The acceleration of the passengers can be determined using the centripetal acceleration formula, which is given by a = v^2 / r.

In this case, the velocity v can be found by dividing the circumference of the circle (2πr) by the time taken to complete one circle (T). The radius r is given as 5.0 m. Plugging in the values, we have:

a = (v^2) / r = ((2πr / T)^2) / r = (4π^2r) / T^2 = (4π^2 * 5.0) / 16.0 = 19.6 m/s^2

To find the new acceleration when the time taken to complete one circle is 3.0 s, we can use the proportional reasoning to determine the relationship between the two accelerations. Since the time is inversely proportional to the acceleration, when T is 3.0 s, the new acceleration arad can be found using the equation:

arad / 19.6 = 4.0 / 3.0

Simplifying the equation, arad = (19.6 * 4.0) / 3.0 = 26.13 m/s^2

For more such questions on acceleration, click on:

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Air enters an adiabatic compressor at 104 kPa and 292 K and exits at a temperature of 565 K. Determine the power (kW) for the compressor if the inlet volumetric flow rate is 0.15 m3/s. Use constant specific heats evaluated at 300 K.