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4 years ago

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The maximum strength of the earth's magnetic field is about 6.9 10-5 T near the south magnetic pole. In principle, this field co

uld be used with a rotating coil to generate 60.0-Hz ac electricity. What is the minimum number of turns (area per turn

1 answer:

Ganezh [65]4 years ago

7 0

Answer:

Number of turns, $N=5\times 10^5$

Explanation:

Given that,

The maximum strength of the Earth's magnetic field, $B=6.9\times 10^{-5}\ T$

This field could be used with a rotating coil to generate 60.0-Hz ac electricity.

Let us assume that the coil must have to produce an rms voltage of 120 V and area per turn is 0.022 meter square.

$\epsilon=\sqrt{2} V_{rms}\\\\\epsilon=\sqrt 2\times 120\\\\\epsilon=169.7\ V$

The emf induced in the coil is given by :

$\epsilon=NAB\omega\\\\\epsilon=2\pi NABf$

N is the number of turns

f is the frequency

$N=\dfrac{\epsilon}{2\pi ABf}\\\\N=\dfrac{169.7}{2\pi 0.022\times 6.9\times 10^{-5}\times 60}\\\\N=2.96\times 10^5$

or

$N=5\times 10^5$

So, there are $5\times 10^5$ turns in the coil.

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When the atom's electrons step down to lower energy levels in a thin cloud of hot gas, what is produced?

Reika [66]

Answer: The correct answer is an emission line spectrum.

Explanation:

When the electrons are excited to the higher energy level, the energy is absorbed in this case.

When the electrons in the atom step down to lower energy levels in a thin cloud of hot gas then the radiation will emit.

The electron will lose energy in this case in the form of radiation. There will be an emission line spectrum.

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3 years ago

Read 2 more answers

A potential difference of 3.27 nV is set up across a 2.16 cm length of copper wire that has a radius of 2.33 mm. How much charge

miskamm [114]

Answer:

Charge = 4.9096 x 10⁻⁷ C

Explanation:

First, we find the resistance of the copper wire.

R = ρL/A

where,

R = resistance = ?

ρ = resistivity of copper = 1.69 x 10⁻⁸ Ω.m

L = Length of wire = 2.16 cm = 0.0216 m

A = Cross-sectional area of wire = πr² = π(0.00233 m)² = 1.7 x 10⁻⁵ m²

Therefore,

R = (1.69 x 10⁻⁸ Ω.m)(0.0216 m)/(1.7 x 10⁻⁵ m²)

R = 2.14 x 10⁻⁵ Ω

Now, we find the current from Ohm's Law:

V =IR

I = V/R

I = 3.27 x 10⁻⁹ V/2.14 x 10⁻⁵ Ω

I = 1.52 x 10⁻⁴ A

Now, for the charge:

I = Charge/Time

Charge = (I)(Time)

Charge = (1.52 x 10⁻⁴ A)(3.23 x 10⁻³ s)

<u>Charge = 4.9096 x 10⁻⁷ C</u>

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4 years ago

ASAP PLEASE! WILL GIVE BRAINLIEST ANSWER! The pictures below show the wavelengths and intensities of electromagnetic radiations

rjkz [21]

As per Weins displacement law the wavelength of light for which we get the peak of the graph is always inversely proportional to the temperature.

So we can say

$/lambda = \frac{b}{T}$

So here if temperature becomes more cool then wavelength will increase

here we know that

$\lambda_1 = 8500 A^o$

$\lambda_2 = 6000 A^o$

$\lambda_3 = 3500 A^o$

It means the hottest star out of all three is star 3

and coolest star is star 1

now if we star 2 becomes cooler then it means its temperature will go near to star 1 and hence it will more look like to star 1.

So correct answer is

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3 years ago

The weight of the heart of a cow whose weight is 1518 lbs. Answer in units of lbs.

valina [46]

Answer:

Weight of cow = 7.59 lbs

Explanation:

Given:

Weight of cow = 1518 lbs

Find:

Weight of heart of a cow

Computation:

Note: It is not given that in the mammals the weight of heart is approximately 0.5% of total body weight

Weight of heart of a cow = [Weight of cow][0.5%]

Weight of cow = [1518][0.5%]

Weight of cow = 7.59 lbs

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3 years ago

Ayden and Steven are playing catch with a football. Ayden throws the football at a velocity of 15 m/s with an angle of 60° above

Pachacha [2.7K]

1) 1.33 s

2) 8.6 m

3) 19.9 m

Explanation:

1)

The motion of the football is a projectile motion, which consists of two independent motions:

- A uniform motion (=constant velocity) along the horizontal direction

- A uniformly accelerated motion (=constant acceleration) along the vertical direction

The hang time of the football is the time it takes for the football to reach its maximum height. It is given by:

$t=\frac{u_y}{g}$

where

$u_y = u sin \theta$ is the initial vertical velocity of the ball, where

u = 15 m/s is the initial velocity

$\theta=60^{\circ}$ is the angle of projection of the ball

$g=9.8 m/s^2$ is the acceleration due to gravity

Substituting, we find the hang time:

$t=\frac{u sin \theta}{g}=\frac{(15)(sin 60^{\circ})}{9.8}=1.33 s$

2)

The motion of the ball along the vertical direction is a uniformly accelerated motion, so we can use the suvat equation:

$s=u_y t - \frac{1}{2}gt^2$

where:

s is the vertical displacement

$u_y = u sin \theta$ is the initial vertical velocity

t is the time

$g=9.8 m/s^2$ is the acceleration due to gravity

In this part, we want to find the maximum height, so the height reached by the ball when the time is

t = 1.33 s

Therefore, by substituting the values in the equation, we can find the maximum height:

$s=usin \theta t-\frac{1}{2}gt^2=(15)(sin 60^{\circ})(1.33)-\frac{1}{2}(9.8)(1.33)^2=8.6 m$

3)

Here we want to find the horizontal range covered by the ball during its flight.

The horizontal range for a projectile is given by the equation

$d=\frac{u^2 sin(2\theta)}{g}$

where

u is the initial velocity

$\theta$ is the angle of projection

g is the acceleration due to gravity

For the ball in this problem we have:

u = 15 m/s

$\theta=60^{\circ}$

$g=9.8 m/s^2$

Substituting into the equation, we find the horizontal distance covered by the ball:

$d=\frac{(15)^2sin(2\cdot 60^{\circ})}{9.8}=19.9 m$

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4 years ago