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3 years ago

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A 1000-turn solenoid is 50 cm long and has a radius of 2.0 cm. It carries a current of 18.0 A. What is the magnetic field inside

the solenoid near its center

1 answer:

vitfil [10]3 years ago

3 0

Answer:

The value is $B = 0.0452 \ T$

Explanation:

From the question we are told that

The number of turns is N = 1000

The length is L = 50 cm = 0.50 m

The radius is r = 2.0 cm = 0.02 m

The current is I = 18.0 A

Generally the magnetic field is mathematically represented as

$B = \mu_o * \frac{N }{L} * I$

Here $\mu_o$ is the permeability of free space with value

$\mu_o = 4\pi * 10^{-7} N/A^2$

So

$B = 4\pi * 10^{-7} * \frac{1000}{0.50} * 18.0$

=> $B = 0.0452 \ T$

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Compare the gravitational acceleration on the following objects compared to the Sun using:

arsen [322]

The gravitational acceleration of White dwarf compared to Sun is 13,675.86.

The gravitational acceleration of Neutron star compared to Sun is 6.79 x 10⁻²⁴.

The gravitational acceleration of Star Betelgeuse compared to Sun is 8.5 x 10¹⁰.

<h3>Mass of the planets</h3>

Mass of sun = 2 x 10³⁰ kg

Mass of white dwarf = 2.765 x 10³⁰ kg

Mass of Neutron star = 5.5 x 10¹² kg

Mass of star Betelgeuse = 2.188 x 10³¹ kg

<h3>Radius of the planets</h3>

Radius of sun = 696,340 km

Radius of white dwarf = 7000 km

Radius of Neutron star = 11 km

Radius of star Betelgeuse = 617.1 x 10⁶ km

<h3>Gravitational acceleration of White dwarf compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.765 \times 10^{30}}{2\times 10^{30}} \times [\frac{696,340,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 13,675.86$

<h3>Gravitational acceleration of Neutron star compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{5.5 \times 10^{12}}{2\times 10^{30}} \times [\frac{11,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 6.79\times 10^{-24}$

<h3>Gravitational acceleration of Star Betelgeuse compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.188 \times 10^{31}}{2\times 10^{30}} \times [\frac{617.1 \times 10^9}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 8.5\times 10 ^{10}$

Learn more about acceleration due to gravity here: brainly.com/question/88039

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3 years ago

A student sits on a rotating stool holding two 3.09-kg masses. When his arms are extended horizontally, the masses are 1.08 m fr

schepotkina [342]

Answer:

a

The New angular speed is $w_f = 2.034 rad/s$

b

The Kinetic energy before the masses are pulled in is $KE_i = 3.101 \ J$

c

The Kinetic energy after the masses are pulled in is $KE_f = 8.192 \ J$

Explanation:

From the we are told that masses are 1.08 m from the axis of rotation, this means that

The radius $r =1.08m$

The mass is $m = 3.09\ kg$

The angular speed $w = 0.770 \ rad/sec$

The moment of inertia of the system excluding the two mass $I = 3.25 \ kg \cdot m^2$

New radius $r_{new} = 0.34m$

Generally the conservation of angular momentum can be mathematical represented as

$w_f = [\frac{I_i}{I_f} ]w_i .....(1)$

Where $w_f$ is the final angular speed

$w_i$ is the initial angular speed

$I_i$ is the initial moment of inertia

$I_f$ is the final moment of inertia

Moment of inertia is mathematically represented as

$I = m r^2$

Where I is the moment of inertia

m is the mass

r is the radius

So the Initial moment of inertia is given as

$I_i = moment \ of \ inertia \ of\ the \ two \ mass \ + 3.25 \ kg \cdot m^2$

$I_i = 2m r^2 + 3.25$

The multiplication by is because we are considering two masses

$I_i = 2 [(3.09)(1.08)^2] +3.25 = 10.46 kg \cdot m^2$

So the final moment of inertia is given as

$I_f = 2[(3.09)(0.34)^2] +3.25 = 3.96 \ kg \cdot m^2$

Substituting these values into equation 1

$w_f = [\frac{10.46}{3.96} ] * 0.77 = 2.034 \ rad/sec$

Generally Kinetic energy is mathematically represented in term of moment of inertia as

$KE = \frac{1}{2} * I * w^2$

Now considering the kinetic energy before the masses are pulled in,

$KE_i = \frac{1}{2} * I_i * w^2_i$

The Moment of inertia would be $I_i = 10.46 \ Kg \cdot m^2$

The Angular speed would be $w_i = 0.77 \ rad/s$

Now substituting these value into the equation above

$KE_i = \frac{1}{2} * (10.46) * (0.770)^2 = 3.101 J$

Now considering the kinetic energy after the masses are pulled in,

$KE_f = \frac{1}{2} * I_f * w^2_f$

The Moment of inertia would be $I_f = 3.96 \ Kg \cdot m^2$

The Angular speed would be $w_f = 2.034 \ rad/s$

Now substituting these value into the equation above

$KE_f= \frac{1}{2} *(3.96)(2.034)^2$

$= 8.192J$

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3 years ago

Why would a rocket fall vertically downwards if it was launched vertically upwards

IgorC [24]

Answer:

It would because the shape of the rocket is designed to be able to slice through the air as smooth as possible and now you may be thinking that air is already smooth but when you try to push something as large and heavy like a rocket then the shape of the rocket will be very important. The bottom of the rocket is flatter then the top so it is not designed to fly smoothly through the air. So the rocket would fall vertically downward(If it was still in one piece)because of it's shape. It is easier for the top of the rocket to go smoothly through the air then the bottom.

Explanation:

I am 90% sure this is correct but if I'm not please tell me

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3 years ago

The acceleration (a) of a particle moving with uniform speed (v) in a circle of radius (r) is found to be propotional to r^n and

KATRIN_1 [288]

Answer:

n=-1 and m=2

Explanation:

A particle is moving uniformly with acceleration "a"

Uniform speed is v

And radius of circle is r

The acceleration is proportional to

r^n and v^m

i.e

a∝ rⁿ

a ∝v^m

Combing the two

Then, a∝rⁿv^m

Let k be constant of proportionality

Then, a=krⁿv^m. Equation 1

So, we know that the centripetal acceleration keeping an object in circular path is given as

a=v²/r

Rearranging

a=v²r^-1. Equation 2

So comparing this to the proportional

Equating equation 1 and 2

krⁿv^m = v²r^-1

This shows that,

k=1

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Then, n =-1

Also, v^m =v²

Then, m=2

Therefore,

n=-1 and m=2

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3 years ago

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aliya0001 [1]

Weight of an arrow is 20 plus 3

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