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sergeinik [125]

3 years ago

10

An ideal solenoid 20 cm long is wound with 5000 turns of very thin wire. What strength magnetic field is produced at the center

of the solenoid when a current of 10 A flows through it?

1 answer:

KIM [24]3 years ago

7 0

<h2>Answer:</h2>

0.31425 Tesla

<h2>Explanation:</h2>

The magnetic field strength of a solenoid can be found by using the Ampere's law as follows;

BL = μ₀ x N x I -------------------(i)

Where;

B = magnetic field strength

L = length of the solenoid

μ₀ = magnetic constant = 1.257 x 10⁻⁶H/m

N = number of turns in the coil of the solenoid

I = current flowing through the coil of the solenoid.

<em>From the question, </em>

L = 20cm = 0.2m

N = 5000 turns

I = 10A

<em>Substitute these values into equation (i) as follows;</em>

B x 0.2 = 1.257 x 10⁻⁶ x 5000 x 10

0.2B = 6.285 x 10⁻²

<em>Solve for B;</em>

B = 6.285 x 10⁻² / 0.2

B = 31.425 x 10⁻²

B = 0.31425 T

Therefore, the magnetic field strength is 0.31425 Tesla

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Answer:

The answer is C "think about the problem first, systematically consider all factors, and form a hypothesis"

Explanation:

In physics there is some basic fomula that sir Isacc Newton proposed under the topic of motion. The three formulas are below;

<em>1) v=u+at</em>

<em>2)v^2=u^2+2as</em>

<em>3)s=ut+(1/2)(at^2)</em>

the variables are explained below;

u= initial velocity of the body

a=acceleration/Speed of the body

t= time taken by the body while travelling

s= displacement of the body.

Therefore to solve keatons problem, the factors(variables) in the formulas above need to be systematically considered. Since the ball was dropped from the top of the building, the initial velocity is 0 because the body was at rest. Also the acceleration will be acceleration due to gravity (9.8m/s^2)

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3 years ago

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Bas_tet [7]

1. +72.0 kg m/s

The momentum of an object is given by:

p = mv

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m is the mass of the object

v is its velocity

Taking "to the right" as positive direction, for Elena we have

m = 60.0 kg is the mass

v = +1.20 m/s is the velocity

So, Elena's momentum is

$p_e=(60.0 kg)(+1.20 m/s)=+72.0 kg m/s$

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Here Madison is moving in the opposite direction of Elena (to the left), so her velocity is

v = -2.50 m/s

while her mass is

m = 65.0 kg

Therefore, her momentum is

$p_m= (65.0 kg)(-2.50 m/s)=-162.5 kg m/s$

3. -90.5 kg m/s

The total momentum of Elena and Madison is equal to the algebraic sum of their momenta; taking into account the correct signs, we have:

$p=p_e + p_m = +72.0 kg m/s - 162.5 kg m/s =-90.5 kg m/s$

4. 0.72 m/s to the left

We can find the final speed of Elena and Madison by using the law of conservation of momentum. In fact, the final momentum must be equal to the initial momentum (before the collision).

The initial momentum is the one calculated at the previous step:

$p_i = -90.5 kg m/s$

while the final momentum (after the collision) is given by

$p_f = (m_e + m_m) v$

where

$m_e$ is Elena's mass

$m_m$ is Madison's mass

v is their final velocity

According to the law of conservation of momentum,

$p_i = p_f\\p_i = (m_e + m_m) v$

So we can find v:

$v=\frac{p_i}{m_e + m_m}=\frac{-90.5 kg m/s}{60.0 kg+65.0 kg}=-0.72 m/s$

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Mashcka [7]

Hello!

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Margaret [11]

Answer:

Speed of the spacecraft right before the collision: $\displaystyle \sqrt{\frac{2\, G\cdot M_\text{e}}{R\text{e}}}$ .

Assumption: the earth is exactly spherical with a uniform density.

Explanation:

This question could be solved using the conservation of energy.

The mechanical energy of this spacecraft is the sum of:

the kinetic energy of this spacecraft, and
the (gravitational) potential energy of this spacecraft.

Let $m$ denote the mass of this spacecraft. At a distance of $R$ from the center of the earth (with mass $M_\text{e}$ ), the gravitational potential energy ( $\mathrm{GPE}$ ) of this spacecraft would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R}$ .

Initially, $R$ (the denominator of this fraction) is infinitely large. Therefore, the initial value of $\mathrm{GPE}$ will be infinitely close to zero.

On the other hand, the question states that the initial kinetic energy ( $\rm KE$ ) of this spacecraft is also zero. Therefore, the initial mechanical energy of this spacecraft would be zero.

Right before the collision, the spacecraft would be very close to the surface of the earth. The distance $R$ between the spacecraft and the center of the earth would be approximately equal to $R_\text{e}$ , the radius of the earth.

The $\mathrm{GPE}$ of the spacecraft at that moment would be:

$\displaystyle \text{GPE} = -\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}$ .

Subtract this value from zero to find the loss in the $\rm GPE$ of this spacecraft:

$\begin{aligned}\text{GPE change} &= \text{Initial GPE} - \text{Final GPE} \\ &= 0 - \left(-\frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\right) = \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \end{aligned}$

Assume that gravitational pull is the only force on the spacecraft. The size of the loss in the $\rm GPE$ of this spacecraft would be equal to the size of the gain in its $\rm KE$ .

Therefore, right before collision, the $\rm KE$ of this spacecraft would be:

$\begin{aligned}& \text{Initial KE} + \text{KE change} \\ &= \text{Initial KE} + (-\text{GPE change}) \\ &= 0 + \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}} \\ &= \frac{G \cdot M_\text{e}\cdot m}{R_\text{e}}\end{aligned}$ .

On the other hand, let $v$ denote the speed of this spacecraft. The following equation that relates $v\!$ and $m$ to $\rm KE$ :

$\displaystyle \text{KE} = \frac{1}{2}\, m \cdot v^2$ .

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$\displaystyle v = \sqrt{\frac{2\, \text{KE}}{m}}$ .

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