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Tems11 [23]
3 years ago
6

You wind a small paper tube uniformly with 153 turns of thin wire to form a solenoid. The tube\'s diameter is 5.11 mm and its le

ngth is 2.47 cm. What is the inductance, in microhenries, of your solenoid
Physics
1 answer:
lina2011 [118]3 years ago
3 0

The concept required to solve this problem is linked to inductance. This can be defined as the product between the permeability in free space by the number of turns squared by the area over the length. Recall that Inductance is defined as the opposition of a conductive element to changes in the current flowing through it. Mathematically it can be described as

L = \frac{\mu N^2 A}{l}

Here,

\mu = Permeability at free space

N = Number of loops

A = Cross-sectional Area

l = Length

Replacing with our values we have,

L = \frac{(4\pi *10^{-7})(153)^2(\pi (\frac{5.11*10^{-3}}{2})^2)}{2.47*10^{-2}}

L = 0.00002442H

L = 24.42\mu H

Therefore the Inductance is 24.42\mu H

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A force of 1.150×103 N pushes a man on a bicycle forward. Air resistance pushes against him with a force of 795 N. If he starts
nalin [4]

Answer:

He has a speed of 16.60m/s after 35.0 meters.

Explanation:

The final velocity can be determined by means of the equations for a Uniformly Accelerated Rectilinear Motion:

v_{f}^{2} = v_{i}^{2} + 2ad        

v_{f} = \sqrt{v_{i}^{2} + 2ad}  (1)

The acceleration can be found by means of Newton's second law:

\sum F_{net} = ma

Where \sum F_{net} is the net force, m is the mass and a is the acceleration.

Fx + Fy = ma  (2)

All the forces can be easily represented in a free body diagram, as it is shown below.

Forces in the x axis:

F_{x} = F - F_{air}  (3)

Forces in the y axis:

F_{y} = 0 (4)

Solving for the forces in the x axis:

F_{x} = F - F_{air}

Where F = 1.150x10^{3} N and F_{air} = 795 N:

F_{x} = 1.150x10^{3} N - 795 N

F_{x} = 355 N

Replacing in equation (2) it is gotten:

Fx + Fy = ma

355 N + 0 N = (90.0 Kg)a

355 N = (90.0 Kg)a

a = \frac{355 N}{90.0Kg}

a = \frac{355 Kg.m/s^{2}}{90.0Kg}

a = 3.94 m/s^{2}

So the acceleration for the cyclist is 3.94 m/s^{2}, now that the acceleration is known, equation (1) can be used:

v_{f} = \sqrt{v_{i}^{2} + 2ad}

However, since he was originally at rest its initial velocity will be zero (v_{i} = 0).

v_{f} = \sqrt{2ad}

v_{f} = \sqrt{2(3.94m/s^{2})(35.0m)}

v_{f} = 16.60m/s

He has a speed of 16.60m/s after 35.0 meters

8 0
3 years ago
Which BEST describes the speed and velocity of the graph?
Neko [114]
I think the answer is A (sorry if it isn't).
6 0
3 years ago
Read 2 more answers
A bond between a positively-charged particle and a negatively-charged particle is a(n) _______ bond.
weeeeeb [17]
The answer is an ionic bond.
6 0
3 years ago
Determine the volume displaced and then calculate the density of this 54 g sample of brass.
inessss [21]

Answer:

DETAILS IN THE QUESTION INSUFFICIENT TO ANSWER

Explanation:

Assuming the liquid to be water ,

the density d_{w} of water is : 1000kgm^{-3}=1gcm^{-3}

Buoyant force exerted by a liquid on an object with V_{imm} of it's volume immersed is :

F_{B}=V_{imm}*d_{l}*g

where ,

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  • d_{l} is the density of the liquid
  • g is the acceleration due to gravity

Thus at equilibrium:

m_{brass}*g=V_{imm}*d_{l}*g\\m_{brass}=V_{imm}*d_{l}\\54=V_{imm}*1\\V_{imm}=54cm^{3}

from these , we get the density of brass to be 1gcm^{-3}

which is not possible

7 0
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gavmur [86]
A solar eclipse occurs when the moon crosses in front of the Sun, blocking some or all of its rays. A lunar eclipse happens when the moon is directly behind the earth, blocking the moon from receiving light. The only light comes from the light on earth's reflected shadow.

You can look at a lunar eclipse because there is very little light or none at all. You can't look at a solar eclipse because you are looking directly at the sun unless it is complete. Before totality, only some of the Sun is blocked, causing your pupils dilate to let in more light. Since they do this, more of the Sun's rays can be let in to the eye, which effectively allows your eyes to burn. 

Some doctors and eye care specialists say that after someone complains of blindness after looking at a solar eclipse unaided, they can see what the Sun and moon looked like at the time that they looked at it, as it is burned onto their retinas. 
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