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cricket20 [7]
2 years ago
10

The resistivity of

Physics
1 answer:
Pachacha [2.7K]2 years ago
5 0

The current density of the copper wire is determined as (7.059 x 10⁸ A/m²)/L

<h3>Current density of the copper wire</h3>

The current density of the copper wire is calculated as follows;

J =1/ρ(E) = 1/ρ(V/L)

where;

  • E is electric field
  • V is voltage
  • L is length of the wire
<h3>Current density the wire</h3>

J =1/ρ(E) = 1/ρ(V/L)

J = (1/1.7 x 10⁻⁸) x (12)/L

J = (7.059 x 10⁸ A/m²)/L

Learn more about current here: brainly.com/question/24858512

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A mass weighing 14 pounds stretches a spring 2 feet. The mass is attached to a dashpot device that offers a damping force numeri
Elodia [21]

Answer:

The motion is over-damped when λ^2 - w^2 > 0 or when b^{2} > 0.86

The motion is critically when λ^2 - w^2 = 0 or when b^{2} = 0.86

The motion is under-damped when λ^2 - w^2 < 0 or when b^{2} < 0.86

Explanation:

Using the newton second law

k is the spring constante

b positive damping constant

m mass attached

m\frac{d^{2} x}{dt^{2}} = - kx - b\frac{dx}{dt}

x(t) is the displacement from the equilibrium position

\frac{d^{2} x}{dt^{2}} +\frac{b}{m}\frac{dx}{dt} + \frac{k}{m}x = 0

Converting units of weights in units of mass (equation of motion)

m = \frac{W}{g} = \frac{14}{32} = 0.43 slug

From hook's law we can calculate the spring constant k

k = \frac{W}{s} = \frac{14}{2} = 7 lb/ft

If we put m and k into the DE, we get

\frac{d^{2} x}{dt^{2}} +\frac{b}{0.43}\frac{dx}{dt} + 16.28x = 0

Denoting the constants

2λ = \frac{b}{m} = \frac{b}{0.43}

λ = b/0.215

w^{2} = \frac{k}{m} = 16.28

λ^2 - w^2 = \frac{b^{2} }{0.046} - 16.28

This way,

The motion is over-damped when λ^2 - w^2 > 0 or when b^{2} > 0.86

The motion is critically when λ^2 - w^2 = 0 or when b^{2} = 0.86

The motion is under-damped when λ^2 - w^2 < 0 or when b^{2} < 0.86

3 0
4 years ago
Science whoever gets this will get a brainlest
Zanzabum

Answer:

I got it right xd

Explanation:

__________

6 0
3 years ago
Read 2 more answers
In which direction does a bag at rest move when a force of 20 newton's is applied from the right?
Olegator [25]

The bag moves to the left.

This is because of Newton's third law of motion that states:

For every action force on a body, there is an opposite and equal reaction force.

Thus pushing the bag from the right makes it move to the left.


8 0
4 years ago
Jonah is trying to move his 22-kg desk by pushing on it with a force of 130 N, but his brother is leaning on it with a downward
Dafna11 [192]

Answer:

0.54

Explanation:

Draw a free body diagram.  There are 5 forces on the desk:

Weight force mg pulling down

Applied force 24 N pushing down

Normal force Fn pushing up

Applied force 130 N pushing right

Friction force Fnμ pushing left

Sum of the forces in the y direction:

∑F = ma

Fn − mg − 24 = 0

Fn = mg + 24

Fn = (22)(9.8) + 24

Fn = 240

Sum of the forces in the x direction:

∑F = ma

130 − Fnμ = 0

Fnμ = 130

μ = 130 / Fn

μ = 130 / 240

μ = 0.54

6 0
3 years ago
Patient is in the ed due to a football hitting his nose when playing tackle football in the park. X-ray shows a displaced nasal
Art [367]

ICD-10-CM codes are -S02.2XXA, W21.01XA, Y93.61, Y92.830


S02.2 for Fracture, Traumatic/Nasal (Bone(s)), ICD-10-CM Alphabetic Index. Both the open fracture code and the dislocation code are not reported. Only the fracture code is provided if a fracture and a dislocation happen at the same place. Search for "dislocation/with fracture" in the alphabetical index to be sent to a doctor. A closed fracture is a fracture with displacement. To report the conditions leading up to the injury, external cause codes are utilized. Look for Struck (accidentally) by/ball (struck) (thrown)/football W21.01 in the ICD-10-CM External Cause of Injuries Index. Seven characters are required in the Tabular List to finish the code. For the first encounter, X is utilized as a stand-in for character number six, and character number seven is given the letter A.

To learn more about ICD-10-CM please visit -brainly.com/question/27932590
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6 0
2 years ago
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