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konstantin123 [22]
3 years ago
5

The magnitude of the magnetic field at a certain distance from a long, straight conductor is represented by B. What is the magni

tude of the magnetic field at twice the distance from the conductor
Physics
1 answer:
masya89 [10]3 years ago
8 0

Answer:

  B/4

Explanation:

The magnetic field strength is inversely proportional to the square of the distance from the current. At double the distance, the strength will be 1/2^2 = 1/4 of that at the original distance:

The field at twice the distance is B/4.

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a uniform disc and hollow right circular cone have the same formula for their moment of inertia when rotating about the central
Romashka-Z-Leto [24]

Answer:

This is as a result that about the central axis a collapsed hollow cone is equivalent to a uniform disc

Explanation:

The integration of the differential mass of the hollow right circular cone yields

I=\int\limits   dmr^2 = \int\limits^a_b {\frac{2Mxr^2}{R^2 +H^2} } \, dx  = \frac{2MR^2dx}{(R^2 +H^2)^2} \frac{(R^2 +H^2)^2}{4} = \frac{1}{2}MR^2

and for a uniform disc

I = 1/2πρtr⁴ = 1/2Mr².

6 0
3 years ago
Please help meeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeeee
Charra [1.4K]

Answer:

try finding on a website

Explanation:

6 0
3 years ago
How does your education contribute to comunity development
Angelina_Jolie [31]
Because everybody in community needs to be smart & have some type of knowledge
4 0
3 years ago
A 2-lb slider is propelled upward at A along the fixed curved bar which lies in a vertical plane. If the slider is observed to h
timofeeve [1]

To develop this problem it is necessary to apply the concepts given in the balance of forces for the tangential force and the centripetal force. An easy way to detail this problem is through a free body diagram that describes the behavior of the body and the forces to which it is subject.

PART A) Normal Force.

F_n = \frac{mv^2}{r}

N+mgcos\theta = \frac{mv^2}{r}

Here,

Normal reaction of the ring is N and velocity of the ring is v

N+mgcos\theta = \frac{mv^2}{r}

N+Wcos\theta = \frac{W}{g} (\frac{v^2}{r})

N+2cos30\° = \frac{2}{32.2}*\frac{10^2}{2}

N = 1.374lb

PART B) Acceleration

F_t = ma_t

-mgsin\theta = ma_t

-W sin\theta = \frac{W}{g} a_t

-2Sin30\° = (\frac{2}{32.2})a_t

a_T = -16.10ft/s^2

Negative symbol indicates deceleration.

<em>NOTE: For the problem, the graph in which the turning radius and the angle of suspension was specified was not supplied. A graphic that matches the description given by the problem is attached.</em>

8 0
3 years ago
Answer the following question​
Ray Of Light [21]

Answer:

A) OA, AB, BC

B) 25m/s^2

C) see explanation

D) 25

E) Rest

Explanation:

From the Velocity time graph shown:

The positive slope = OA ; This is positive because, it is the point of uniform acceleration on the graph.

Constant slope = AB, the slope here is constant because, AB on the graph is the point of constant velocity.

-ve slope = BC

B) Acceleration of body in path OA.

Acceleration = change in Velocity / time

Acceleration = (150 - 0) / 6

Acceleration = 150/6 = 25m/s^2

C) Path AB is Parallel to the because it marks the period of constant velocity (that is Velocity does not increase or decrease during the time interval).

D) Length of BC

BC corresponds to the distance moved, that velocity / time

Velocity = 150 ; time = 6

Therefore Distance (BC) = 150/6 = 25

E.) Velocity =0 ; Hence body is at rest

5 0
3 years ago
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