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igomit [66]
3 years ago
12

There is a current I flowing in a clockwise direction in a square loop of wire that is in the plane of the paper. If the magneti

c field is toward the right, and if each side of the loop has length L, the net magnetic torque acting on the loop is:
Physics
1 answer:
MArishka [77]3 years ago
4 0

Answer:

IBL²

Explanation:

Given that

Current flowing in the wire, is I

Magnetic field towards the right, is B

Length of each side of the loop, is L

Number of windings on the loop, is N

Torque exerted on the wire, is T

The torque, T is usually given by the formula

Torque, T = NBIAsinθ,

and if N = 1 and sinθ = 1 also, then we have

Torque, T = BIA

And we know that Area A from this particular question will be given as = L². If we then substitute A for L², we then have

Torque, T = IBL²

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A typical person's eye is 2.5 cm in diameter and has a near point (the closest an object can be and still be seen in focus) of 2
Paraphin [41]

Answer:

2.27 cm

2.5 cm

Explanation:

u = Object distance =  25 cm

v = Image distance = 2.5 cm

f = Focal length

Lens Equation

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}=\frac{1}{25}+\frac{1}{2.5}\\\Rightarrow \frac{1}{f}=\frac{11}{25}\\\Rightarrow f=\frac{25}{11}=2.27\ cm

The minimum effective focal length of the focusing mechanism of the typical eye is 2.27 cm

when u=\infty

\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\\\Rightarrow \frac{1}{f}=\frac{1}{\infty}+\frac{1}{2.5}\\\Rightarrow \frac{1}{f}=\frac{1}{2.5}\\\Rightarrow f=\frac{2.5}{1}=2.5\ cm

The maximum effective focal length of the focusing mechanism of the typical eye is 2.5 cm

3 0
3 years ago
What is an internal resistance?
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Answer:

Internal resistance refers to the opposition to the flow of current offered by the cells and batteries themselves resulting in the generation of heat. Internal resistance is measured in Ohms. ... Example: 1 The potential difference across the cell when no current flows through the circuit is 3 V.

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9. A wave front has the form of a
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A wave front has the form of a surface of a sphere
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A 1.80 m radius playground merry-go-round has a mass of 120 kg and is rotating with an angular velocity of 0.330 rev/s. what is
maxonik [38]
 Solution.
Initial angular  velocity ωi = 0.330 × 2π rad/sec
                                         = 2.0737 rad/sec
The initial moment of inertia li = 0.5×M×r²
                                                = 0.5 × 120 × 1.8²
                                                = 194.4 kgm²
But the angular momentum is given by the product of moment of inertia and the angular velocity.
  = li × ωi
  = 2.0737 × 194.4
  = 403.13 kg.m²/s which is constant 
Child moment of inertia (point mass) = m × r² 
                                                          = 27.5 × 1.8²
                                                         =  27.5 × 3.24
                                                         =  89.1 kgm²
Thus, the final moment of inertia if = 194.4 + 89.1 =  283.5 kgm²
Therefore; the final speed: =403.13/ 283.5 
                                            = 1.422 rad/sec
This is equivalent to;
 1.422/2π = 0.2263 rev/s
                                       
4 0
3 years ago
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