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Marina86 [1]
3 years ago
11

A 2 m long piece wire of attached to a 3 V battery has a magnetic force of 8 N acting on it as it

Physics
2 answers:
Reptile [31]3 years ago
6 0

Answer:

<h2>R = 3.6Ω</h2>

Explanation:

Before we can calculate the resistance of the wire, we need to first calculate the Force experienced by the wire in the magnetic field. The force experienced by  the wire is expressed as;

F = BIL

B is the magnetic field measured in Tesla = 4.8T

I is the current in the wire

L is the length of the wire = 2m

F is magnetic force acting on the wire = 8N

from the formula I = F/BL

I = 8/4.8*2

I = 8/9.6

I = 0.83Amperes

According to ohms law;

V=IR

R = V/I

given Voltage = 3V

R = 3/0.83

R = 3.6Ω

The resistance of the wire to nearest tenth is 3.6Ω to nearest tenth

Ira Lisetskai [31]3 years ago
4 0

Answer:

The resistance of the wire is 3.6 Ohm's.

Explanation:

The force on a conductor in a magnetic field is given as;

         F = BILSinθ

where: F is the force, i is the current through the conductor, L is the length of the conductor and θ is the angle between the conductor and field.

From the question, F = 8N, B = 4.8T, L = 2m, I = ? and θ = 90^{0}.

So that,

    8 = 4.8 × I × 2 × Sin90^{0}

    8 = 9.6I

    I = \frac{8}{9.6}

    I = 0.8333A

The current flowing through the wire is 0.83A.

to determine the resistance of the wire, Ohm's law states that:

     V = IR

where v is the electromotive force, I is the current flowing and R is the resistance of the conductor.

V = 3V, I = 0.8333, then;

   R = \frac{V}{I}

      = \frac{3}{0.8333}

     = 3.6001Ohm's

The resistance of the wire is 3.6 Ohm's.

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A fan at a rock concert is 50.0 m from the stage, and at this point the sound intensity level is 114 dB. Sound is detected when
Marianna [84]

Answer:

A) P=13.92\ J.s^{-1}

B) v=3730.9912\ m.s^{-1}

C) v=74.44\ mm.s^{-1}

D) mosquitoes speed in part B is very much larger than that of part C.

Explanation:

Given:

  • Distance form the sound source, s=50\ m
  • sound intensity level at the given location, \beta=114\ dB
  • diameter of the eardrum membrane in humans, d=8.4 \times 10^{-3}\ m
  • We have the minimum detectable intensity to the human ears, I_0=10^{-12}\ W.m^{-2}

(A)

<u>Now the intensity of the sound at the given location is related mathematically as:</u>

\beta=10\ log(\frac{I}{I_0} ) ..........................................(1)

114=10\ log\ (\frac{I}{10^{-12}} )

11.4=log\ I+12\ log\ 10

I=0.2512\ W.m^{-2}

<em>As we know :</em>

I=\frac{P}{A}

0.2512=\frac{P}{\pi\times \frac{8.4^2}{4} }

P=13.92\ J.s^{-1} is the energy transferred to the  eardrums per second.

(B)

mass of mosquito, m=2\times 10^{-6}\ kg

<u>Now the velocity of mosquito for the same kinetic energy:</u>

KE=\frac{1}{2} m.v^2

13.92=\frac{1}{2}\times 2\times 10^{-6}\times v^2

v=3730.9912\ m.s^{-1}

(C)

Given:

  • Sound intensity, \beta = 20\ dB

<u>Using eq. (1)</u>

20=10\ log\ (\frac{I}{10^{-12}} )

2=log\ I+12\ log\ 10

I=10^{-10}\ W.m^{-2}

Now, power:

P=I.A

P=10^{-10}\times \pi\times \frac{8.4^2}{4}

P=5.54\times 10^{-9}\ J.s^{-1}

Hence:

KE=\frac{1}{2} m.v^2

5.54\times 10^{-9}=0.5\times 2\times 10^{-6}\times v^2

v=0.07444\ m.s^{-1}

v=74.44\ mm.s^{-1}

(D)

mosquitoes speed in part B is very much larger than that of part C.

7 0
3 years ago
A sphere of mass of 1.55 kg is accelerated upwards by a string to which the sphere is attached. Its speed increases from 2.81 m/
Mama L [17]

Answer:

The tension in the string is 16.24 N

Explanation:

Given;

mass of the sphere, m = 1.55 kg

initial velocity of the sphere, u = 2.81 m/s

final velocity of the sphere, v = 4.60 m/s

duration of change in the velocity, Δt = 2.64 s

The tension of the string is calculated as follows;

T =  ma  + mg\\\\T = m(a + g)\\\\where;\\\\a \ is \ upward \ acceleration \ of \ the \ sphere\\\\g \ is \ acceleration \ due \ to \ gravity =9.8 \ m/s^2\\\\a = \frac{\Delta V}{\Delta t} = \frac{v- u}{ t} = \frac{4.6 - 2.81 }{2.64} = 0.678 \ m/s^2

T = 1.55(0.678 + 9.8)

T = 1.55(10.478)

T = 16.24 N

Therefore, the tension in the string is 16.24 N

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