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MariettaO [177]
2 years ago
5

A wire as a length of 1.50m, diameter 0.60mm and resistance of 2ohms. calculate the resistance R of a wire of the same materials

of length 0.50m and diameter 0.3mm​
Physics
1 answer:
Otrada [13]2 years ago
3 0
  1. R=ρ×L/A

ρ=R×A/L

A=pie(r²)=pie(0.6²)=pie(0.36)=1.13

ρ=2ohm×1.13mm²/1500mm

ρ=0.0015ohm.mm

2. we were told that the wires were made of the same substance so the resistivity(ρ)is the same for both wires so:

R=ρ×L/A

R=0.0015ohm.mm×500mm/0.09mm²

R=8.3'ohm

so our resistance for the second wire is :

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>R</em><em>=</em><em>8</em><em>.</em><em>3</em><em>'</em><em>ohm</em>

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A wheel has a constant angular acceleration of 4.5 rad/s2. during a certain 5.0 s interval, it turns through an angle of 128 rad
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3 0
3 years ago
51. Suppose you measure the terminal voltage of a 3.200-V lithium cell having an internal resistance of 5.00Ω by placing a 1.00-
Tanya [424]

Explanation:

Given that,

Terminal voltage = 3.200 V

Internal resistance r= 5.00\ \Omega

(a). We need to calculate the current

Using rule of loop

E-IR-Ir=0

I=\dfrac{E}{R+r}

Where, E = emf

R = resistance

r = internal resistance

Put the value into the formula

I=\dfrac{3.200}{1.00\times10^{3}+5.00}

I=3.184\times10^{-3}\ A

(b). We need to calculate the terminal voltage

Using formula of terminal voltage

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Put the value into the formula

V=3.200-3.184\times10^{-3}\times5.00

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(c). We need to calculate the ratio of the terminal voltage of voltmeter equal to emf

\dfrac{Terminal\ voltage}{emf}=\dfrac{3.18}{3.200 }

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3 years ago
An elastic conducting material is stretched into a circular loop of 9.65 cm radius. It is placed with its plane perpendicular to
Nadya [2.5K]

Answer:

The induced emf in the coil is 0.522 volts.                        

Explanation:

Given that,

Radius of the circular loop, r = 9.65 cm

It is placed with its plane perpendicular to a uniform 1.14 T magnetic field.

The radius of the loop starts to shrink at an instantaneous rate of 75.6 cm/s , \dfrac{dr}{dt}=-0.756\ m/s

Due to the shrinking of radius of the loop, an emf induced in it. It is given by :

\epsilon=\dfrac{-d\phi}{dt}\\\\\epsilon=\dfrac{-d(BA)}{dt}\\\\\epsilon=B\dfrac{-d(\pi r^2)}{dt}\\\\\epsilon=2\pi rB\dfrac{dr}{dt}\\\\\epsilon=2\pi \times 9.65\times 10^{-2}\times 1.14\times 0.756\\\\\epsilon=0.522\ V

So, the induced emf in the coil is 0.522 volts.                                

8 0
3 years ago
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