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elena-14-01-66 [18.8K]
3 years ago
6

A certain wire has a resistance of 110 Ω. What is the resistance of a second wire, made of the same material, that is 1/4 as lon

g and has 1/3 the diameter?

Physics
2 answers:
Karo-lina-s [1.5K]3 years ago
7 0

Answer: 247.5 ohms

Explanation: Since the two wires are of the same materials, their resistivity will be the same. Find solution attached.

Svet_ta [14]3 years ago
4 0

Answer:

New Resistance = 247.5 ohm

Explanation:

Resistance = resistivity * length / area

Since resistivity for the material is constant, resistance is directly proportional to (length/area).

This means that if (length/area) decreases or increases by any ratio, then resistance will increase or decrease by the same ratio.

So let's find the change in length/area

New length = 0.25 old length

New area = (1/9) old area                                 (This is because area equation contains a square of the diameter. if diameter decreases by 1/3, area decreases by (1/3)^2   )

So we now get length /area:

New length / new area = ( 0.25 old length) / (1/9 of old area)

New length / new area = 9*0.25 (old length / old area)

New length / new area = 2.25 (old length / old area)

To get the new resistance, we simply multiply it by the ratio we just found.

This equals:

110 * 2.25 = 247.5 ohm

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Answer:

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7\times 10^{-5}\ rad/s

Explanation:

If a satellite is in sync with Earth then the period of each satellite is 24 hours.

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Two steamrollers begin 115 mm apart and head toward each other, each at a constant speed of 1.10 m/sm/s . At the same instant, a
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Answer:

The distance of fly travel is 115.06 m.

Explanation:

Given that,

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Speed = 1.10 m/s

Speed of fly = 2.20 m/s

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Using formula of relative speed

v=v_{1}+v_{2}

Put the value into the formula

v=1.10+1.10

v=2.20\ m/s

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Using formula of time

t=\dfrac{d}{v}

Put the value into the formula

t=\dfrac{115}{2.20}

t=52.3\ sec

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Using formula of distance

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d=2.20\times52.3

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3 years ago
Light of wavelength 608.0 nm is incident on a narrow slit. The diffraction pattern is viewed on a screen 88.5 cm from the slit.
Effectus [21]

Answer:

The width of the slit is 0.167 mm

Explanation:

Wavelength of light, \lambda=608\ nm=608\times 10^{-9}\ m

Distance from screen to slit, D = 88.5 cm = 0.885 m

The distance on the screen between the fifth order minimum and the central maximum is 1.61 cm, y = 1.61 cm = 0.0161 m

We need to find the width of the slit. The formula for the distance on the screen between the fifth order minimum and the central maximum is :

y=\dfrac{mD\lambda}{a}

where

a = width of the slit

a=\dfrac{mD\lambda}{y}

a=\dfrac{5\times 0.885\ m\times 608\times 10^{-9}\ m}{0.0161\ m}

a = 0.000167 m

a=1.67\times 10^{-4}\ m

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So, the width of the slit is 0.167 mm. Hence, this is the required solution.

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Answer:

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