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3 years ago

A wire of length L and cross-sectional area A has resistance R.

Physics

1 answer:

Leona [35]3 years ago

6 0

Answer:

The resistance R stretched of the wire if it is stretched to twice its original length is [Δ(2L) / A ]* 2

Explanation:

Original Resistance R = ΔL / A

Given length is doubled

New length L' = 2L

New Volume is constant, resistivity and density of material is also assumed to be constant

AL = A'L'

AL = A'(2L)

A' = AL / 2L

A' = A / 2

∴ new Resistance R' = ΔL' / A'

= Δ(2L) / A/2

= [Δ(2L) / A ]* 2

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cupoosta [38]

Answer:

$F = 1.5 N$

Explanation:

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So here the change in the momentum of the hails is given as

$\Delta P = mv - (-mv)$

$\Delta P = 2mv$

now we know by Newton's 2nd Law

$F = \frac{\Delta P}{\Delta t}$

here we have

$F = \frac{\Delta m}{\Delta t} (2v)$

now plug in all data

$F = 0.050(2\times 15)$

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3 years ago

Which energy does a car travelling 30 m/ph as it slows have:

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Answer:

c) kinetic energy

Explanation:

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3 years ago

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(e)

Hatshy [7]

Answer:

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Explanation:

Density=Mass/Volume

11200kg/12.0= 933.3333kg/m^3

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3 years ago

A 1.2 x 103 kg racecar, with a velocity of 8 m/s, collides with an unsuspecting 80 kg honey badger who is standing

aalyn [17]

Answer: 90 m/s

Explanation:

Given

mass of racecar $M=1.2\times10^3\ kg$

velocity of racecar $u=8\ m/s$

mass of still honeybadger $m=80\ kg$

after collision race car is traveling at a speed of $v_1=2\ m/s$

conserving linear momentum

$Mu+m\times0=Mv_1+ mv_2\quad[v_2=\text{velocity of honeybadger after colllision}]$

$1.2\times10^3\times8+0=1.2\times10^3\times2+80\times v_2$

$1.2\times10^3(8-2)=80v_2\\v_2=\frac{7.2\times10^3}{80}=90\ m/s$

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3 years ago

A 0.20 kg mass is oscillating at a small angle from a light string with a period of 0.78 s.

Scorpion4ik [409]

Answer:

L = 15 cm

Explanation:

T = 2π√(L/g)

L = g(T/2π)²

L = 9.8(0.78/2π)²

L = 0.151027... m

L = 15 cm

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3 years ago

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