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3 years ago

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

Physics

2 answers:

lara31 [8.8K]3 years ago

4 0

Answer:

The resistance will be twice the original resistance

Explanation:

This is a fairly simple question, the formular for the resistance of a wire is given as

R = rho *length / area

Where R = resistance

Rho = resistivity

L = length

A = area

Since the density and area are constant I.e they do not change

R/ length = rho/ area

The initial length is given as L, when this length is stretched to twice its original length ,it becomes 2×L = 2L

Let x represent the resistance when the length is doubled

R/ L = x / 2L

x = 2LR / L ; dividing by L

We have that x = 2R ; twice the resistance

soldi70 [24.7K]3 years ago

4 0

Answer:

Explanation:

Given:

L2 = 2 × L1

Using the formula,

Resistivity, d = (R × A)/L

Where,

R = resistance

A = area

L = length

1. d1 = (R1 × A1)/L1

2. d2 = (R2 × A2)/L2

Equating both 1 and 2 together,

(R2 × A2)/L2 = (R1 × A1)/L1

(R2 × A2)/(2 × L1) = (R1 × A1)/L1

Assume A1 = A2,

R2 = [(R1 × A1) × 2 × L1]/(A1 × L1)

R2 = 2 × R1

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An automobile tire is inflated with air originally at 10.0°C and normal atmospheric pressure. During the process, the air is com

solong [7]

Answer:

(a) $3.81\times 10^5\ Pa$

(b) $4.19\times 1065\ Pa$

Explanation:

Given:

$T_1$ = The first temperature of air inside the tire = $10^\circ C =(273+10)\ K =283\ K$

$T_2$ = The second temperature of air inside the tire = $46^\circ C =(273+46)\ K= 319\ K$

$T_3$ = The third temperature of air inside the tire = $85^\circ C =(273+85)\ K=358 \ K$

$V_1$ = The first volume of air inside the tire

$V_2$ = The second volume of air inside the tire = $30\% V_1 = 0.3V_1$

$V_3$ = The third volume of air inside the tire = $2\%V_2+V_2= 102\%V_2=1.02V_2$

$P_1$ = The first pressure of air inside the tire = $1.01325\times 10^5\ Pa$

Assume:

$P_2$ = The second pressure of air inside the tire

$P_3$ = The third pressure of air inside the tire
n = number of moles of air

Since the amount pof air inside the tire remains the same, this means the number of moles of air in the tire will remain constant.

Using ideal gas equation, we have

$PV = nRT\\\Rightarrow \dfrac{PV}{T}=nR = constant\,\,\,(\because n,\ R\ are\ constants)$

Part (a):

Using the above equation for this part of compression in the air, we have

$\therefore \dfrac{P_1V_1}{T_1}=\dfrac{P_2V_2}{T_2}\\\Rightarrow P_2 = \dfrac{V_1}{V_2}\times \dfrac{T_2}{T_1}\times P_1\\\Rightarrow P_2 = \dfrac{V_1}{0.3V_1}\times \dfrac{319}{283}\times 1.01325\times 10^5\\\Rightarrow P_2 =3.81\times 10^5\ Pa$

Hence, the pressure in the tire after the compression is $3.81\times 10^5\ Pa$ .

Part (b):

Again using the equation for this part for the air, we have

$\therefore \dfrac{P_2V_2}{T_2}=\dfrac{P_3V_3}{T_3}\\\Rightarrow P_3 = \dfrac{V_2}{V_3}\times \dfrac{T_3}{T_2}\times P_2\\\Rightarrow P_3 = \dfrac{V_2}{1.02V_2}\times \dfrac{358}{319}\times 3.81\times 10^5\\\Rightarrow P_3 =4.19\times 10^5\ Pa$

Hence, the pressure in the tire after the car i driven at high speed is $4.19\times 10^5\ Pa$ .

8 0

3 years ago

Examples of uniform velocity

slava [35]

Explanation:

When a truck travels in equal distances in equal intervals of time then we say that the body has got a uniform velocity. In the above example a truck is traveling at 5 miles in all the positions at A, B, and C and all in the intervals of 5 minutes each.

8 0

3 years ago

Jaclyn plays singles for South's varsity tennis team. During the match against North, Jaclyn won the sudden death tiebreaker poi

Nataly [62]

Answer

given,

mass of ball, m = 57.5 g = 0.0575 kg

velocity of ball northward,v = 26.7 m/s

mass of racket, M = 331 g = 0.331 Kg

velocity of the ball after collision,v' = 29.5 m/s

a) momentum of ball before collision

P₁ = m v

P₁ = 0.0575 x 26.7

P₁ = 1.535 kg.m/s

b) momentum of ball after collision

P₂ = m v'

P₂ = 0.0575 x (-29.5)

P₂ = -1.696 kg.m/s

c) change in momentum

Δ P = P₂ - P₁

Δ P = -1.696 -1.535

Δ P = -3.231 kg.m/s

d) using conservation of momentum

initial speed of racket = 0 m/s

M u + m v = Mu' + m v

M x 0 + 0.0575 x 26.7 = 0.331 x u' + 0.0575 x (-29.5)

0.331 u' = 3.232

u' = 9.76 m/s

change in velocity of the racket is equal to 9.76 m/s

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3 years ago

PLSS HELP

trasher [3.6K]

Answer:

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In an experiment in space, one proton is held fixed and another proton is released from rest a distance of 1.00 mm away. part a

mihalych1998 [28]

We can use Coulomb's law to find the force F acting on the proton that is released. F = k x Q1 x Q2 / r^2 k = 9 x 10^9 Q1 is the charge on one proton which is 1.6 x 10^{-19} C Q2 is the same charge on the other proton r is the distance between the protons F = (9x10^9) x (1.6 x 10^{-19} C) x (1.6 x 10^{-19} C) / (10^{-3})^2 F = 2.304 x 10^{-22} N We can use the force to find the acceleration. F = ma a = F / m a = (2.304 x 10^{-22} N) / (1.67 x 10^{-27} kg) a = 1.38 x 10^5 m/s^2 The initial acceleration of the proton is 1.38 x 10^5 m/s^2

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3 years ago