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Sonbull [250]
2 years ago
10

10 points

Physics
1 answer:
e-lub [12.9K]2 years ago
4 0

Answer:

e = 0.46 m

Explanation:

From the laws of friction, frictional force, F is proportional to normal reaction, R.

F₁ = μR

where μ is coefficient of friction; R = mg and g = 9.8 ms⁻²

Also, from Hooke's law, extension, e, in an elastic spring is proportional to applied force.

F₂ = Ke

where K is force constant of the spring

Since the box is just about to move, the coefficient of friction involved is static friction.

The force on the spring equals the frictional force experienced by the box the box; F₁ = F₂

Ke = μR

e = μR/K

where μ = 0.65; R = 18 kg * 9.8 ms⁻²; K = 250 N/m

e = (0.65 * 18 * 9.8)/250

e = 0.46 m

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Read 2 more answers
Four copper wires of equal length are connected in series. Their cross-sectional areas are 0.7 cm2 , 2.5 cm2 , 2.2 cm2 , and 3 c
Travka [436]

Answer:

22.1 V

Explanation:

We are given that

A_1=0.7 cm^2=0.7\times 10^{-4} m^2

A_2=2.5 cm^2=2.5\times 10^{-4} m^2

A_3=2.2 cm^2=2.2\times 10^{-4} m^2

A_4=3 cm^2=3\times 10^{-4} m^2

Using 1cm^2=10^{-4} m^2

We know that

R=\frac{\rho l}{A}

In series

R=R_1+R_2+R_3+R_4

R=\frac{\rho l}{A_1}+\frac{\rho l}{A_2}+\frac{\rho l}{A_3}+\frac{\rho l}{A_4}

R=\frac{\rho l}{\frac{1}{A_1}+\frac{1}{A_2}+\frac{1}{A_3}+\frac{1}{A_4}}

Substitute the values

R=\rho A(\frac{1}{0.7\times 10^{-4}}+\frac{1}{2.5\times 10^{-4}}+\frac{1}{2.2\times 10^{-4}}+\frac{1}{3\times 10^{-4}})

R=\rho l(2.62\times 10^4)

V=145 V

I=\frac{V}{R}=\frac{145}{\rho l(2.62\times 10^4)}

Voltage across the 2.5 square cm wire=IR=I\times \frac{\rho l}{A_2}

Voltage across the 2.5 square cm wire=\frac{145}{\rho l(2.62\times 10^4)}\times \frac{\rho l}{2.5\times 10^{-4}}=22.1 V

Voltage across the 2.5 square cm wire=22.1 V

6 0
3 years ago
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