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3 years ago

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Two wires with the same resistance have the same diameter but different lengths. If wire 1 has length L 1 and wire 2 has length

L 2, how do L 1 and L 2 compare if wire 1 is made from copper and wire 2 is made from aluminum? The resistivity of copper is 1.7 × 10 –5 Ω·m and the resistivity of aluminum is 2.82 × 10 –5 Ω·m.

1 answer:

SVEN [57.7K]3 years ago

7 0

Answer with Explanation:

We are given that

Length of wire 1= $L_1$

Length of wire 2= $L_2$

Resistivity of copper wire= $\rho_1=1.7\times 10^{-5}\Omega-m$

Resistivity of aluminum wire= $\rho_2=2.82\times 10^{-5}\Omega-m$

Wire 1=Copper wire

Wire 2=Aluminum wire

Diameter of both wires are same and resistance of both wires are also same.

We know that

Resistance = $\frac{\rho l}{A}$

When diameter of wires are same then their cross section area are also same .

$l=\frac{RA}{\rho}$

When resistance and area are same then the length of wire depend upon the resistivity of wire .

The length of wire is inversely proportional to resistivity.

When resistivity is greater then the length of wire will be short and when the resistivity is small then the length of wire will be large.

$\rho_1$

Therefore, $L_1>L_2$

Hence, the length of wire 1 (copper wire) is greater than the length of wire 2 (aluminum).

$\frac{L_1}{L_2}=\frac{\frac{RA}{1.7\times 10^{-5}}}{\frac{RA}{2.82\times 10^{-5}}}=1.66$

$L_1=1.66L_2$

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The figure of the problem is missing: find in attachment.

(a) 1.64 s

The ball follows a projectile motion path. The horizontal displacement is given by

$x(t) = v_0 cos \theta t$

where

$v_0$ is the initial speed

t is the time

$\theta=32.0^{\circ}$ is the angle below the horizontal

We can rewrite this equation as

$t=\frac{x(t)}{v_0 cos \theta}$ (1)

The vertical displacement instead is given by

$y(t) = -v_0 sin \theta t - \frac{1}{2}gt^2$ (2)

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$y(t) = -45 m$

Substituting everything into the equation, we can find the time of flight:

$\frac{1}{2}gt^2=-y -x tan \theta\\t=\sqrt{\frac{2(-y-xtan \theta)}{g}}=\sqrt{\frac{2(-(-45)-50.8 tan 32.0^{\circ})}{9.8}}=1.64 s$

(b) 36.5 m/s

We can now find the initial speed directly by using the equation for the horizontal displacement:

$x(t) = v_0 cos \theta t$

where we have

x = 50.8 m

$\theta=32.0^{\circ}$

Substituting the time of flight,

t = 1.64 s

We find:

$v_0 = \frac{x}{t cos \theta}=\frac{50.8}{(1.64)(cos 32.0^{\circ})}=36.5 m/s$

(c) 47.1 m/s at 48.8 degrees below the horizontal

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$v_x = v_0 cos \theta = (36.5)(cos 32.0^{\circ})=31.0 m/s$

The initial vertical velocity is instead

$u_y = -v_0 sin \theta = -(36.5)(sin 32.0^{\circ})=-19.3 m/s$

And it changes according to the equation

$v_y = u_y -gt$

So at t = 1.64 s (when the ball hits the ground),

$v_y = -19.3 - (9.8)(1.64)=-35.4 m/s$

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$v=\sqrt{v_x^2+v_y^2}=\sqrt{(31.0)^2+(-35.4)^2}=47.1 m/s$

While the direction is:

$\theta=tan^{-1}(\frac{v_y}{v_x})=tan^{-1}(\frac{-35.4}{31.0})=-48.8^{\circ}$

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