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3 years ago

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A copper wire has radius 0.800 mm and carries current I at 20.0°C. A silver wire with radius 0.500 mm carries the same current a

nd is also at 20.0°C. Let Ecu be the electric field in the copper wire and Eag be the electric field in the silver wire. Use the resistivities at room temperature for copper. PCu = 1.72 x 10^-8.m and for silver PAE= 1.47 x 10^-8.m. What is the ratio Ecu/Eng?

1 answer:

IgorC [24]3 years ago

7 0

Answer:

Ecu/Eag = 0.46

Explanation:

E = PI/A

Ecu = Pcu × I/A

Pcu = 1.72×10^-8 ohm-meter

r = 0.8 mm = 0.8/1000 = 8×10^-4 m

A = πr^2 = π×(8×10^-4)^2 = 6.4×10^-7π

Ecu = 1.72×10^-8I/6.4×10^-7π = 0.026875I/1

Eag = Pag × I/A

Pag = 1.47×10^-8 ohm-meter

r = 0.5 mm = 0.5/1000 = 5×10^-4 m

A = πr^2 = π × (5×10^-4)^2 = 2.5×10^-7π

Eag = 1.47×10^-8I/2.5×10^-7π = 0.0588I/π

Ecu/Eag = 0.026875I/π × π/0.0588I = 0.46

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Gravitational potential energy, the energy exhibited by an object at rest due to the influence of gravitational force. So the increase in distance of object from the surface of earth leads to increase in the gravitational potential energy. Thus,

$\text {Gravitational potential energy}=m \times \text { Acceleration } \times \text { Distance of bird from bottom }$

So, as the gravitational potential energy is given as 2033 J and the position of bird placed on the tall tower is 639 m away from the bottom, then the mass (m) of the bird can be found as below.

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A cord is used to vertically lower an initially stationary block of mass M = 3.6 kg at a constant downward acceleration of g/7.

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Answer:

(a) $W_c=127.008 J$

(b) $W_g=148.176 J$

(c) K.E. = 21.168 J

(d) $v=3.4293m.s^{-1}$

Explanation:

Given:

mass of a block, M = 3.6 kg

initial velocity of the block, $u=0 m.s^{-1}$

constant downward acceleration, $a_d= \frac{g}{7}$

$\Rightarrow$ That a constant upward acceleration of $\frac{6g}{7}$ is applied in the presence of gravity.

∴ $a=- \frac{6g}{7}$

height through which the block falls, d = 4.2 m

(a)

Force by the cord on the block,

$F_c= M\times a$

$F_c=3.6\times (-6)\times\frac{9.8}{7}$

$F_c=-30.24 N$

∴Work by the cord on the block,

$W_c= F_c\times d$

$W_c= -30.24\times 4.2$

We take -ve sign because the direction of force and the displacement are opposite to each other.

$W_c=-127.008 J$

(b)

Force on the block due to gravity:

$F_g= M.g$

∵the gravity is naturally a constant and we cannot change it

$F_g=3.6\times 9.8$

$F_g=35.28 N$

∴Work by the gravity on the block,

$W_g=F_g\times d$

$W_g=35.28\times 4.2$

$W_g=148.176 J$

(c)

Kinetic energy of the block will be equal to the net work done i.e. sum of the two works.

mathematically:

$K.E.= W_g+W_c$

$K.E.=148.176-127.008$

K.E. = 21.168 J

(d)

From the equation of motion:

$v^2=u^2+2a_d\times d$

putting the respective values:

$v=\sqrt{0^2+2\times \frac{9.8}{7}\times 4.2 }$

$v=3.4293m.s^{-1}$ is the speed when the block has fallen 4.2 meters.

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