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qaws [65]
3 years ago
7

A carbon resistor is 4 mm long and has a constant cross section of 0.2 mm2. The conductivity of carbon at room temperature is 3

× 104 per ohm·m. In a circuit its potential at one end of the resistor is 12 volts relative to ground, and at the other end the potential is 16 volts. Calculate the resistance and the current .

Physics
2 answers:
tangare [24]3 years ago
7 0

Answer:

The resistance is found to be 6Ω

The current is found to be 0.66 A

Explanation:

The resistance of a conductor in terms of its dimensions is given as:

R = ρL/A

where,

R = resistance = ?

ρ = resistivity = 3 x 10⁴ Ω.m

L = Length = 4 mm = 0.004 m

A = Cross-sectional area = 0.2 mm² = 0.2 x 10⁻⁶ m²

Therefore,

R = (3 x 10⁴ Ω.m)(0.004 m)/(0.2 x 10⁻⁶ m²)

<u>R = 6 Ω</u>

Now, the potential difference between both ends of the resistor is:

ΔV = 16 V - 12 V = 4 V

Now, from Ohm's Law:

V = IR

I = V/R

I = 4 V/ 6 Ω

<u>I = 0.66 A</u>

melamori03 [73]3 years ago
7 0

Answer:

Resistance (R) = 0.6667 ohm and Current (I) = 6A

Explanation: Please see the attachments below

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Answer:

Please find the answer in the explanation.

Explanation:

Given that two porters are available to carry a long timber wood.out of them one is weak. how do you make less load to the weak one?

We can make the weak one to carry less load through two different ways or means.

First, if we can locate the centre of gravity and centre of mass of the long timbre wood, the week one can carry the other end away from the center of gravity and centre of mass.

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The position of a particle moving along the x-axis depends on the time according to the equation x = ct2 - bt3, where x is in me
Sav [38]

Answer:

(a):  \rm meter/ second^2.

(b):  \rm meter/ second^3.

(c):  \rm 2ct-3bt^2.

(d):  \rm 2c-6bt.

(e):  \rm t=\dfrac{2c}{3b}.

Explanation:

Given, the position of the particle along the x axis is

\rm x=ct^2-bt^3.

The units of terms \rm ct^2 and \rm bt^3 should also be same as that of x, i.e., meters.

The unit of t is seconds.

(a):

Unit of \rm ct^2=meter

Therefore, unit of \rm c= meter/ second^2.

(b):

Unit of \rm bt^3=meter

Therefore, unit of \rm b= meter/ second^3.

(c):

The velocity v and the position x of a particle are related as

\rm v=\dfrac{dx}{dt}\\=\dfrac{d}{dx}(ct^2-bt^3)\\=2ct-3bt^2.

(d):

The acceleration a and the velocity v of the particle is related as

\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(2ct-3bt^2)\\=2c-6bt.

(e):

The particle attains maximum x at, let's say, \rm t_o, when the following two conditions are fulfilled:

  1. \rm \left (\dfrac{dx}{dt}\right )_{t=t_o}=0.
  2. \rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Applying both these conditions,

\rm \left ( \dfrac{dx}{dt}\right )_{t=t_o}=0\\2ct_o-3bt_o^2=0\\t_o(2c-3bt_o)=0\\t_o=0\ \ \ \ \ or\ \ \ \ \ 2c=3bt_o\Rightarrow t_o = \dfrac{2c}{3b}.

For \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6\cdot 0=2c

Since, c is a positive constant therefore, for \rm t_o = 0,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}>0

Thus, particle does not reach its maximum value at \rm t = 0\ s.

For \rm t_o = \dfrac{2c}{3b},

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}=2c-6bt_o = 2c-6b\cdot \dfrac{2c}{3b}=2c-4c=-2c.

Here,

\rm \left ( \dfrac{d^2x}{dt^2}\right )_{t=t_o}

Thus, the particle reach its maximum x value at time \rm t_o = \dfrac{2c}{3b}.

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