Physics, 24.07.2019 17:30 abbymcd05

Arectangular solid made of carbon has sides of lengths 1.0 cm, 2.0 cm, and 4.0 cm resistivity is ρ = 3.5×10−5 ω⋅m . determine the resistance for current that passes through the solid in -"g. com"

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Answer from: hunttj

Missing part in the text of the question. Found the rest on internet:
"determine the resistance for current that passes through the solid in (a) the x direction (b) the y direction and (c) the z direction"

Solution

The relationship between resistance and resitivity is given by
$R= \frac{\rho L}{A}$
where R is the resistance
$\rho$ is the resistivity of the material (for carbon, as in this problem, $\rho=3 \cdot 10^{-5} \Omega m$ )
L is the length of the conductor
A is its cross-sectional area.

We can solve the 3 parts of the problem by using the same formula, but every time L and A will be different (because they will depend on the direction of the current)

(a) the current is in the x-direction, so the length of the conductor is
L=1.0 cm=0.01 m

while the cross sectional area is
$A=(2.0 cm)(4.0 cm)=8.0 cm^2 = 8.0 \cdot 10^{-4}m^2$
So the resistance in this case is
$R= \frac{\rho L}{A} = \frac{(3 \cdot 10^{-5}\Omega m)(0.01 m)}{8.0 \cdot 10^{-4} m^2} = 3.8 \cdot 10^{-4}\Omega$

(b) the current is in the y-direction, so the length of the conductor is
L=2.0 cm=0.02 m

while the cross sectional area is
$A=(1.0 cm)(4.0 cm)=4.0 cm^2 = 4.0 \cdot 10^{-4}m^2$
So the resistance in this case is
$R= \frac{\rho L}{A} = \frac{(3 \cdot 10^{-5}\Omega m)(0.02 m)}{4.0 \cdot 10^{-4} m^2} = 1.5 \cdot 10^{-3}\Omega$

(c) the current is in the z-direction, so the length of the conductor is
L=4.0 cm=0.04 m

while the cross sectional area is
$A=(1.0 cm)(2.0 cm)=2.0 cm^2 = 2.0 \cdot 10^{-4}m^2$
So the resistance in this case is
$R= \frac{\rho L}{A} = \frac{(3 \cdot 10^{-5}\Omega m)(0.04 m)}{2.0 \cdot 10^{-4} m^2} = 6.0 \cdot 10^{-3}\Omega$

Answer from: Quest

resonant frequency

Answer from: Quest

air exerts forces on falling objects near earth’s surface.

explanation:

the term 'free falls' refer to object that are falling towards the earth, in a situation in which gravity is the only force acting on the object.

for an object in free fall, the acceleration is constant (g, the gravitational acceleration), and so the velocity of the object constantly increases.

for objects near earth's surface, there is always another force acting on the object: the resistance due to the air. this force is opposed to the direction of fall of the object, and it is proportional to the velocity of the object, so as the object falls down, this force increases up to a point when it balances the force of gravity, and the object continues its motion at constant velocity (called terminal velocity). since in this case gravity is not the only force acting on the object, we are not in a situation of free fall.

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