Ask question

Ask question

All categories

3 years ago

7

A wire has a length of 0.50 m and measures about 0.50 mm in its cross-sectional radius. At normal temperature, what is its resis

tance in Ohms, if Aluminum has a resistivity of 2.82x10^-8 Ohms*meter

1 answer:

miv72 [106K]3 years ago

4 0

Answer:

Explanation:

For resistance , the expression is as follows .

R = ρ L / S where ρ is specific resistance , L is length of wire and S is cross sectional area .

cross sectional area = π x ( .5 x 10⁻³ )²

S = .785 x 10⁻⁶ m²

Putting the values

R = 2.82 x 10⁻⁸ x .50 / .785 x 10⁻⁶

= 1.796 x 10⁻² ohm .

You might be interested in

WILL MARK AS BRAINLIEST

kumpel [21]

The answer is either C or D.

8 0

3 years ago

Consider a composite cube made of epoxy with fibers aligned along one axis of the cube (the fibers are parallel to four of the t

coldgirl [10]

The question is incomplete. The complete question is :

Consider a composite cube made of epoxy with fibers aligned along one axis of the cube (the fibers are parallel to four of the twelve cube edges). If the cube can only be loaded in axial tension such that the force is uniformly applied over - and is normal to - a cube face, what is the lowest possible positive length change the cube can experience under this tension? The applied tensile force is 102 KN. The unloaded cube edge length is 56 mm. The glass fibers have an elastic modulus of 200 GPa. The epoxy has an elastic modulus of 38 GPa. The cube is comprised of 18 vol% epoxy (the balancing vol % is glass fiber). Hint: The loading axis is intentionally unspecified. Answer Format: Lowest possible length increase (change of length) under tension.

Solution :

Given :

$E_{glass fibre}$ = 200 GPa

$V_{glass fibre} = 82\%$

$E_{epoxy}$ = 38 GPa

$V_{epoxy} = 82\%$

Edge length = 56 mm

Cube is loaded in axial tension such that the force is uniformly applied over a cube face.

$E_{\text{composite}}=\frac{E_{glass fibre} \times E_{epoxy}}{(E_{glass fibre .E_{epoxy}})+(E_{fibre}.V_{glass fibre})}$

$E_{composite} = \frac{200 \times 38}{(200 \times 0.18)+(38\times 0.82)}$

$=113.16$ GPa

Applied stress $=\frac{\text{applied load}}{\text{area}}$

$\sigma=\frac{102 \times 10^3 \ N}{56 \times 56 \times 10^{-6} \ m^2}$

= 32.5 MPa

By Hooke's law

$\sigma = E . \epsilon$

$\sigma = E. \frac{\Delta l}{l}$

$\Delta l = \frac{\sigma}{E}\times l$

Length change, $\Delta l =\frac{32.5 \times 10^6 \ Pa}{113.16 \times 10^9 \ Pa}\times 56 \times 10^{-2} \ m$

$\Delta l = \frac{32.5 \times 56}{113.16} \times 10^{-3} \ mm$

= 0.016 mm

7 0

3 years ago

Consider a system of two particles with spin quantum numbers 1andS2respectivelyand with no orbital angular momentum. We have see

o-na [289]

Answer:

The Clebsh-Gordan Coefficients to relate the coupled and uncoupled bases.

Explanation:

In quantum mechanics, two different sources of angular momentum eigenstates, show their widening through the Clebsch–Gordan (CG) coefficients, first in an uncoupled product bases.

- The uncoupled basis writes the state as eigenstates of the z-components of the two particles:

IS1+S2ms1ms2)

- The coupled basis writes the state as eigenstates of the two particles:

IS1S2Stotms)

5 0

3 years ago

An insulating sphere of radius 13 cm has a uniform charge density throughout its volume. 13 cm 21.6 cm 7.5 cm p If the magnitude

sdas [7]

Answer:

The value is $E_1 = 49224.1 \ N/C$

Explanation:

From the question we are told that

The radius is $r = 13 \ cm = 0.13 \ m$

The electric field is $E = 78400 \ N/C$ at a distance $d = 7.5 \ cm = 0.075 \ m$

Generally the electric field at a distance $d = 7.5 \ cm = 0.075 \ m$ is mathematically represented as

$E = \frac{k * q * d}{r^3}$ Note: the reason we are using this

formula is because $d < r$

Here k is the coulomb constant with value $k = 9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.$

=> $78400= \frac{9*10^{9} * q * 0.075}{0.13^3 }$

Generally the electric field at a distance $d = 21.6 \ cm = 0.216 \ m$ is mathematically represented as

$E_1 = \frac{k * q }{^2}$ Note: the reason we are using this

formula is because $d > r$

Now dividing $E_1\ \ by \ \ E$

$\frac{E_1}{78400} = \frac{\frac{9*10^9 * q}{0.216^2} }{\frac{9*10^9 * q * 0.075}{0.13^3} }$

=> $E_1 = \frac{0.13 ^3}{ 0.075 * 0.216^2} * 78400$

=> $E_1 = 49224.1 \ J$

4 0

3 years ago

A pendulum of mass 240 g undergoes simple harmonic motion when acted upon by a force of 14 N. The pendulum crosses the point of

andrey2020 [161]

Answer:

4.3J

Explanation:

The energy associated with a simple pendulum is the potential energy and the kinetic energy. At rest the kinetic energy is zero while the potential energy is maximize, also at the center of the oscillation, the potential energy is zero while the kinetic energy is maximize.

Hence from the question given, we can conclude that the only associated energy is the Kinetic energy which is expressed as

$Kinetic Energy = \frac{1}{2}mv^{2} \\$

since mass,m=240g=0.24kg,

Velocity V=6m/s

If we substitute values we arrive at

$Kinetic Energy = \frac{1}{2}mv^{2} \\Kinetic Energy = \frac{1}{2} *0.24*6^{2}\\Kinetic Energy =4.3J$

8 0

3 years ago