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2 years ago

Compute the resistance in ohms of a copper block 5.0 cm long and 0.10 cm2 in cross-sectional area. (ρ = 1.77 × 10-6 ohm-cm)

1 answer:

soldier1979 [14.2K]2 years ago

4 0

R=ρ l/a
l=5 cm
a=0.10 cm²
ρ=1.77×10∧-6 Ωcm
R=(1.77×10∧-6) ×5/0.10
=8.85 ×10∧-5 Ω

resistance is directly proportional to length and inversly proportional to area of cross section is given by above equation

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A pin fin of uniform, cross-sectional area is fabricated of an aluminum alloy (k = 160 W/m-K). The fin diameter is D = 4 mm, and

frozen [14]

Answer:

Given that

D= 4 mm

K = 160 W/m-K

h=h = 220 W/m²-K

ηf = 0.65

We know that

$m=\sqrt{\dfrac{hP}{KA}}$

For circular fin

$m=\sqrt{\dfrac{4h}{KD}}$

$m=\sqrt{\dfrac{4\times 220}{160\times 0.004}}$

m = 37.08

$\eta_f=\dfrac{tanhmL}{mL}$

$0.65=\dfrac{tanh37.08L}{37.08L}$

By solving above equation we get

L= 36.18 mm

The effectiveness for circular fin given as

$\varepsilon =\dfrac{2\ tanhmL}{\sqrt{\dfrac{hD}{K}}}$

$\varepsilon =\dfrac{2\ tanh(37.08\times 0.03618)}{\sqrt{\dfrac{220\times 0.004}{160}}}$

ε = 23.52

5 0

2 years ago

Read 2 more answers

The function of a capacitor in an electric circuit is to?

Tema [17]

Well let's see:

A). No. A capacitor doesn't measure anything.

B). No. The power delivered to the circuit is determined by
      the battery or power supply and all the things in
the circuit that dissipate energy. A capacitor doesn't
do any of these things.

C). No. If any current actually flows between its plates,
   the capacitor is shot and can't do its job, and
   must be replaced.

D). Yes. A capacitor stores charges on its plates, and
electrical energy in the field between its plates.

4 0

3 years ago

Following Newton's Third Law, when a cannon goes off the cannon ball exerts a force on the cannon and the cannon exerts the same

bagirrra123 [75]

Answer:

First choice: Because the mass of the cannon ball is much less than the cannon

Explanation:

Indeed, Newton's Third Law, i.e. the action-reaction law, states that any action (force) will have a reaction (force) of same magnitude but opposite direction.

That means that when a cannon goes off the cannon ball exerts a force on the cannon and the cannon exerts the same force back on the cannon ball.

To find out how much the cannon ball and the cannon itsel move, you must consider Newton's second law.

F = m×a (force equal mass times acceleration).

Clearing the acceleration you get:

a = F / m

Then, since the mass is in the denominator and both the force that the cannon ball exerts on the cannon and the cannon exerts on the cannon ball are equal in magnitude, then the body that has the smaller mass (the cannon ball) will experience a greater acceleration, which is stated by the first choice: because the mass of the cannon ball is much less than the cannon.

6 0

2 years ago

A wheel of radius 25cm has eight spokes. It is mounted on a fixed axle and is rotating at a constant angular speed w. You shoot

spayn [35]

Explanation:

We will assume that the rim of the wheel is also very thin, like the spokes. The distance s between the spokes along the rim is

$s = \frac{1}{8}C = \frac{1}{8}(2\pi)(0.25\:\text{m}) = 0.196\:\text{m}$

The 20-cm arrow, traveling at 6 m/s, will travel its length in

$t = \dfrac{0.2\:\text{m}}{6\:\text{m/s}} = \dfrac{1}{30}\:\text{s}$

The fastest speed that the wheel can spin without clipping the arrow is

$v = \dfrac{s}{t} = \dfrac{0.196\:\text{m}}{\left(\dfrac{1}{30\:\text{s}}\right)} = 5.9\:\text{m/s}$

The angular velocity $\omega$ of the wheel is given by

$\omega = \dfrac{v}{r} = \dfrac{5.9\:\text{m/s}}{0.25\:\text{m}} = 23.6\:\text{rad/s}$

In terms of rev/s, we can convert the answer above as follows:

$23.6\:\dfrac{\text{rad}}{\text{s}}×\dfrac{1\:\text{rev}}{2\pi\:\text{rad}} = 3.8\:\text{rev/s}$

As you probably noticed, I did the calculations based on the assumption that I'm aiming for the edge of the wheel because this is the part of the wheel where a point travels a longer linear distance compared to ones closer to the axle, thus giving the arrow a better chance to pass through the wheel without getting clipped by the spokes. If you aim closer to the axle, then the wheel needs to spin slower to allow the arrow to get through without hitting the spokes.

3 0

2 years ago

Consider a disk of radius 2.6 cm with a uni- formly distributed charge of +6.7.1C. Compute the magnitude of the electric field a

MrRissso [65]

Answer:

E=15.3*10¹³ N/C : approximate

Explanation:

We use the following formula to calculate the electric field due to a disk with uniform surface charge at a point P that is along the central perpendicular axis of the disk and at a distance x from the center of the disk:

E= 2*π*k*σ*F Fórmula ( 1 )

$F= 1-\frac{x}{\sqrt{R^{2}+x^{2} } }$

E: Electric field at point P (N/C)

σ: surface charge density (C/m²)

R: disk radio (m)

x :distance from the center of the disk to the point P located on the axis of the disk (m)

K: Coulomb constant ( N*m²/C)

Equivalences

1cm = 10⁻²m

1mm = 10⁻³m

Data

R =2.6 cm= 2.6*10⁻²m = 0.026m

x=3.7 mm = 3.7* 10⁻³m = 0,0037 m

Q=+6.71C.

k= 8.98774 * 10⁹ N* m²/C

Calculation of surface charge density (σ )

σ= Q/A

Q: uniformly distributed charge (C)

A: disk area (m²) = π*R²

σ= +6.71C/π*(2.6*10⁻²)²m² = 3159.56C/m²

Calculation of the electric field at point P

We apply formula (1) and replace data

E= 2*π*3159.56*8.98774*10⁹ *F

$F= 1-\frac{x}{\sqrt{R^{2}+x^{2} } }$

E=15.3*10¹³ N/C : approximate

3 0

2 years ago