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2 years ago

14 gauge copper wire has a diameter of 1.6 mm. what length of this wire has a resistance of 4.8ω?

1 answer:

4 0

The relationship between resistance R and resistivity $\rho$ is
$R= \frac{\rho L}{A}$
where L is the length of the wire and A its cross section.

The radius of the wire is half the diameter:
$r= \frac{d}{2}= \frac{1.6 mm}{2}=0.8 mm=8\cdot 10^{-4} m$
and the cross section is
$A=\pi r^2 = \pi (8\cdot 10^{-4} m)^2=2.01\cdot 10^{-6} m^2$

From the first equation, we can then find the length of the wire when $R=4.8 \Omega$ (copper resistivity: $\rho = 1.724 \cdot 10^{-8} \Omega m$ )
$L= \frac{AR}{\rho}= \frac{(2.01\cdot 10^{-6} m^2)(1.724 \cdot 10^{-8} \Omega m)}{4.8 \Omega}=7.21 \cdot 10^{-15} m$

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750 million years ago, there was enough Uranium-235 present in Oklo, Gabon to have a natural fission reactor occur and generate

Komok [63]

<span>With a half-life of 700 million years, U-235 would have had twice as much mass at a time 700 MYA. This would have put the mass at 100kg at that time. Going back another 50 million years would be (50/700) or 1/14 of the half-life, or (1/2 * 1/14), or 1/28 of the total mass. 1/28 of 100kg is 3.57kg, so the amount present at the 750MYA mark would be approximately 103.57kg of U-235.</span>

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2 years ago

Read 2 more answers

In a "worst-case" design scenario, a 2000 kg elevator with broken cables is falling at 4.00 m/s when it first contacts a cushion

Whitepunk [10]

Answer:

A. V =3.65m/s

B. a = 4m/s^2

Explanation:

Determine force of gravity (f) on the elevator.

f = mg

(m = 2000kg, g = 9.8m/s

2000kg × 9.8m/s^2= 19600N

Given,

Force of opposing friction clampforce of gravity = 17000N

the Net force on the elevator

= force of gravity - Force of opposing friction clamp

=19600 - 17000

= 2600 N

Lets determine the kinetic energy of the elevator at the point of contact with the spring

K.E = 1/2 m v^2

(m = 2000kg, v = 4.00m/s)

= (1/2) × 2000kg × (4m/s)^2

= 16000J

kinetic energy and energy gain will be absorbed by the spring across the next 2m

Therefore,

E = K.E + P.E

K.E = 16000J,

P.E of spring = net force absorbed × distance at compression

net force absorbed = 2600N and distance at compression = 2.0m)

P.E = 5200J

E = 16000J + 5200J

E = 21200 J

Note, spring constant wasn't given

Lets determine it's value

Using,

E = (1/2) × k × (x)^2

Where:

E = energy = 21200J, K = ?, X = 2m

21200J=(1/2) × k × (2m)^2

21200J × 2 =(4m)k

K = 42400J/4m

K = 10600 N/m

Therefore,

acceleration at 1m compression = ?

Using F = K × X

(F is force provided by the spring = 10600N/m, K = 10600 N/m and X = 1m)

= 10600N/m × 1m = 10600 N ( upward)

A. The speed of the elevator after it has moved downward 1.00 {\rm m} from the point where it first contacts a spring?

Using.

original Kinetic energy + net force on the elevator = final kinetic energy + spring energy

16000N + 2600N = (1/2)mv^2 + (1/2)k x^2

18600 = (1/2)(2000)(v^2) + (1/2)(10600N)(1^2)

18600 = 1000(v^2) + 5300

18600 - 5300 = 1000(v^2)

13300 = 1000(v^2)

V^2 = 13.300

V =3.65m/s

The acceleration of the elevator is 1.00 {\rm m} below point where it first contacts a spring

Spring constant = net force on the elevator + resultant force

(Spring constant = 10600N, net force on the elevator = 2600N, resultant force = ?)

10600N = 2600N + resultant force

resultant force = 10600N - 2600N

=8000N

Therefore

F = ma

a = f/m

(a = ?, f =8000N and m =2000kg)

= 8000 / 2000

a = 4m/s^2

(It's accelerating upward, since acceleration is positive

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2 years ago

Read 2 more answers

An electric air heater consists of a horizontal array of thin metal strips that are each 10 mm long in the direction of an airst

sweet-ann [11.9K]

Answer:

see explanation below

Explanation:

Given that,

$T_1 =$ 500°C

$T_2$ = 25°C

d = 0.2m

L = 10mm = 0.01m

U₀ = 2m/s

Calculate average temperature

$\\T_{avg} = \frac{T_1 + T_2}{2} \\\\T_{avg} = \frac{500 + 25}{2} \\\\T_{avg} = 262.5$

262.5 + 273

= 535.5K

From properties of air table A-4 corresponding to $T_{avg}$ = 535.5K $\approx 550K$

k = 43.9 × 10⁻³W/m.k

v = 47.57 × 10⁻⁶ m²/s

$P_r = 0.63$

Number for the first strips is equal to

$R_e_x = \frac{u_o.L}{v}$

$R_e_x = \frac{2\times 0.01}{47.57 \times 10^-^6 }\\\\= 420.4$

Calculating heat transfer coefficient from the first strip

$h_1 = \frac{k}{L} \times 0.664 \times R_e_x^1^/^2 \times P_r^1^/^3$

$h_1 = \frac{43.9 \times 10^-^3}{0.01} \times 0.664\times420 \times 4^1^/^2 \times 0.683^1^/^3\\\\= 52.6W/km^2$

The rate of convection heat transfer from the first strip is

$q_1 = h_1\times(L\times d)\times(T_1 - T_2)\\\\q_1 = 52.6 \times (0.01\times0.2)\times(500-25)\\\\q_1 = 50W$

The rate of convection heat transfer from the fifth trip is equal to

$q_5 = (5 \times h_o_-_5-4\times h_o_-_4) \times(L\times d)\times (T_1 -T_2)$

$h_o_-_5 = \frac{k}{5L} \times 0.664 \times (\frac{u_o\times 5L}{v} )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.05} \times0.664\times (\frac{2 \times 0.05}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 25.5W/Km^2$

Calculating $h_o_-_4$

$h_o_-_4 = \frac{k}{4L} \times 0.664 \times (\frac{u_o\times 4L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.04} \times0.664\times (\frac{2 \times 0.04}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 26.3W/Km^2$

The rate of convection heat transfer from the tenth strip is

$q_1_0 = (10 \times h_o_-_1_0-9\times h_o_-_9) \times(L\times d)\times (T_1 -T_2)$

$h_o_-_1_0 = \frac{k}{10L} \times 0.664 \times (\frac{u_o\times 10L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.1} \times0.664\times (\frac{2 \times 0.1}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 16.6W/Km^2$

Calculating

$h_o_-_9 = \frac{k}{9L} \times 0.664 \times (\frac{u_o\times 9L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.09} \times0.664\times (\frac{2 \times 0.09}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 17.5W/Km^2$

Calculating the rate of convection heat transfer from the tenth strip

$q_1_0 = (10 \times h_o_-_1_0-9\times h_o_-_9) \times(L\times d)\times (T_1 -T_2)\\\\q_1_0 = (10 \times 16.6 -9\times 17.5) \times(0.01\times 0.2)\times (500 -25)\\\\=8.1W$

The rate of convection heat transfer from 25th strip is equal to

$q_2_5 = (25 \times h_o_-_2_5-24\times h_o_-_2_4) \times(L\times d)\times (T_1 -T_2)$

Calculating $h_o_-_2_5$

$h_o_-_2_5 = \frac{k}{25L} \times 0.664 \times (\frac{u_o\times 25L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.25} \times0.664\times (\frac{2 \times 0.25}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 10.5W/Km^2$

Calculating $h_o_-_2_4$

$h_o_-_2_4 = \frac{k}{24L} \times 0.664 \times (\frac{u_o\times 24L}{v } )^1^/^2\times Pr^1^/^3\\\\= \frac{43.9\times10^-^3}{0.24} \times0.664\times (\frac{2 \times 0.24}{47.57 \times 10^-^6} )^1^/^2\times 0.683^1^/^3\\\\= 10.7W/Km^2$

Calculating the rate of convection heat transfer from the tenth strip

$q_2_5 = (25 \times h_o_-_2_5-24\times h_o_-_2_4) \times(L\times d)\times (T_1 -T_2)\\\\q_1_0 = (25 \times 10.5 -24\times 10.7) \times(0.01\times 0.2)\times (500 -25)\\\\=5.4W$

6 0

2 years ago

Arrange the following substances from the lightest to the heaviest:

storchak [24]

molecular weights are written in the picture.

CH4<NH3<H2O<Cl2

6 0

1 year ago

A green ball has a mass of 0.525 kg and a blue ball has a mass of 0.482 kg. A croquet player strikes the green ball and it gains

Gelneren [198K]

Answer:

v' = 1.21 m/s

Explanation:

Mass of a green ball, m = 0.525 kg

Mass of a blue ball, m' = 0.482 kg

Initial velocity of green ball, u = 2.26 m/s

Initial velocity of blue ball, u' = 0 (at rest)

After the collision,

The final velocity of the green ball, v = 1.14 m/s

We need to find the final velocity of the blue ball after the collision if the collision is head on. Let v' is the final velcity of the blue ball. Using the conservation of momentum to find it :

$mu+m'u'=mv+m'v'\\\\0.525 (2.26)+0=0.525 (1.14)+0.482v'\\\\0.588=0.482v'\\\\v'=\dfrac{0.588}{0.482}\\\\v'=1.21\ m/s$

So, the final velocity of the blue ball is 1.21 m/s.

4 0

2 years ago