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Kryger
2 days ago
7

A thin electrical heating element provides a uniform heat flux qo" to the outer surface of a duct through which air flows. The d

uct wall has a thickness of 10 mm and a thermal conductivity of 20 W/m·K. (a) At a particular location, the air temperature is 30°C and the convection heat transfer coefficient between the air and inner surface of the duct is 100 W/m2·K. What heat flux is required to maintain the inner surface of the duct at Ti = 58°C? (b) For the conditions of part (a), what is the temperature To of the duct surface next to the heater?
Physics
1 answer:
Ostrovityanka [942]2 days ago
5 0

Answer:

a) q = 2800 W/m²

b) To = 59.4°C

Explanation:

Given the following parameters:

L = 10 mm

K = 20 W/m·K

T = 30°C

h = 100 W/m²K

Ti = 58°C

a)

Calculating heat flux q:

q = h ΔT

q = 100 x (58 - 30)

q = 2800 W/m²

b)

According to Fourier's law, heat transfer is defined as:

Q = K A ΔT/L

Assuming the outer temperature is To:

To = Ti + qL/K

Substituting the given values:

To = Ti + qL/K

To= 58 + 2800 \times \dfrac{ 0.01}{20}

To = 59.4°C

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Two frictionless lab carts start from rest and are pushed along a level surface by a constant force. Students measure the magnit
Yuliya22 [1153]

Answer:

The kinetic energy is higher for the first cart.

Explanation:

For the second cart, its mass is 2kg and the momentum measured is 10kg m/s, which leads to

(2kg)v = 10kg\: m/s

resulting in

v = 5m/s.

Consequently, the kinetic energy for the 3kg cart ends up as

K.E.  = \dfrac{1}{2}mv^2

= \dfrac{1}{2}(2kg)(5m/s)^2 =  25J

\boxed{K.E = 25J}

indicating it is less than that of the 1kg cart so it follows that the first cart possesses greater kinetic energy.

3 0
15 days ago
the minute hand on a clock is 9 cm long and travels through an arc of 252 degrees every 42 minutes. To the nearest tenth of a ce
Ostrovityanka [942]

Answer:

The distance covered by the minutes hand is 39.60 cm.

Explanation:

Note: A clock has a circular shape, where the minutes hand acts as the radius, and its motion creates an arc.

Length of an arc is calculated as ∅/360(2πr)

L = ∅/360(2πr).................... Equation 1π

Here, L represents the arc’s length, ∅ is the angle made by the arc, and r is the arc’s radius.

Given: ∅ = 252°, r = 9 cm, π = 3.143.

Upon substituting these values into equation 1,

L = 252/360(2×3.143×9)

L = 0.7×2×3.143×9

L = 39.60 cm.

Thus, the distance traversed by the minutes hand is 39.60 cm.

4 0
14 days ago
A cave diver enters a long underwater tunnel, when her displacement with respect to the entry point is 20m,she accidentally drop
Maru [1053]

Answer:

3x864/y488bjehdksuwiieirjr

4 0
7 days ago
A transformer is to be designed to increase the 30 kV-rms output of a generator to the transmission-line voltage of 345 kV-rms.
inna [987]

Answer:

n_s = 920 \turns

Explanation:

Given,

Voltage of the primary coil (V_p) = 30 kV-rms

Voltage of the secondary coils (V_s) = 345 kV-rms

number of turns in the primary coil (n_p) = 80 turns

number of turns in the secondary coil (n_s) =?

the ratio of turns between primary and secondary coils

     \dfrac{n_p}{n_s} = \dfrac{V_p}{V_s}

     \dfrac{n_s}{n_p} = \dfrac{V_s}{V_p}

     n_s = n_p \dfrac{V_s}{V_p}

     n_s = 80\times \dfrac{345}{30}

     n_s = 80\times 11.5

     n_s = 920 \turns

The number of turns in the secondary coil is equal to n_s = 920 \turns

4 0
1 day ago
2. On January 21 in 1918, Granville, North Dakota, had a surprising change in temperature. Within 12 hours, the temperature chan
Yuliya22 [1153]

Respuesta:

El cambio de temperatura en Celsius equivale a 46°C.

El cambio de temperatura en Fahrenheit equivale a 82.8°F.

Explicación:

Un grado Celsius es equivalente a un grado Kelvin; por lo tanto,

\Delta C = \Delta K = 283K-237K\\\\ \boxed{\Delta C = 46^oC}

Un grado Fahrenheit es 1.8 veces un grado Celsius; por lo tanto

\Delta F = 1.8(46^o)

\boxed{\Delta F = 82.8^oF}

Por lo tanto, el cambio en Celsius es de 46°C y en Fahrenheit es de 82.8°F.

7 0
11 days ago
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