Ask question

Ask question

All categories

2 years ago

7

An electric double oven can draw up to 34 amps at 240 volts when heating both ovens. What is the power demand, and how much ener

gy is used if the oven operates at that level for 2 hours?

1 answer:

MrRa [10]2 years ago

7 0

Answer:

8.16 kW
16.32 kWh

Explanation:

Power is the product of volts and amps:

P = VI = (240 V)(34 A) = 8160 W = 8.16 kW

Energy is the product of power and time:

(8.16 kW)(2 h) = 16.32 kWh

You might be interested in

The rectangular frame is composed of four perimeter two-force members and two cables AC and BD which are incapable of supporting

Rama09 [41]

Answer:

Your question is lacking some information attached is the missing part and the solution

A) AB = AD = BD = 0, BC = LC

AC = $\frac{5L}{3}T, CD = \frac{4L}{3} C$

B) AB = AD = BC = BD = 0

AC = $\frac{5L}{3} T, CD = \frac{4L}{3} C$

Explanation:

A) Forces in all members due to the load L in position A

assuming that BD goes slack from an inspection of Joint B

AB = 0 and BC = LC from Joint D, AD = 0 and CD = 4L/3 C

B) steps to arrive to the answer is attached below

AB = AD = BC = BD = 0

AC = $\frac{5L}{3} T, CD = \frac{4L}{3}C$

7 0

2 years ago

The closed tank of a fire engine is partly filled with water, the air space above being under pressure. A 6 cm bore connected to

skelet666 [1.2K]

Answer:

The air pressure in the tank is 53.9 $kN/m^{2}$

Solution:

As per the question:

Discharge rate, Q = 20 litres/ sec = $0.02\ m^{3}/s$

(Since, 1 litre = $10^{-3} m^{3}$ )

Diameter of the bore, d = 6 cm = 0.06 m

Head loss due to friction, $H_{loss} = 45 cm = 0.45\ m$

Height, $h_{roof} = 2.5\ m$

Now,

The velocity in the bore is given by:

$v = \frac{Q}{\pi (\frac{d}{2})^{2}}$

$v = \frac{0.02}{\pi (\frac{0.06}{2})^{2}} = 7.07\ m/s$

Now, using Bernoulli's eqn:

$\frac{P}{\rho g} + \frac{v^{2}}{2g} + h = k$ (1)

The velocity head is given by:

$\frac{v_{roof}^{2}}{2g} = \frac{7.07^{2}}{2\times 9.8} = 2.553$

Now, by using energy conservation on the surface of water on the roof and that in the tank :

$\frac{P_{tank}}{\rho g} + \frac{v_{tank}^{2}}{2g} + h_{tank} = \frac{P_{roof}}{\rho g} + \frac{v_{tank}^{2}}{2g} + h_{roof} + H_{loss}$

$\frac{P_{tank}}{\rho g} + 0 + 0 = \0 + 2.553 + 2.5 + 0.45$

$P_{tank} = 5.5\times \rho \times g$

$P_{tank} = 5.5\times 1\times 9.8 = 53.9\ kN/m^{2}$

4 0

2 years ago

The 8-mm-thick bottom of a 220-mm-diameter pan may be made from aluminum (k = 240 W/m ⋅ K) or copper (k = 390 W/m ⋅ K). When use

Artemon [7]

Answer:

For aluminum 110.53 C

For copper 110.32 C

Explanation:

Heat transmission through a plate (considering it as an infinite plate, as in omitting the effects at the borders) follows this equation:

$q = \frac{k * A * (th - tc)}{d}$

Where

q: heat transferred

k: conduction coeficient

A: surface area

th: hot temperature

tc: cold temperature

d: thickness of the plate

Rearranging the terms:

d * q = k * A * (th - tc)

$\frac{d * q}{k * A} = th - tc$

$th = \frac{d * q}{k * A} + tc$

The surface area is:

$A = \frac{\pi * d^2}{4}$

$A = \frac{\pi * 0.22^2}{4} = 0.038 m^2$

If the pan is aluminum:

$th = \frac{0.008 * 600}{240 * 0.038} + 110 = 110.53 C$

If the pan is copper:

$th = \frac{0.008 * 600}{390 * 0.038} + 110 = 110.32 C$

7 0

2 years ago

Given an array of words representing your dictionary, you test words to see if it can be made into another word in dictionary. T

katen-ka-za [31]

Answer:

Detailed solution is given below:

8 0

2 years ago

Which of these is true about airbag charts?

Arlecino [84]

A should be it or b

4 0

2 years ago