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anzhelika [568]

3 years ago

5

An object whose mass is 251 kg is located at an elevation of 24 m above the surface of the earth. For g-9.78 ms, determine the g

ravitational potential energy of the object, in kJ, relative to the surface of the earth.

1 answer:

Harlamova29_29 [7]3 years ago

4 0

Answer:

Gravitational Potential =58.914 KJ

Explanation:

We know that

$Gravitational Potential Energy = mass\times g\times Height$

Given mass = 251 kg

Height= 24 m

g is acceleration due to gravity = $9.78m/s^{2}$

Applying values in the equation we get

$Gravitational Potential Energy=251X9.78X24 Joules$

$Gravitational Potential Energy=58914.72 Joules$

$Gravitational Potential Energy =\frac{58914.72}{1000}KJ= 58.914KJ$

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Given resistance 30ohms Inductance 200mH is connected to a 230v,50hZ supply. Impedance 69.6ohms Calculate current consumed?

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Answer:

the current consumed is 3.3 A

Explanation:

Given;

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inductance, L = 200 mH

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frequency of the coil, f = 50 Hz

impedance, Z = 69.6 Ohms

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3 years ago

A building window pane that is 1.44 m high and 0.96 m wide is separated from the ambient air by a storm window of the same heigh

sdas [7]

Answer:

the rate of heat loss by convection across the air space = 82.53 W

Explanation:

The film temperature

$T_f = \frac{T_1+T_2}{2} \\\\= \frac{20-10}{2}\\\\= \frac{10}{2}\\\\= 5^0\ C$

to kelvin = (5 + 273)K = 278 K

From the " thermophysical properties of gases at atmospheric pressure" table; At $T_f$ = 278 K ; by interpolation; we have the following

$\frac{278-250}{300-250}= \frac{v-11.44(10^{-6})}{15.89(10^{-6})-11.44(10^{-6})}$ → v 13.93 (10⁻⁶) m²/s

$\frac{278-250}{300-250}= \frac{k-22.3(10^{-3}}{26.3(10^{-3}-22.3(10^{-3})}$ → k = 0.0245 W/m.K

$\frac{278-250}{300-250}= \frac{\alpha - 15.9(10^{-6})}{22.5(10^{-6}-15.9(10^{-6})}$ → ∝ = 19.6(10⁻⁶)m²/s

$\frac{278-250}{300-250}= \frac{Pr-0.720}{0.707-0.720}$ → Pr = 0.713

$\beta = \frac{1}{T_f} \\=\frac{1}{278} \\ \\ = 0.00360 \ K ^{-1}$

The Rayleigh number for vertical cavity

$Ra_L = \frac{g \beta (T_1-T_2)L^3}{\alpha v}$

= $\frac{9.81*0.00360(20-(-10))*0.06^3}{19.6(10^{-6})*13.93(10^{-6})}$

= $8.38*10^5$

$\frac{H}{L}= \frac{1.44}{0.06} \\ \\= 24$

For the rectangular cavity enclosure , the Nusselt number empirical correlation:

$Nu_L = 0.42(8.38*10^5)^{\frac{1}{4}}(0.713)^{0.012}(24){-0.3}$

$NU_L= \frac{hL}{k}= 4.878$

$\frac{hL}{k}= 4.878$

$\frac{h*0.06}{0.0245}= 4.878$

$h = \frac{4.878*0.0245}{0.06}$

h = 1.99 W/m².K

Finally; the rate of heat loss by convection across the air space;

q = hA(T₁ - T₂)

q = 1.99(1.4*0.96)(20-(-10))

q = 82.53 W

3 0

3 years ago

Fresh cut potato strips with a moisture content of 50% (w/w) on wet basis are fried in peanut oil to produce French fries.

Rufina [12.5K]

Answer:

Percentage of oil uptake by the potato = 5.57% on a wet basis (including the moisture content)

Percentage of oil uptake by the potato = 7.24% on a dry basis (excluding the moisture content)

Explanation:

Starting with 1000kg of fresh potatoes,

It is given that this consists of 50% (w/w) of water.

Meaning that, 1000 kg of fresh potatoes consists of (1000 × 50%) of water = 500 kg of water.

If 1000 kg of potatoes consist of 500 kg of water, then the remaining 500 kg is pure potatoes (since it is stated that freshly cut potatoes have no oil content)

After frying, the weight of the French fries is 700 kg now.

But of this 700 kg, 23 % of its weight is moisture content, i.e. water,

23% of 700 kg = 161 kg.

This means that the amount of potatoes and oil in the French fries is 700 - 161 = 539 kg

But, recall, that the amount of potatoes in this process from the start is 500 kg. This amount doesn't change even on frying the fresh potatoes into French fries.

So, amount of oil in the French fries = 539 - 500 = 39 kg

Percentage of oil uptake by the potato = 100% × (amount of oil intake)/(total mass of potatoes) = 100% × 39/700 = 5.57% on a wet basis (including the moisture content)

And 100% × 39/539 = 7.24% on a dry basis (excluding the moisture content)

4 0

3 years ago