When did michael faraday invent wind powered electrical generation?

If a building is too humid, what harmful substance may be stored there?

DedPeter [7]

Answer:

carbondioxide

Explanation:

4 0

4 years ago

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The lab technician you recently hired tells you the following: Boss, an undisturbed sample of saturated clayey soil was brought

harkovskaia [24]

Answer:

The water of the saturated clayed soil is 66.67 %.

Explanation:

Given;

mass of saturated clayed soil, $M_s$ = 600 g

mass of dry soil sample, $M_d$ = 200 g

mass of water content, $M_w$ = $M_s$ - $M_d$ = 600 g - 200 g = 400 g

The water content is determined as;

$M_w(\%) = \frac{M_s - M_d}{M_s} *100\%\\\\M_w(\%) = \frac{600-200}{600} *100 \% \\\\M_w(\%) = 66.67 \%$

Therefore, the water of the saturated clayed soil is 66.67 %.

6 0

3 years ago

1. A copper block of volume 1 L is heat treated at 500ºC and now cooled in a 200-L oil bath initially at 20◦C. Assuming no heat

MaRussiya [10]

Answer:

final temperature T = 24.84ºC

Explanation:

given data

copper volume = 1 L

temperature t1 = 500ºC

oil volume = 200 L

temperature t2 = 20ºC

solution

Density of copper $\rho$ cu = 8940 Kg/m³

Density of light oil $\rho$ oil = 889 Kg/m³

Specific heat capacity of copper Cv = 0.384 KJ/Kg.K

Specific heat capacity of light oil Cv = 1.880 KJ/kg.K

so fist we get here mass of oil and copper that is

mass = density × volume ................1

mass of copper = 8940 × 1 × $10^{-3}$ = 8.94 kg

mass of oil = 889 × 200 × $10^{-3}$ = 177.8 kg

so we apply here now energy balance equation that is

$M(cu)\times Cv \times (T-T1)_{cu} + M(oil) \times Cv\times (T-T2)_{oil}$ = 0

put here value and we get T2

$8.94\times 0.384 \times (T-500) + 177.8 \times 1.890\times (T-20)$ = 0

solve it we get

T = 24.84ºC

7 0

3 years ago

Argon is compressed in a polytropic process with n = 1.2 from 100 kPa and 30°C to 1200 kPa in a piston–cylinder device. Determin

gulaghasi [49]

Answer:

181 °C

Explanation:

Initial pressure $P_{1}$ = 100 kPa

Initial temperature $T_{1}$ = 30 °C = 30 + 273 K = 303 K

Final pressure $P_{2}$ = 1200 kPa

Final temperature $T_{2}$ = ?

n = 1.2

For a polytropic process, we use the relationship

( $T_{2}$ / $T_{1}$ ) = ( $P_{2}$ / $P_{1}$ )^γ

where γ = (n-1)/n

γ = (1.2-1)/1.2 = 0.1667

substituting into the equation, we have

( $T_{2}$ /303) = (1200/100)^0.1667

$T_{2}$ /303 = 12^0.1667

$T_{2}$ /303 = 1.513

$T_{2}$ = 300 x 1.513 = 453.9 K

==> 453.9 - 273 = 180.9 ≅ 181 °C

5 0

3 years ago

Calculate the equivalent capacitance of the three series capacitors in Figure 12-1 A) 0.060 uF B) 0.8 uF C) 0.58 uF D) 0.01 uF

Naddik [55]

Answer:

The correct answer is option (A) 0.060 uF

Note: Kindly find an attached image of the complete question below

Sources: The complete question was well researched from Quizlet.

Explanation:

Solution

Given that:

C₁ = 0.1 μF

C₂ =0.22 μF

C₃ = 0.47 μF

In this case, C₁, C₂ and C₃ are in series

Thus,

Their equivalent becomes:

1/Ceq = (1/C₁ + 1/C₂ +1/C₃

1/Ceq =[ (1/0.1 + 1/0.22 +1/0.47)]

1/Ceq =[(0.22 * 0.47) + (0.1 * 0.47) + (0.1 * 0.22)/(0.1 * 0.22 *0.47)]

1/Ceq =[(0.1034 + 0.047 + 0.022)/(0.01034)

1/Ceq =[(0.1724)/(0.01034)]

1/Ceq = [(16.67)]

1/Ceq =(1/16.67) = 0.059μf

Ceq = 0.059μf ≈ 0.060μf

Therefore the equivalent capacitance of the three series capacitors is 0.060μf

4 0

3 years ago