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3 years ago

How much energy (in joules) is released when 0.06 kilograms of mercury is condensed to a liquid at the same temperature?

Physics

2 answers:

viva [34]3 years ago

8 0

The energy released when a certain amount of a substance condenses to a liquid is given by:

$Q=m \lambda_v$

where m is the mass of the substance and $\lambda_v$ is the latent heat of vaporization.

In this problem, the mass is m=0.06 kg, while the latent heat of vaporization of mercury is $\lambda_v=296 kJ/ kg$ . Therefore, if we apply the formula we get:

$Q=(0.06 kg)(296 kJ/ kg)=17.76 kJ=17760 J$

So, the correct answer is

C. 17,705.1 J

zvonat [6]3 years ago

7 0

The answer is .... C. 17,705.1 J

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A wheel moves in the xy plane in such a way that the location of its center is given by the equations xo = 12t3 and yo = R = 2,

Stella [2.4K]

Answer:

the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s is $P = 104.04 \hat{i} -314.432 \hat{j}$

Explanation:

The free-body diagram below shows the interpretation of the question; from the diagram , the wheel that is rolling in a clockwise directio will have two velocities at point P;

the peripheral velocity that is directed downward $(-V_y)$ along the y-axis

the linear velocity $(V_x)$ that is directed along the x-axis

Now;

$V_x = \frac{d}{dt}(12t^3+2) = 36 t^2$

$V_x = 36(1.7)^2\\\\V_x = 104.04\ ft/s$

Also,

$-V_y = R* \omega$

where $\omega$ (angular velocity) = $\frac{d\theta}{dt} = \frac{d}{dt}(8t^4)$

$-V_y = 2*32t^3)\\\\\\-V_y = 2*32(1.7^3)\\\\-V_y = 314.432 \ ft/s$

∴ the velocity of the point P located on the horizontal diameter of the wheel at t = 1.4 s is $P = 104.04 \hat{i} -314.432 \hat{j}$

3 0

3 years ago

The half-wave rectifier circuit of vs(t) = 170 sin(377t) V and a load resistance R = 15Ω. Determine: a. The average load current

irina1246 [14]

Answer:

a average load current = 11.33 A

b rms load current = 8.02A

c true power =962.64 W

d apparent power =962.64 W

e. power factor cosθ =1

Explanation:

Vs (t) = 170 sin(377t) v

Vm =170v

Vrms = 170/√2 =120.23 v

Im = Vm/R = 170/15 = 11.33 A

Irms = Im/ √2 = 11.33/√2 =8.02A

Resistors are electronic components that consume energy

the power in a resistor is given by P =IVcosθ ; in a resistor cosθ =1

P =IV

The electrical power consumed by a resistance, (R) is called the true or real power

and is obtained by multiplying the rms voltage with the rms current.

P= Vrms × Irms

120.03×8.02

P= 962.64 W ; true power

apparent power = Vrms × Irms

=120.03×8.02

= 962.64W ; apparent power

power factor cosθ = true power/ apparent power

cosθ = 962.64/962.64

cosθ = 1

For the purely resistive circuit, the power factor is 1 , because the reactive power is equal to zero (0).

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3 years ago

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Degger [83]

Yes, it's true.

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That's what I think. Hope it's right, all the best.

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