All categories

3 years ago

Which of the following is true about ALL energy conversions?

Physics

2 answers:

iogann1982 [59]3 years ago

8 0

Answer:

d I think

Explanation:

hope this helps

zavuch27 [327]3 years ago

5 0

B or D i think it is B

You might be interested in

An ocean thermal energy conversion system is being proposed for electric power generation. Such a system is based on the standar

defon

Answer:

Explanation:

Dear Student, this question is incomplete, and to attempt this question, we have attached the complete copy of the question in the image below. Please, Kindly refer to it when going through the solution to the question.

To objective is to find the:

(i) required heat exchanger area.

(ii) flow rate to be maintained in the evaporator.

Given that:

water temperature = 300 K

At a reasonable depth, the water is cold and its temperature = 280 K

The power output W = 2 MW

Efficiency $\zeta$ = 3%

where;

$\zeta = \dfrac{W_{out}}{Q_{supplied }}$

$Q_{supplied } = \dfrac{2}{0.03} \ MW$

$Q_{supplied } = 66.66 \ MW$

However, from the evaporator, the heat transfer Q can be determined by using the formula:

Q = UA(L MTD)

where;

$LMTD = \dfrac{\Delta T_1 - \Delta T_2}{In (\dfrac{\Delta T_1}{\Delta T_2} )}$

Also;

$\Delta T_1 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_1 = 300 -290 \\ \\ \Delta T_1 = 10 \ K$

$\Delta T_2 = T_{h_{in}}- T_{c_{out}} \\ \\ \Delta T_2 = 292 -290 \\ \\ \Delta T_2 = 2\ K$

$LMTD = \dfrac{10 -2}{In (\dfrac{10}{2} )}$

$LMTD = \dfrac{8}{In (5)}$

LMTD = 4.97

Thus, the required heat exchanger area A is calculated by using the formula:

$Q_H = UA (LMTD)$

where;

U = overall heat coefficient given as 1200 W/m².K

$66.667 \times 10^6 = 1200 \times A \times 4.97 \\ \\ A= \dfrac{66.667 \times 10^6}{1200 \times 4.97} \\ \\ \mathbf{A = 11178.236 \ m^2}$

The mass flow rate:

$Q_{H} = mC_p(T_{in} -T_{out} ) \\ \\ 66.667 \times 10^6= m \times 4.18 (300 -292) \\ \\ m = \dfrac{ 66.667 \times 10^6}{4.18 \times 8} \\ \\ \mathbf{m = 1993630.383 \ kg/s}$

3 0

3 years ago

A pendulum oscillates 12 times in 4 seconds. what is the length of the pendulum?

seraphim [82]

Answer:

L = 2.8 cm

Explanation:

Period T = 4 / 12 = 1/3 s

T = 2π√(L/g)

L = (T/2π)²g

L = ((1/3)/2π)²9.8 = 0.02758... ≈ 2.8 cm

6 0

2 years ago

You stand on a straight desert road at night and observe a vehicle approaching. This vehicle is equipped with two small headligh

spayn [35]

To solve this problem we will apply the concepts related to Reyleigh's criteria. Here the resolution of the eye is defined as 1.22 times the wavelength over the diameter of the eye. Mathematically this is,

$\theta = \frac{1.22 \lambda }{D}$

Here,

D is diameter of the eye

$D = \frac{1.22 (539nm)}{5.11 mm}$

$D= 1.287*10^{-4}m$

The angle that relates the distance between the lights and the distance to the lamp is given by,

$Sin\theta = \frac{d}{L}$

For small angle, $sin\theta = \theta$

$sin \theta = \frac{d}{L}$

Here,

d = Distance between lights

L = Distance from eye to lamp

For small angle $sin \theta = \theta$

Therefore,

$L = \frac{d}{sin\theta}$

$L = \frac{0.691m}{1.287*10^{-4}}$

$L = 5367m$

Therefore the distance is 5.367km.

4 0

3 years ago

A 20.0-N weight slides down a rough inclined plane which makes an angle of 30 degree with the horizontal. The weight starts from

Ulleksa [173]

Answer:

$1270.64\ \text{J}$

Explanation:

m = Mass of object = $\dfrac{mg}{g}$

mg = Weight of object = 20 N

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

v = Final velocity = 15 m/s

u = Initial velocity = 0

d = Distance moved by the object = 150 m

$\theta$ = Angle of slope = $30^{\circ}$

f = Force of friction

fd = Work done against friction

The force balance of the system is

$\dfrac{1}{2}m(v^2-u^2)=(mg\sin\theta-f)d\\\Rightarrow \dfrac{1}{2}mv^2=mg\sin\theta d-fd\\\Rightarrow fd=mg\sin\theta d-\dfrac{1}{2}mv^2\\\Rightarrow fd=20\times \sin 30^{\circ}\times 150-\dfrac{1}{2}\times \dfrac{20}{9.81}\times 15^2\\\Rightarrow fd=1270.64\ \text{J}$

The work done against friction is $1270.64\ \text{J}$ .

8 0

3 years ago

How can you use a magnet to induce a current?

Furkat [3]

A coil of wire with a current flowing thru it becomes a magnet

3 0

3 years ago

Read 2 more answers