Ask question

Ask question

All categories

3 years ago

9

A panel of solar cells has an efficiency of 0.15. The total power input to the panel of solar cells is 3.2 kW. Calculate the use

ful power output of this panel of solar cells in kW.

1 answer:

charle [14.2K]3 years ago

3 0

Answer:

0.48 kW

Explanation:

Given;

Efficiency of solar panel = 0.15

Total power input of the solar panel = 3.2kW

The useful power output of the solar panel is the amount of power transmitted by the solar panel for use. It can be expressed as a function of its efficiency and total power input.

Useful power output = efficiency × Total power input

Substituting the given values;

Useful power output = 0.15 × 3.2 kW = 0.48 kW

You might be interested in

0.16 mol of argon gas is admitted to an evacuated 70 cm^3 container at 30°C. The gas then undergoes an isothermal expansion to a

Semmy [17]

Answer:

The final pressure of the gas is 9.94 atm.

Explanation:

Given that,

Weight of argon = 0.16 mol

Initial volume = 70 cm³

Angle = 30°C

Final volume = 400 cm³

We need to calculate the initial pressure of gas

Using equation of ideal gas

$PV=nRT$

$P_{i}=\dfrac{nRT}{V}$

Where, P = pressure

R = gas constant

T = temperature

Put the value in the equation

$P_{i}=\dfrac{0.16\times8.314\times(30+273)}{70\times10^{-6}}$

$P_{i}=5.75\times10^{6}\ Pa$

$P_{i}=56.827\ atm$

We need to calculate the final temperature

Using relation pressure and volume

$P_{2}=\dfrac{P_{1}V_{1}}{V_{2}}$

$P_{2}=\dfrac{56.827\times70}{400}$

$P_{2}=9.94\ atm$

Hence, The final pressure of the gas is 9.94 atm.

3 0

3 years ago

Which of the objects below has the greatest acceleration? *

lesya692 [45]

$\\ \bull\sf\dashrightarrow F=ma$

$\\ \bull\sf\dashrightarrow a=\dfrac{F}{m}$

Now

$\\ \bull\sf\dashrightarrow a\propto\dfrac{1}{m}$

$\\ \bull\sf\dashrightarrow a\propto F$

Lower mass=Higher acceleration
Lower Force=Lower Acceleration

Option B has lowest mass and highest force hence its correct

8 0

3 years ago

Một vật dao động điều hòa với phương trình x= 4cos(6πt) cm. Kể từ khi vật bắt đầu dao động, thời điểm vật qua a) vị trí cân bằng

dezoksy [38]

Answer:

what are you saying.........

......

7 0

3 years ago

A horizontal pipe of diameter 1.03m has a smooth constriction to a section of diameter 0.618 m. The density of oil flowing in th

vodka [1.7K]

Velocity of the oil in the pipe: 0.76 m/s, in the constricted section: 5.87 m/s

Explanation:

We can solve this problem by using Bernoulli's equation:

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho v_2^2$ (1)

where

$p_1 = 7340 N/m^2$ is the pressure in the pipe

$p_2 = 5505 N/m^2$ is the pressure in the constricted section

$\rho = 821 kg/m^3$ is the density of the oil

$v_1$ is the velocity of the oil in the pipe

$v_2$ is the velocity of the oil in the constricted section

Also, according to the continuity equation,

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-sectional area of the pipe, with

$r_1 = \frac{1.03}{2}=0.515 m$ is the radius

$A_2 = \pi r_2^2$ is the cross-sectional area of the constricted section, with

$r_2=\frac{0.618}{2}=0.309 m$ is the radius

So the equation becomes

$r_1^2 v_1 = r_2^2 v_2$

So we can write

$v_2=\frac{r_1^2}{r_2^2}v_1$

Substituting into eq.(1),

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho (\frac{r_1^2}{r_2^2}v_1)^2$

And solving the equation for $v_1$ :

$p_1 + \frac{1}{2}\rho v_1^2 = p_2 + \frac{1}{2}\rho \frac{r_1^4}{r_2^4}v_1^2\\v_1=\sqrt{\frac{p_2-p_1}{\frac{1}{2}\rho-\frac{1}{2}\rho \frac{r_1^4}{r_2^4}}}=0.76 m/s$

And the velocity in the constricted section is

$v_2=\frac{r_1^2}{r_2^2}v_1=5.87 m/s$

Learn more about flow rate:

brainly.com/question/9805263

#LearnwithBrainly

7 0

3 years ago

A roller coaster at the Six Flags Great America amusement park in Gurnee, Illinois, incorporates some clever design technology a

Zigmanuir [339]

Answer:

5.95 m

Explanation:

Given that the biggest loop is 40.0 m high. Suppose the speed at the top is 10.8 m/s and the corresponding centripetal acceleration is 2g

For the car to stick to the loop without falling down, at the top of the ride, the centripetal force must be equal to the weight of the car. That is,

(MV^2) / r = mg

V^2/ r = centripetal acceleration which is equal to 2g

2 × 9.8 = 10.8^2 / r

r = 116.64 /19.6

r = 5.95 m

3 0

3 years ago