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3 years ago

PV System for the Smith Family answers

1 answer:

3 0

System B: The amorphous silicon solar modules have an efficiency of 6%. The dimensions of the solar modules amount to 0.5m by 1.0m. The output of each module is 30 Wp. The modules cost 20€ each. The advanage of the amorphous silicon solar modules is that they perform better on cloudy days in which there is no direct sunlight. Installed in the Netherlands, this system gives, on a yearly basis, 10% more output per installed Wp than the multicrystalline silicon modules. 
System A: The efficiency of the multicrystalline silicon module amounts to 15%. The dimensions of the solar module are 0.5m by 1.0m. Each module has 75 Wp output. The modules cost 60€ each.

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An empty beaker is placed on a top-pan balance. Some water is now poured into the beaker.What is the weight of the water? A. 0.0

mario62 [17]

Answer:

A. 0.044 kg

Explanation:

We need to subtract the sum of (beaker+water - empty beaker) which is 0.106 kg - 0.062 kg = 0.044 kg. The answer will not be written in Newton because this unit is used for force only and in this question w have to find the weight.

Hope it is enough.

Please mark me as brainliest.

6 0

2 years ago

HELPP ILL GIVE BRIANLISG

STatiana [176]

Answer:

31. Respiratory System

32. Excretory System

33. Nervous System

34. Reproductive system

35. Skeleton System

Explanation:

Hope it helps you.^_^

6 0

2 years ago

Read 2 more answers

A silver wire 2.6 mm in diameter transfers a charge of 420 C in 80 min. Silver contains 5.8 x 10- free electrons per cubic meter

kifflom [539]

Answer:

a). 87.5 mA or $87.5 x10^{-3}$ A

b). 1.78 $\frac{m}{s}$

Explanation:

$d=2.6 mm \\Q=420C\\t=80min\\n=5.8x10^{28} \\q=1.6x10^{-19}$

n the number of free electrons is 28 in text reference and if they don't give q is take as the charge of electron.

a).

$I=\frac{Q}{t}\\ I= \frac{420 C}{80 min}*\frac{1min}{60 s} =\frac{420 C}{4800s}\\ I=87.5 x10^{-3}$ A

b).

$I=n*abs (q)*V_{d}*A$

$A= \pi * (\frac{d}{2})^{2} \\A=\pi (*\frac{2.6x10^{-3} m}{2})^{2} \\A=5.309x10^{-6}$

$V_{d} =\frac{I}{n*abs(q)*A} \\V_{d}=\frac{87.5 x10^{-2} }{5.8x^10{28} *1.6x^{-19} *5.3x^{6} }\\V_{d}=1.78 \frac{m}{s}$

8 0

2 years ago

In a lab experiment a light flexible string is wrapped around a solid cylinder with mass M = 9.50 kg and radius R. The cylinder

Lunna [17]

Answer: $3.79 m/s^2$

Explanation:

Given

Mass $M=9.5 kg$

$m=3 kg$

Net Force is equivalent to $\sum F=ma$

with tension T in the string

For mass $m$

$mg-T=ma$

$T=mg-ma--------1$

For cylinder

$T\cdot R=I\times \alpha$

I for solid cylinder is $\frac{2}{5}MR^2 , and \alpha =\frac{a}{R}$

thus $T=\frac{Ma}{2}----2$

Substitute the value of T we get

$\frac{Ma}{2}=mg-ma$

$a(\frac{M}{2}+m)=mg$

$a=\frac{mg}{\frac{M}{2}+m}$

$a=3.79 m/s^2$

5 0

2 years ago

4. In a object filled with a type of gas, which of the following actions would decrease the pressure

ExtremeBDS [4]

Answer:

Decrease the number of gas particles, increase the object's area, and reduce the temperature of the gas

Explanation:

TEMPERATURE:

Decreasing the temperature will slow down the molecules. Hence, less no. of collisions will take place between walls of object and molecules. This will result in decrease of pressure.

Therefore, the pressure of a gas can be decreased by increasing its temperature.

NUMBER OF GAS PARTICLES:

Decreasing the number of particles will result in less no. of collisions, hence decreasing the pressure.

Therefore, the pressure of a gas can be decreased by decreasing its no. of molecules or no. of particles.

AREA OF OBJECT:

The pressure is given by the formula:

$P = \frac{F}{A}\\\\For constant Force (F):\\P\ \alpha\ \frac{1}{A}$

where,

A = Area of Object

Therefore, the pressure of a gas can be decreased by increasing area of object.

So, the correct option is:

Decrease the number of gas particles, increase the object's area, and reduce the temperature of the gas

7 0

2 years ago