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astra-53 [7]
2 years ago
15

A geothermal plant has been built in a location. Which of these is the most likely benefit of the geothermal plant in this locat

ion?
A. Decrease in geothermal fuels consumption
B. Improvement to hot springs and geysers
C. Long-term cost savings for energy users
D. Increase in emission of greenhouse gases
Physics
1 answer:
kramer2 years ago
4 0

Answer:

C

Explanation:

A and B are not true and D is a disadvantage

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Difference between reproducibility and replicability
victus00 [196]
Suppose you are doing an experiment where you determine the value of one parameter, say density of a liquid. You have two methods in doing this. By finding the mass and volume, and by using a densitometer. Reproducibility is when you get the same value of density for both methods. Replicability is when you have similar results in one method. So, replicability is a measure of precision, while reproducibility is a measure of accuracy. 
8 0
3 years ago
Evaporation of sweat cools the skin because
Law Incorporation [45]

Answer: c. the molecules with the highest energy evaporate first, lowering the temperature of the sample

Explanation:

The process by which liquid starts to change into vapor phase at any temperature is known as evaporation.

During evaporation , the molecules which possess higher energies escape from the upper layer into vapor phase. the molecules which escape draw energy from surroundings and thus decrease the energy of the surroundings and hence lead to decrease in temperature.

As temperature of the system is directly proportional to the energy of the system , thus decrease in energy leads to decrease in temperature.

K.E=\frac{3RT}{2}

K.E. = Kinetic energy

T = temperature

R= gas constant

6 0
3 years ago
What is anything that has mass and volume?
Reika [66]
Matter, substance. Material howya call it.
3 0
3 years ago
The starter motor of a car engine draws a current of 140 A from the battery. The copper wire to the motor is 4.20 mm in diameter
GenaCL600 [577]

Answer:

(a)106.4C

b)0.5676mm

Explanation:

(a)To get the charge that have passed through the starter then The current will be multiplied by the duration

I= current

t= time taken

Q= required charge

Q= I*t = 140*0.760 = 106.C

(b) b. How far does an electron travel along the wire while the starter motor is on?(mm)

diameter of the conductor is 4.20 mm

But Radius= diameter/2= 4.20/2=

The radius of the conductor is 2.1mm, then if we convert to metre for consistency same then

radius of the conductor is 0.0021m.

We can now calculate the area of the conductor which is

A = π*r^2

= π*(0.0021)^2 = 13.85*10^-6 m^2

We can proceed to calculate the current density below

J = 140/13.85*10^-6 = 10108303A/m

According to the listed reference:

Where e= 1.6*10^-19

n= 8.46*10^28

Vd = J/(n*e) = 10108303/ ( 8.46*10^28 * 1.6*10^-19 ) =0.0007468m/s=0 .7468 mm/s

Therefore , the distance traveled is:

x = v*t = 0.7468 * 0.760 = 0.5676mm

7 0
3 years ago
How to solve it? Three capacitors with capacities of 600 pF, 300 pF, 200 pF are connected in series. The 60 V voltage is applied
adell [148]

Answer:

1. Voltage across 600 pF is 10 V.

2. Voltage across 300 pF is 20 V.

3. Voltage across 200 pF is 30 V.

Explanation:

We'll begin by calculating the total capacitance of capacitor. This can be obtained as follow:

Capicitance 1 (C₁) = 600 pF

Capicitance 2 (C₂) = 300 pF

Capicitance 3 (C₃) = 200 pF

Total capacitance (Cₜ) =?

1/Cₜ = 1/C₁ + 1/C₂ + 1/C₃

1/Cₜ = 1/600 + 1/300 + 1/200

1/Cₜ = 1 + 2 + 3 / 600

1/Cₜ = 6/600

1/Cₜ = 1/100

Cₜ = 100 pF

Next, we shall convert 100 pF to Farad (F). This can be obtained as follow:

1 pF = 1×10¯¹² F

Therefore,

100 pF = 100 pF × 1×10¯¹² F / 1 pF

100 pF = 1×10¯¹⁰ F

Thus, 100 pF is equivalent to 1×10¯¹⁰ F.

Next, we shall determine the charge. This can be obtained as follow:

Voltage (V) = 60 V

Capicitance (C) = 1×10¯¹⁰ F

Charge (Q) =?

Q = CV

Q = 60 × 1×10¯¹⁰ F

Q = 6×10¯⁹ C

1. Determination of the voltage across 600 pF.

Capicitance 1 (C₁) = 600 pF = 6×10¯¹⁰ F

Charge (Q) = 6×10¯⁹ C

Voltage 1 (V₁) =?

Q = C₁V₁

6×10¯⁹ = 6×10¯¹⁰ × V₁

Divide both side by 6×10¯¹⁰

V₁ = 6×10¯⁹ / 6×10¯¹⁰

V₁ = 10 V

2. Determination of the voltage across 300 pF.

Capicitance 2 (C₂) = 300 pF = 3×10¯¹⁰ F

Charge (Q) = 6×10¯⁹ C

Voltage 2 (V₂) =?

Q = C₂V₂

6×10¯⁹ = 3×10¯¹⁰ × V₂

Divide both side by 3×10¯¹⁰

V₂ = 6×10¯⁹ / 3×10¯¹⁰

V₂ = 20 V

3. Determination of the voltage across 200 pF.

Capicitance 3 (C₃) = 200 pF = 2×10¯¹⁰ F

Charge (Q) = 6×10¯⁹ C

Voltage 3 (V₃) =?

Q = C₃V₃

6×10¯⁹ = 2×10¯¹⁰ × V₃

Divide both side by 2×10¯¹⁰

V₃ = 6×10¯⁹ / 2×10¯¹⁰

V₃ = 30 V

7 0
3 years ago
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