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Liula [17]
3 years ago
10

In hydroelectric dams, gravitational potential energy is used to produce electrical energy. What is the mass of water that must

fall over a dam that is 12 m high in order to produce the same amount of energy in an AA battery (10,800 J)?
Physics
1 answer:
ddd [48]3 years ago
5 0

What mass of water must fall 12m in order to lose 10,800J of gravitational potential energy ?

Potential energy = (mass) x (gravity) x (height)

10,800J = (mass) x (9.8 m/s²) x (12m)

Mass = (10,800J) / (9.8 m/s² x 12m)

Mass = (10,800 / 9.8 x 12) (Joule-sec²/m²)

(Units:  Joule-sec²/m² = N-m-sec²/m² = (kg-m/s²)-m-sec²/m² = <u><em>kg </em></u>)

<u>Mass = 91.8 kg</u>

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Suppose 3 mol of neon (an ideal monatomic gas) at STP are compressed slowly and isothermally to 0.19 the original volume. The ga
Radda [10]

Answer:

a. 273 K b. 90.1 K c. 5.26 atm d. 0.33 atm

Explanation:

For isothermal expansion PV = constant

So, P₁V₁ = P₂V₂ where P₁ = initial pressure of gas = 1 atm (standard pressure), V₁ = initial volume of gas, P₂ = final pressure of gas and V₂ = final volume of gas,

So, P₁V₁ = P₂V₂

P₂ = P₁V₁/V₂

Since V₂/V₁ = 0.19,

P₂ = P₁V₁/V₂

P₂ = 1 atm (1/0.19)  

P₂ = 5.26 atm

For an adiabatic expansion, PVⁿ = constant where n = ratio of molar heat capacities = 5/3 for monoatomic gas

So, P₂V₂ⁿ = P₃V₃ⁿ where P₂ = initial pressure of gas = 5.26 atm, V₂ = initial volume of gas, P₃ = final pressure of gas and V₃ = final volume of gas,

So, P₂V₂ⁿ = P₃V₃ⁿ

P₃ = P₂V₂ⁿ/V₃ⁿ

P₃ = P₂(V₂/V₃)ⁿ

Since V₃ = V₁ ,V₂/V₃ = V₂/V₁ = 0.19

1/0.19,

P₃ = P₂(V₂/V₃)ⁿ

P₃ = 5.26 atm (0.19)⁽⁵/³⁾

P₃ = 5.26 atm × 0.0628

P₃ = 0.33 atm

Using the ideal gas equation

P₃V₃/T₃ = P₄V₄/T₄ where P₃ = pressure after adiabatic expansion = 0.33 atm , V₃ = volume after adiabatic expansion, T₃ = temperature after adiabatic expansion  P₄ = initial pressure of gas = P₁ = 1 atm , V₄ = initial volume of gas = V₁ and T₄ = initial temperature of gas = T₁ = 273 K (standard temperature)

P₃V₃/T₃ = P₄V₄/T₄

T₃ = P₃V₃T₄/P₄V₄    

T₃ = (P₃/P₄)(V₃/V₄)T₂

Since V₃ = V₄ = V₁ and P₄ = P₁

V₃/V₄ = 1 and P₃/P₄ = P₃/P₁

T₃ = (P₃/P₁)(V₃/V₄)T₂

T₃ = (0.33 atm/1 atm)(1)273 K  

T₃ = 90.1 K

So,

a. The highest temperature attained by the gas is T₁ = 273 K

b. The lowest temperature attained by the gas = T₃ = 90.1 K

c. The highest pressure attained by the gas is P₂ = 5.26 atm

d. The lowest pressure attained by the gas is P₃ = 0.33 atm

6 0
3 years ago
Warm air is more dense than cool air. Warm air is less dense than cool air Warm air and cool air have the same density
Wewaii [24]

Answer:

You are exactly right. The molecules in hot air are moving faster than the molecules in cold air. Because of this, the molecules in hot air tend to be further apart on average, giving hot air a lower density. That means, for the same volume of air, hot air has fewer molecules and so it weighs less

7 0
3 years ago
Calculate the momentum of a 4 kg Bolling ball being thrown at a speed of 3 m/s
bonufazy [111]

We are given –

  • Mass of boiling ball is, m = 4 kg
  • Speed is, v = 3 m/s
  • Momentum, P =?

As we know –

↠Momentum = Mass × Speed(Velocity)

↠Momentum = 4 × 3 kgm/s

↠Momentum = 12 kgm/s

  • Henceforth,Momentum will be 12 kgm/s.
7 0
2 years ago
A biker can ride at 12 m/s on a level road when there is no wind and at 7.5 m/s on a level road when there is a head wind of 5 m
alexandr402 [8]

Answer:

The answer is "1.94 \ m^2".

Explanation:

Formula:

F= \frac{1}{2} \times C_D \times density \times area \times  velocity^2\\\\Power (P) = F \times velocity \\\\P = \frac{C_D}{2} \times density \times area \times  velocity^3\\

Given value:

\ P = 10.25 \ W \\\\\ density = 1.2 \ kg/m^3 \\\\\ velocity= 2.5 \\\\

10.25 = \frac{C_D A}{2} \times 1.2 \times 2.5^3\\\\C_D A= \frac{10.25 \times 2 }{1.2 \times 2.5^3}\\\\C_D A = 1.94 m^2\\

8 0
3 years ago
For a science project, you would like to horizontally suspend an 8.5 by 11 inch sheet of black paper in a vertical beam of light
liubo4ka [24]

Answer:

I = 3.9 x 10⁷ W/m²

Explanation:

given,

Sheet of black paper dimension = 8.5 x 11 inch

Area of sheet = 8.5 x 11 = 93.5 inch^2

1 inch =0.0254 m

Area = 0.06032 m²

mass of sheet = 0.80 g

Force = m g =  0.8 x 9.8 x 10⁻³ N

                    =  7.84 x 10⁻³ N

speed of light = c = 3 x 10⁸ m/s

Using equation

F = \dfrac{IA}{c}

where I is the intensity of light

7.84 \times 10^{-3} = \dfrac{I\times 0.06032}{3 \times 10^8}

2.352 \times 10^{6} = I\times 0.06032

I = \dfrac{2.352 \times 10^{6}}{0.06032}

I = 3.9 x 10⁷ W/m²

Intensity of the light is equal to I = 3.9 x 10⁷ W/m²

4 0
3 years ago
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