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True [87]
3 years ago
7

Consider a household that uses 14,000 kWh of electricity and 900 gal of fuel oil, per year, during a heating season. The average

amount of CO₂ produced is 26.4 lbm/gal of fuel oil and 1.54 lbm/kWh of electricity. If this household reduces its oil and electricity usage by 21 percent by implementing some energy conservation measures, determine the reduction in the amount of CO₂ emissions by that household per year.
Physics
1 answer:
MissTica3 years ago
3 0

Answer:

reduction in the amount of CO₂ emissions by that household per year is 9517.2 lbm per year

Explanation:

given data

electricity consume = 14000 kWh

fuel consume = 900 gal

CO₂ produced of fuel = 26.4 lbm/gal

CO₂ produced of electricity  = 1.54 lbm/kWh

oil and electricity usage = 21 percent

to find out

the reduction in the amount of CO₂ emissions

solution

we calculate the amount of CO₂ produce here that is

amount of CO₂ produce = ( electricity consume×CO₂ produce electricity + fuel consume × CO₂ consume fuel )    ........................1

put here value

amount of CO₂ produce = ( 14000 × 1.54 + 900 × 26.4 )  

amount of CO₂ produce = 45320 lbm/yr

we know reduction is 21%

so

reduction in amount of CO₂ produced is

reduction in CO₂ produced = 45320 × 21%

reduction in CO₂ produced = 9517.2 lbm per year

so reduction in the amount of CO₂ emissions by that household per year is 9517.2 lbm per year

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2. Two toy cars are involved in a race. Car A has mass m while car B has mass 2m. a. The two cars have the same force applied to
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Answer:

a) The kinetic energy of the two cars is the same

the moment of car 2 is greater than the moment of car 1

b)  the kinetic energy of car 1 is greater than that of car 2

the moment of the two cars is the same

Explanation:

a) to know the kinetic energy of each car, we must find the speed, use Newton's second law to find the acceleration

Car 1

     F = m a

    a = F / m

Let's use kinematics to find the velocity after x = 1 m

       v² = v₀² + 2 a x

The initial speed is zero

       v = √ (2 F/m  x)

For the distance of x = 1 m

        v₁ = √ (2 F / m)

Car 2

      F = 2m a

      a = F / 2m

      v² = 2 a x

      v = √ (F/m  x)

 For x = 1 m

       v₂ = √(F / m)

Let's calculate the kinetic energy of each car

Car 1

      K₁ = ½ m v₁²

      K₁ = ½ m 2F / m

      K₁ = F

Car 2

      K₂ = ½ 2m v₂²

      K₂ = ½ 2m F / m

      K₂ = F

The kinetic energy of the two cars is the same

Let's calculate the moment

Car 1

   P₁ = m v₁

   P₁ = m √ (2F / m)

Car 2

    P₂ = 2m v²

    P₂ = 2m √(F / m)

We see that the moment of car 2 is greater than the moment of car 1

b) in this part the force is applied by t = 10 s

Acceleration is the same, let's find the speed

Car1

          v = v₀ + a t

          v = F / m t

          v₁ = F / m 10

Car 2

           v₂ = F / 2m 10

           v₂ = F / m 5

Let's calculate the kinetic energy of each car

Car 1

           K₁ = ½ m v₁²

           K₁ = ½ m (F / m 10)²

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Car2

         K₂ = ½ 2m v₂²

         K₂ = m (F / m 5)²

         K₂ = 25 F² / m

In this case we see that the kinetic energy of car 1 is greater than that of car 2

Let's calculate the moment

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         P₁ = 10 F

 

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        P₂ = 2m v₂

        P₂ = 2m F / m 5

        P₂ = 10 F

In this case the moment of the two cars is the same

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