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Ann [662]
3 years ago
14

In the ENGR 10 lab (E391), there are 50 long light bulbs (P=100 W) and 30 regular bulbs (P=60 W). How much energy is consumed li

ghting the lab during the 3 hour lab period. Calculate the amount of energy saved if all bulbs are replaced by CFL (spiral) bulbs (P=25W).
Physics
1 answer:
Alenkinab [10]3 years ago
6 0

Answer:

Total energy saving will be 0.8 KWH

Explanation:

We have given there are 50 long light bulbs of power 100 W so total power of 50 bulb = 100×50 = 5000 W = 5 KW

30 bulbs are of power 60 W

So total power of 30 bulbs = 30×60 = 1800 W = 1.8 KW

Total power of 80 bulbs = 1.8+5 = 6.8 KW

Total time = 3 hour

We know that energy E=power\times time=6.8\times 3=20.4KWH

Now power of each CFL bulb = 25 W

So power of 80 bulbs = 80×25 = 2000 W = 2 KW

Energy of 80 bulbs = 2×3 = 6 KWH

So total energy saving = 6.8-6 = 0.8 KWH

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Imma say go with C (:
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Here on Earth you hang a mass from a vertical spring and start it oscillating with amplitude 1.9 cm. You observe that it takes 3
Vlada [557]

Answer:

T = 3.23 s

Explanation:

In the simple harmonic movement of a spring with a mass the angular velocity is given by

               w = √ K / m

With the initial data let's look for the ratio k / m

The angular velocity is related to the frequency and period

           w = 2π f = 2π / T

            2π / T = √ k / m

            k₀ / m₀ = (2π / T)²

            k₀ / m₀ = (2π / 3.0)²

            k₀ / m₀ = 4.3865

The period on the new planet is

          2π / T = √ k / m

           T = 2π √ m / k

In this case the amounts are

           m = 6 m₀

           k = 10 k₀

We replace

            T = 2π√6m₀ / 10k₀

            T = 2π √6/10 √m₀ / k₀

            T = 2π √ 0.6  √1 / 4.3865

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5 0
3 years ago
The rate at which heat enters an air conditioned building is often roughly proportional to the difference in temperature between
erma4kov [3.2K]

Answer:

Considering first question

    Generally the coefficient of performance of the air condition  is mathematically represented as

   COP  =  \frac{T_i}{T_o - T_i}

Here T_i is the inside temperature

while  T_o is the outside temperature

What this coefficient of performance represent is the amount of heat the air condition can remove with 1 unit of electricity

So it implies that the air condition removes   \frac{T_i}{T_o - T_i} heat with 1 unit of electricity

Now from the question we are told that the rate at which heat enters an air conditioned building is often roughly proportional to the difference in temperature between inside and outside. This can be mathematically represented as

         Q \ \alpha \ (T_o - T_i)

=>        Q= k (T_o - T_i)

Here k is the constant of proportionality

So  

    since  1 unit of electricity  removes   \frac{T_i}{T_o - T_i}  amount of heat

   E  unit of electricity will remove  Q= k (T_o - T_i)

So

      E =  \frac{k(T_o - T_i)}{\frac{T_i}{ T_h - T_i} }

=>   E = \frac{k}{T_i} (T_o - T_i)^2

given that  \frac{k}{T_i} is constant

    =>  E \  \alpha  \  (T_o - T_i)^2

From this above equation we see that the  electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square of the temperature difference.

 Considering the  second question

Assuming that  T_i   =  30 ^oC

 and      T_o  =  40 ^oC

Hence  

     E = K (T_o - T_i)^2

Here K stand for a constant

So  

        E = K (40 -  30)^2

=>      E = 100K

Now if  the  T_i   =  20 ^oC

Then

       E = K (40 -  20)^2

=>      E = 400 \ K

So  from this see that the electricity require (cost of powering and operating the air conditioner)when the inside temperature is low  is  much higher than the electricity required when the inside temperature is higher

Considering the  third question

Now in the case where the  heat that enters the building is at a rate proportional to the square-root of the temperature difference between inside and outside

We have that

       Q = k (T_o - T_i )^{\frac{1}{2} }

So

       E =  \frac{k (T_o - T_i )^{\frac{1}{2} }}{\frac{T_i}{T_o - T_i} }

=>   E =  \frac{k}{T_i} * (T_o - T_i) ^{\frac{3}{2} }

Assuming \frac{k}{T_i} is a constant

Then  

     E \ \alpha \ (T_o - T_i)^{\frac{3}{2} }

From this above equation we see that the  electricity required(cost of powering and operating the air conditioner) is approximately proportional to the square root  of the cube of the  temperature difference.

   

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2 years ago
Which warning sign suggests being bullied?
Vikentia [17]

Answer:

when you tell someone over and over to stop bothering you and they dont so i think you should tell a teacher

Explanation:

6 0
3 years ago
A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 11.1 rad/s in 2.99 s.(a) fi
elena55 [62]
The angular acceleration of a rotating object is given by
\alpha =  \frac{\omega_f - \omega_i}{\Delta t}
where
\omega_f is the final angular speed of the object
\omega_i is its initial angular speed
\Delta t is the time taken to accelerate

For the wheel in our problem, \omega_f=11.1 rad/s, \omega_i = 0 and \Delta t=2.99 s, so its angular acceleration is
\alpha= \frac{11.1 rad/s-0}{2.99 s}=3.71 rad/s^2
8 0
2 years ago
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