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Tema [17]
3 years ago
10

One way to conserve energy is to replace incandescent light bulbs with compact fluorescent bulbs. The fluorescent bulb typically

uses 25% of the energy of an incandescent bulb of comparable brightness typically lasts about 12 times longer. a)How much would you save by replacing a 100-watt incandescent bulb with a compact fluorescent bulb over the 12,000 hour lifetime of the bulb if the electricity cost 0.08$ per kwh (kilowatt hour)
Physics
1 answer:
NemiM [27]3 years ago
4 0

Answer:

     cos to = $ 24

Explanation:

When replacing the bulb only 25% of the energy is used, therefore

          W = 0.25 100

          W = 25 W

Let's look for the energy in the life of the bulb

          E = 25 10⁻³ 12000

          E = 300 Kwh

now we can calculate the cost using a direct proportion rule.

          Cost = 0.08 300

          cos to = $ 24

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Weight of an object various replaces on the earth.why?​
german

Answer:

Weight change from place to place because it is a strong function of gravity. Recall that mass is universally constant. Hence weight varies becuas gravity varies. ... Other reason for variation of weight across the earth's surface are, rotation of the earth, altitude and local topography of the area.

5 0
3 years ago
If we increase the amount of work being done, and all other factors remain the same, the amount of power would
BlackZzzverrR [31]
The amount of power would increase
P=W/t
P is directly proportional to W
4 0
4 years ago
Please HELP me label these symbols!!!!
Xelga [282]

Answer:

A is a battery

B is a light

C is a resistor

D is a closed switch

thank you for this question, I love circuit diagrams

Explanation:

4 0
3 years ago
In the Bohr model of the hydrogen atom, an electron orbits a proton in a circular orbit of radius 0.53 X 10^-10 m. What is elect
maxonik [38]

Answer:

a) 27.2 V

b)27.2 V

Explanation:

Charge of the electron =charge of the proton = q = 1.6 × 10⁻¹⁹ C

Radius = r = 0.53×10⁻¹⁰ m

Electric Potential = V = k q/r

k = 9 ×10⁹ N m²/C² = Coulomb's constant.

V = (9 ×10⁹)(1.6 × 10⁻¹⁹)/( 0.53×10⁻¹⁰) = 27.2 V

b) Potential Energy of the electron =  k q × q / r

   = [(9 ×10⁹)(1.6 × 10⁻¹⁹)(1.6 × 10⁻¹⁹) / (0.53×10⁻¹⁰)] / (1.6 × 10⁻¹⁹) eV,

since 1 electron volt = (1.6 × 10⁻¹⁹)joules

   = 27.2 eV

7 0
4 years ago
What is the net force (magnitude and direction) of the system below?
Crank

The net force (magnitude and direction) of the system given in the question is 40 N horizontal to the right

<h3>How to determine the net force</h3>

Case 1 (Net force between up and downward force)

  • Force up (Fu) = 50 N
  • Force down (Fd) = 30 N
  • Net force 1 (F1) = ?

F1 = Fu - Fd

F1 = 50 - 30

F1 = 20 N up

Case 2 (Net force between right and left)

  • Force right (Fr) = 60 N
  • Force left (Fl) = 20 N
  • Net force 2 (F2) = ?

F2 = Fr - Fl

F2 = 60 - 20

F2 = 40 N right

SUMMARY

  • Net force between up and down = 20 N up
  • Net force between right and left = 40 N right

From the above, the net force between right and left (i.e 40 N) is greater than the net force between up and down (i.e 20 N)

Thus, the net force of the system will be 40 N horizontal to the right

Learn more about force:

brainly.com/question/14361879

#SPJ1

4 0
2 years ago
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