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omeli [17]
3 years ago
6

one deterrent to burglary is to leave your front porch light on all the time if your fixture contains the 23 watt compact fluore

scent bulb at 120 volts and your local power utility cell's energy at ten cents per kilowatt-hour how much will it cost to leave the bulb on for the entire month
Physics
1 answer:
sweet [91]3 years ago
6 0

Answer:

Cost = 165.6 cents = $1.656

Explanation:

First, we will calculate the total energy consumed by the bulb in a month:

E=Pt\\

where,

E = Energy Consumed = ?

P = power = 23 W

t = time = (1 month)(30 days/1 month)(24 h/1 day) = 720 h

Therefore,

E = (23\ W)(720\ h)\\E = 16560\ Wh = 16.56\ Kilowatt-hour

Now, the cost will be:

Cost = P(Unit\ Cost)\\Cost = (16.56\ kilowatt-hour)(10\ cents/kilowatt-hour)

<u>Cost = 165.6 cents = $1.656</u>

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Which of the following statements is consistent with Thomson’s and Millikan’s work with cathode rays and electrons? (A) Cathode
victus00 [196]

Answer:

Dalton's ideas proved foundational to modern atomic theory. However, one of his underlying assumptions was later shown to be incorrect. Dalton thought that atoms were the smallest units of matter-−minustiny, hard spheres that could not be broken down any further. This assumption persisted until experiments in physics showed that the atom was composed of even smaller particles. In this article, we will discuss some of the key experiments that led to the discovery of the electron and the nucleus.

J.J. Thomson and the discovery of the electron

In the late 19^{\text{th}}19  

th

19, start superscript, start text, t, h, end text, end superscript century, physicist J.J. Thomson began experimenting with cathode ray tubes. Cathode ray tubes are sealed glass tubes from which most of the air has been evacuated. A high voltage is applied across two electrodes at one end of the tube, which causes a beam of particles to flow from the cathode (the negatively-charged electrode) to the anode (the positively-charged electrode). The tubes are called cathode ray tubes because the particle beam or "cathode ray" originates at the cathode. The ray can be detected by painting a material known as phosphors onto the far end of the tube beyond the anode. The phosphors spark, or emit light, when impacted by the cathode ray.

A diagram of a cathode ray tube.

A diagram of a cathode ray tube.

A diagram of J.J. Thomson's cathode ray tube. The ray originates at the cathode and passes through a slit in the anode. The cathode ray is deflected away from the negatively-charged electric plate, and towards the positively-charged electric plate. The amount by which the ray was deflected by a magnetic field helped Thomson determine the mass-to-charge ratio of the particles. Image from Openstax, CC BY 4.0.

To test the properties of the particles, Thomson placed two oppositely-charged electric plates around the cathode ray. The cathode ray was deflected away from the negatively-charged electric plate and towards the positively-charged plate. This indicated that the cathode ray was composed of negatively-charged particles.

Thomson also placed two magnets on either side of the tube, and obser

Explanation:

5 0
3 years ago
A family leaves home at 1:00pm and arrives at their vacation spot 200 miles away at 5:00pm on the same day. What is their averag
Shkiper50 [21]
The answer is 40 because you just have to do 200 divided by 5
4 0
3 years ago
A 10.0 V battery is connected across two resistors in series. One resistor has resistance of 840.0 Ω and the other has resistanc
REY [17]

Answer:

Explanation:

There are a couple of ways you could do this.

The easiest is to use E*R1/(R1 + R2)

  • E = 10 volts
  • R1 = 590 ohms
  • R2 = 840 ohms

So the result would be

E_590 = 10 * 590/(590 + 840)

E_590 = 10 * 590/ (1430)

E_590 = 4.13 volts rounded.

You could do this a slightly longer way.

R = 1430 (total ohms in series.

E = 10 volts

I = ???

I = E/R

I = 10 / 1430

I = 0.00699

Now use this current to figure out the voltage drop.

E = I * R

I = 0.00699 amps

R = 590 ohms

E = 0.00699 * 590

E = 4.13 volts

Pick the way of doing it you like best.

4 0
4 years ago
Substance of an atom of 2 or more elements
horrorfan [7]
H. O is the answer hope this helps
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8 0
4 years ago
A bullet glider and a target glider both have a mass of 0.200 kg. The bullet glider is moving 0.450 m/s
Romashka [77]

Answer:

the two gliders collide, the mobile glider will transfer a bit of time to the fixed glider, which is why it comes out with a speed that is smaller than that of the bullet glider.

Explanation:

When the two gliders collide, the mobile glider will transfer a bit of time to the fixed glider, which is why it comes out with a speed that is smaller than that of the bullet glider.

Changes can occur that the gliders unite and move with a cosecant speed less than the initial one.

The whole process must be analyzed using conservation of the moment.

             p₀ = m v₀

celestines que clash case

             p_f = (m + M) v

             po = pf

             m v₀ = (n + M) v

             v = \frac{m}{m+M}

calculemos

            v= \frac{0.200}{0.200+M} 0.450

            v= 0.09 m/s

elastic shock case

           p₀ = m v₀

           p_f = m v₁ +M v₂

           p₀ = p_f

           m v₀ = m v₁ + m v₂

6 0
3 years ago
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