That is FALSE. The equation to calculate the charges has a distance component that is in the denominator which means that it is inversely proportional (as the distance os greater the force is smaller)
Answer:
E
Explanation:
all are proof Eeeeeeeeeee
D) Partial charge cannot be removed, because charge is a discrete quantity that may exist only at certain values
Explanation:
The electric charge of an object is a property of the object that is related to the ability of the object to experience/exert an electric force: if the object is electrically charge, then it is attracted or repelled by other electrically charged object.
The electric charge of an object depends on the amount of charged particles it has on it. In particular, the fundamental particles that carry electric charge are:
- Protons: they carry electric charge of +e
- Electrons: they carry electric charge of -e
Where "e" is the fundamental charge (
). Therefore, one proton carry a charge of +e and one electron carry a charge of -e.
An electron is a fundamental particle: this means that it cannot be divided into smaller particles. This also means that it is not possible to remove part of the charge of the electron: in fact, it is said that electric charge exists only as discrete values, being a multiple of
. Therefore, the correct statement is
D) Partial charge cannot be removed, because charge is a discrete quantity that may exist only at certain values
Learn more about particles:
brainly.com/question/2757829
#LearnwithBrainly
Answer:
See the answers below.
Explanation:
We will solve this problem by calculating each part separately.
A 500 W hair dyer is used to dry hair for 6 minutes a day for 3 days.
Energy can be calculated by multiplying the value of the power of the equipment by the amount of time of use.
![500 [W]*[\frac{6min}{1day} ]*[\frac{1day}{24hr} ]*[\frac{1hr}{60min} ]=2.083 [W]](https://tex.z-dn.net/?f=500%20%5BW%5D%2A%5B%5Cfrac%7B6min%7D%7B1day%7D%20%5D%2A%5B%5Cfrac%7B1day%7D%7B24hr%7D%20%5D%2A%5B%5Cfrac%7B1hr%7D%7B60min%7D%20%5D%3D2.083%20%5BW%5D)
The cots of electricity is 5.6 cents per kWh. How much would it cost to operate the laptop for 24 hours a day for one week?
We know that the power of the latop is 75 [W], then we can calculate the cost, multiplying the value of the power by the value of the cost by the time of use of the computer.
![0.075[kW]*5.6[\frac{cents}{kw*h}}]*[\frac{24hr}{1day}]*[1week]*[\frac{7days}{1week} ]=70.56 [cents]](https://tex.z-dn.net/?f=0.075%5BkW%5D%2A5.6%5B%5Cfrac%7Bcents%7D%7Bkw%2Ah%7D%7D%5D%2A%5B%5Cfrac%7B24hr%7D%7B1day%7D%5D%2A%5B1week%5D%2A%5B%5Cfrac%7B7days%7D%7B1week%7D%20%5D%3D70.56%20%5Bcents%5D)
A toaster oven is 85% efficient. It uses 1200 J of energy. How much thermal energy is it producing?
Efficiency is defined as the relationship between the energy obtained on the energy delivered. Almost always the energy delivered is greater than the energy obtained (first law of thermodynamics).
Therefore.
![Effic = E_{obtained}/E_{delivered}\\0.85=E_{obtained}/1200\\E_{obtained}=1020[J]](https://tex.z-dn.net/?f=Effic%20%3D%20E_%7Bobtained%7D%2FE_%7Bdelivered%7D%5C%5C0.85%3DE_%7Bobtained%7D%2F1200%5C%5CE_%7Bobtained%7D%3D1020%5BJ%5D)
A solar eclipse will be visible over a wide area of the north polar region
on Friday, March 20.
England is not in the path of totality, but it's close enough so that a large
part of the sun will be covered, and it will be a spectacular sight.
For Londoners, the eclipse begins Friday morning at 8:25 AM,when the
moon just begins to eat away at the sun's edge. It advances slowly, as more
and more of the sun disappears, and reaches maximum at 9:31 AM. Then
the obscured part of the sun begins to shrink, and the complete disk is
restored by the end of the eclipse at 10:41AM, after a period of 2 hours
16 minutes during which part of the sun appears to be missing.
The catch in observing the eclipse is:
<em><u>YOU MUST NOT LOOK AT THE SUN</u></em>.
Staring at the sun for a period of time can cause permanent damage to
your vision, even though <em><u>you don't feel it while it's happening</u></em>.
This is not a useful place to try and give you complete instructions or
suggestions for observing the sun over a period of hours. Please look
in your local newspaper, or search online for phrases like "safe eclipse
viewing".