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Contact [7]
3 years ago
9

A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 10.5 cm, giving it a cha

rge of -22.3 µC.. . (a) Find the magnitude of the electric field just inside the paint layer. (b) Find the magnitude of the electric field just outside the paint layer. (c) Find the magnitude of the electric field 5.00 cm outside the surface of the paint layer.

Physics
2 answers:
soldier1979 [14.2K]3 years ago
8 0
The formula that will be used in this problem is E = q/ 4pi*r^2 z where z is the elctric charge constant equal to 8.854  *10 ^-12. The magnitude using r equal to 0.0525 m and q equal to -22.3 *10^-6 C is equal to -22.3 *10^-6/ 4pi*(0.0525)^2 *8.854  *10 ^-12 or equal to -7.272 *10 ^7. The magnitude 5 cm outside the surface is -22.3 *10^-6<span>/ 4pi*(0.0525+0.05)^2 *8.854  *10 ^-12 equal to -1.908 *10^7. 

</span>
olya-2409 [2.1K]3 years ago
8 0

Since it is hard to write the mathematics equations here so i would like to share it in a picture

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Dima020 [189]

Explanation:

Commercially available batteries use a variety of metals and electrolytes. Anodes can be made of zinc, aluminum, lithium, cadmium, iron, metallic lead, lanthanide, or graphite. Cathodes can be made of manganese dioxide, mercuric oxide, nickel oxyhydroxide, lead dioxide or lithium oxide. Potassium hydroxide is the electrolyte used in most battery types, but some batteries use ammonium or zinc chloride, thionyl chloride, sulfuric acid or lithiated metal oxides. The exact combination varies by battery type. For example, common single-use alkaline batteries use a zinc anode, a manganese dioxide cathode, and potassium hydroxide as the electrolyt

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2 years ago
The graph below shows the speed of a car as it drives along a racetrack.
nordsb [41]

Answer:

Average speed = distance/time

From 1 to 9 seconds:

Distance covered = 1 - 0.2 = 0.8 km

Time = 9 - 1 = 8 sec

Average speed = 0.8 km / 8 sec

Average speed = 0.1 km/s .

The average speed for the whole test is 1.6 km / 20 sec = 0.08 km/sec. A graph of speed vs time would average out as a horizontal line at 0.08 km/sec from 1 sec to 21 sec. The area under it would be (0.08 km/s) x (20 sec) = 1.6 km.

Surprise surprise ! The area under a speed/time graph is the distance covered during that time !

In closing, I want to express my gratitude for the gracious bounty of 3 points with which I have been showered. Moreover, the green breadcrust and tepid cloudy water have also been refreshing.

Explanation:

4 0
2 years ago
A force acts on a 5kg object at rest. How fast will the object accelerate on a frictionless surface?
MakcuM [25]

Answer: The answer is C.) 25 m/s^2.

Explanation: If you input 5 as s, you would have to use the exponent 2.  This means that you have to multiply 5 by 5.  5 x 5= 25.  

Edit: Also, because the surface is frictionless, it will make the object go faster too.  Nothing can really slow it down unless something blocks it.

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3 years ago
When the air resistance can be ignored the velocity of an object dropped initially from rest is given by the following equation
ad-work [718]

Answer:

I am confused of your question. Do you want final velocity? To get final velocity, use (initial V)+(Gravity*Time)

Explanation:

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3 years ago
Read 2 more answers
if you are in Birmingham AL, and you want to use your cell phone to talk to your cousin in Houston, TX, what must occur in order
lesantik [10]

Answer:

For communication to be communicated between you and your cousin who is in Houston,Texas, when a call is made by you, a request is made to the specific phone, and the telephone tower will get the request from the mobile phone. Then a signal is sent via a transmitter underground, by this the satellite communicates with the local receiver in Houston, that is linked to the local tower over there, the tower would request for your cousin's number and connects the two of you, once a link is set.

Explanation:

From the example stated, what is required for such for a far distance, is a communication satellite link.

When a call is made by you, the a connection request is sent to the specified phone.The telephone tower receives the request from The mobile phone. The local tower(Birmingham,Al) is linked to a ground transmitter by the means of a Fiber optical cable.

A signal is sent to satellite via the ground transmitter.The satellite then set's off the local receiver in (Houston,Texas) which on it's end is connected to the local tower there. This tower then ask for your cousin's mobile for a call that will be incoming, a link is set, once he/she receives the call, from there a conversation can be done.

3 0
3 years ago
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