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Zarrin [17]
3 years ago
11

Which of the following statements about Gaussʹs law are correct? (There may be more than one correct choice.) Question 3 options

: Gaussʹs law is valid only for symmetric charge distributions, such as spheres and cylinders. If there is no charge inside of a Gaussian surface, the electric field must be zero at points of that surface. Only charge enclosed within a Gaussian surface can produce an electric field at points on that surface. If a Gaussian surface is completely inside an electrostatic conductor, the electric field must always be zero at all points on that surface.
Physics
1 answer:
irina1246 [14]3 years ago
3 0

Answer:

If a Gaussian surface is completely inside an electrostatic conductor, the electric field must always be zero at all points on that surface.

Explanation:

Option A is incorrect because, given this case, it is easier to calculate the field.

Option B is incorrect because, in a situation where the surface is placed inside a uniform field, option B is violated

Option C is also incorrect because it is possible to be a field from outside charges, but there will be an absence of net flux through the surface from these.

Hence, option D is the correct answer. "If a Gaussian surface is completely inside an electrostatic conductor, the electric field must always be zero at all points on that surface."

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Answer:

True

Explanation:

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3 years ago
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Final exam answers for physical science apex?
Tju [1.3M]
First table:

1. For height, we usually use cm. If it is a must to use the conversion to solve for number 1, we can first convert the ft into m then to cm and then convert the in to cm and then add the values.

5ftx\frac{0.305m}{1ft}  = 1.525mx\frac{100cm}{1m} =152.5cm

then we convert the remaining 2in into cm and add it up:
2inx\frac{2.54cm}{1in}=5.08cm

152.5cm + 5.08cm = 157.58cm or 157.6cm

Another way to do this is to convert the ft to in and add it up to the remaining 2 in then convert to cm. The answer would be more or less the same. 

2. For weight, we will convert lbs to kg. 

105lbsx\frac{1kg}{2.2lbs} =  \frac{105kg}{2.2} = 47.7kg

3. For distance we will convert mi to km. 

1.1mix\frac{1.61km}{1mi} = 1.8km

4. The unit mph is miles per hour and we will want to convert that into kph or kilometers per hour. You will use the same method that you did above to get it. We do not need to convert hours anymore, so it will stay as is.

65 \frac{mi}{hr}x\frac{1.61km}{1mi} = 104.7\frac{km}{hr}

5. Now for the next one it is very straightforward because 1V = 1V no matter where you go. So if the given is 220V then that means that the converted value is also 220V. 

6. Now to convert Fahrenheit to Celsius, then all you have to do is use the formula given above and fill in what you know. 

The given is  70°F

°C = \frac{5}{9} (F - 32)
     = \frac{5}{9} (70 - 32)
     = \frac{5}{9} (38)
     = \frac{190}{9}
     = 21.1°C

The second table is very easy, all you need to do is convert the values to scientific notation or the other way around. All you need to do is move the decimal till you have a coefficient that is between 1 to 10 and then count how many times you moved the decimal. That will be the number or the exponent of 10. 

1. $150,000 = $1.5 x 10^5

If you moved the decimal to the left, the exponent will be positive. 

Now the other way around. To convert scientific notation to standard form, just move the decimal the number of places indicated by the exponent of 10. If it is positive, you will move to the right and if it is negative, you will move to the left. 

2. 2.59x10^6 = 2,590,000m

Now notice that the next value is less than 1. This means you will be moving the the decimal to the right to make it have a coefficient of 1 to 10. You need to move it 3 times to the right so that means the exponent will be negative. 

3. 0.005in = 5.0 x 10^(-3)in

Now do not forget to put in your units. In math or even in science the unit is EXTREMELY IMPORTANT. 

Hope this helped you and good luck!
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A rock dropped into a pond produces a wave that takes 11.3 s to reach the opposite shore, 26.5 m away. the distance between cons
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f= \frac{v}{\lambda}= \frac{2.35 m/s}{3 m}=0.78 Hz
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