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2 years ago

5

How much work is required to move a 1 nC charge from an electric potential of 0 V to a potential of 18V?

1 answer:

jolli1 [7]2 years ago

3 0

ni work is required by an external force because this is the natural movement of the charge

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Ok plz answer and tell me how to do it

kirza4 [7]

Answer: 25N

method: total force in the right hand direction is 100N and total force in the left hand direction is 125N. To get the net force, we add forces if they are in the same direction and substract if they are in opposite directions. since 100N and 125N are in opposite directions, we substract the larger value from the smaller value. Then we get 25N in the left hand direction as the final answer.

4 0

3 years ago

386.5<br>- 129.18 what is it

algol [13]

Answer:

257.32

Explanation:

I jus worked it out on paper. Brainliest please?

5 0

3 years ago

Read 2 more answers

Why can't a hot air balloon go to higher layers?

Elena L [17]

If you go to high you’ll run out of oxygen and possibly be blown off due to high winds.

5 0

3 years ago

It would really help if anyone could answers please and thanks

mojhsa [17]

well it would be A because 55 degrees is going strait well 75 is going literally straight up

4 0

2 years ago

Read 2 more answers

A boat race runs along a triangular course marked by buoys A, B, and C. The race starts with the boats headed west for 3700 mete

ale4655 [162]

Answer:

The last two bearings are

49.50° and 104.02°

Explanation:

Applying the Law of cosine (refer to the figure attached):

we have

x² = y² + z² - 2yz × cosX

here,

x, y and z represents the lengths of sides opposite to the angels X,Y and Z.

Thus we have,

$cos X=\frac{x^2-y^2-z^2}{-2yz}$

or

$cos X=\frac{y^2 + z^2-x^2}{2yz}$

substituting the values in the equation we get,

$cos X=\frac{2900^2 + 3700^2-1700^2}{2\times 2900\times 3700}$

or

$cos X=0.8951$

or

X = 26.47°

similarly,

$cos Y=\frac{1700^2 + 3700^2-2900^2}{2\times 1700\times 3700}$

or

$cos Y=0.649$

or

Y = 49.50°

Consequently, the angel Z = 180° - 49.50 - 26.47 = 104.02°

The bearing of 2 last legs of race are angels Y and Z.

7 0

3 years ago