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siniylev [52]
3 years ago
6

What is the x-component of the electric field at (x, y) = (0 m, 4 m)??

Mathematics
1 answer:
Elza [17]3 years ago
8 0
(a)Two charges are placed on the x-axis: one is placed at x = 3 m and the other is at x = -3 m. The magnitude of both charges is 2.3 µC but the blue one (at x = -3 m) is positive while the red one (at x = +3 m) is negative.

What are the x- and y-components of the electric field at (x, y) = (0 m, +4 m)?

(b) Now the positive and negative charge switch places. The magnitude of the charges is still 2.3 µC where the blue one (now at x = +3 m) is positive and the red one (now at x = -3 m) is negative.

What are the x- and y-components of the electric field at (x, y) = (0 m, +4 m)?

(c) Now both charges (still at x = -3 m and x = +3 m) are positive. The magnitude of both charges is still 2.3 µC.

What are the x- and y-components of the electric field at (x, y) = (0 m, +4 m)?

(d) Finally, both charges (still at x = -3 m and x = +3 m) are negative. The magnitude of both charges is still 2.3 µC.
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Firdavs [7]
The answer is 4.

The formula for the area of a rectangle is A = lw. Substitute, so then it's 28 = (3x-5) * x, and you evaluate.

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8 0
3 years ago
Quickly
Gemiola [76]
B(-4; 1)

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6 0
3 years ago
In the kite , AK=9 , JK=15 , and AM=16 .
Veronika [31]

Answer:

The Length of JM is 20.

Step-by-step explanation:

Given,

JKLM is a kite in which JL and KM are the diagonals that intersect at point A.

Length of AK = 9    

Length of JK = 15  

Length of AM = 16

Solution,

Since JKLM is a kite. And JL and KM are the diagonals.

And we know that the diagonals of a kite perpendicularly bisects each other.

So, JL ⊥ KM.

Therefore ΔJAK is aright angled triangle.

Now according to Pythagoras Theorem which states that;

"The square of the hypotenuse is equal to the sum of the square of base and square of perpendicular".

JK^2=KA^2+AJ^2

On putting the values, we get;

(15)^2=9^2+AJ^2\\\\225=81+AJ^2\\\\AJ^2=225-81=144

On taking square root onboth side, we get;

\sqrt{AJ^2} =\sqrt{144}\\\\AJ=12

Again By Pythagoras Theorem,

AJ^2+AM^2=JM^2

On putting the values, we get;

JM^2=(12)^2+(16)^2\\\\JM^2=144+256=400

On taking square root onboth side, we get;

\sqrt {JM^2}=\sqrt{400}\\\\JM=20

Hence The Length of JM is 20.

4 0
3 years ago
Is it true that a line with slope 1 always passes through the origin? Explain your reasoning.
elena55 [62]

Well, yes and no.

Yes, because the straight line needs to pass through a number greater than 0, and 1 is obviously greater than 0.

However, y = x + 1 is not the same as y = x.

hope this helps.! let me know if it's in any way confusing..

I'd be happy to help!

5 0
3 years ago
find the consecutive even integers such that three times the larger number is 30 more than the smaller one. help :')
Minchanka [31]

Answer:

12 and 14

Step-by-step explanation:

Let the even integers be x and x+2

three times the larger number is expressed as 3(x+2)

30 more than the smaller one is x + 30

Equating both expressions

3(x+2) = x +30

Find x

3x+6 = x+30

3x-x = 30-6

2x = 24

x = 12

The second integer is 12+2 = 14

Hence the required integers are 12 and 14

4 0
3 years ago
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