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11111nata11111 [884]
3 years ago
7

HELP PLEASE!!!! What is the net force on this object?

Physics
2 answers:
Ede4ka [16]3 years ago
4 0

The 20N straight up exactly cancels the 20N straight down.  So we can see that there's no net <em><u>vertical</u></em> force on the object, and if there's ANY net force on it at all, it's going to be horizontal.

The 14N force toward the left can't cancel the whole 22N toward the right.  It can only cancel 14N of it, and then there's still another 8N more toward the right.

So the net force on the object is <em>8N toward the right</em>.

rusak2 [61]3 years ago
3 0

the answer to this question is 8 newtons

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Calculate the electric field at one corner of a square 50 cm on a side if the other corners are occupied by 250x10-7C (charges)
SIZIF [17.4K]

The electric field at one corner of a square is 1614217 N/C.

Explanation:

The distance between x and y direction diagonals.

As per the given details the distance between diagonals is calculated as

0.5² + 0.5² = c²  =>  c = 0.707 m

Charge to the right:  In x direction

In order to find the electric charge towards x direction

we use e = kq/r² formula

As 'k' is coulomb's constant it's value is 9 x 10^{9} N m²/C²

e = (9 x 10^{9})(250 x 10^{-7}) / (0.5)²

e = 9 x 10^{5} N/C

Charge diagonal:

e = kq/r²

e = [(9 x 10^{9})(250 x 10^{-7}) / (0.707)²] cos 45

e = 225000√2 N/C

X direction sum = 1218198 N/C.

Similarly as shown in x direction the charge is same for y direction also

Charge below:  For y direction

e = kq/r²

e = (9 x 10^{9})(250 x 10^{-7}) / (0.5)²

e = 9 x 10^{5} N/C

Charge diagonal:

e = kq/r²

e = [(9 x 10^{9})(250 x 10^{-7}) / (0.5)²] sin 45

e = 159099 N/C

Y direction sum = 1059099 N/C

Resultant electric field strength:

1218198 ² + 1059099² = e²

e = 1614217 N/C [45 degrees below the horizontal]

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Answer:

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Explanation:

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