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4 years ago

Please help!

Physics

2 answers:

Afina-wow [57]4 years ago

7 0

Answer: The correct answer is C.

Hi, here you can see that the forces are in the same line, but they are opposite to each other.

So if you want to obtain the net force, you must do a sum of all the forces. in this case we have 250 N to east, and 150 N to west, the first step will be have the same axis, in this case West is the opposite of East (tooking east as the positive direction here), then 150N west = -150N to east.

Now doing the sum we obtain Ft = F1 + F2 = 250 N - 150 N = 100N to the east.

so the correct answer is C.

Ad libitum [116K]4 years ago

6 0

$250 - 150 = 10 \\ so \: the \: net \: force \: is \: 100n \: east$

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Three equal charge 1.8*10^-8 each are located at the corner of an equilateral triangle ABC side 10cm.calculate the electric pote

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Answer:

If all these three charges are positive with a magnitude of $1.8 \times 10^{-8}\; \rm C$ each, the electric potential at the midpoint of segment $\rm AB$ would be approximately $8.3 \times 10^{3}\; \rm V$ .

Explanation:

Convert the unit of the length of each side of this triangle to meters: $10\; \rm cm = 0.10\; \rm m$ .

Distance between the midpoint of $\rm AB$ and each of the three charges:

$d({\rm A}) = 0.050\; \rm m$ .
$d({\rm B}) = 0.050\; \rm m$ .
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Electric potential due to the charge at $\rm B$ : $\displaystyle \frac{k\, q}{d({\rm B})}$ .

Electric potential due to the charge at $\rm A$ : $\displaystyle \frac{k\, q}{d({\rm C})}$ .

While forces are vectors, electric potentials are scalars. When more than one electric fields are superposed over one another, the resultant electric potential at some point would be the scalar sum of the electric potential at that position due to each of these fields.

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3 years ago

Ten high-technology batteries are tested for 200 hours each. One failed at 20 hours; another failed at 140 hours; all others com

Bas_tet [7]

Answer:

Failure rate = 20%

MTBF = 880 hours

Explanation:

given data

batteries = 10

tested = 200 hours

one failed = 20 hours

another fail at = 140 hours

solution

we know that Mean Time between Failures is express as = (Total up time) ÷ (number of breakdowns) ....................1

so here Total up time will be

Total up time = 200 × 10

Total up time = 2000

and here

Number of breakdown = 1 at 20 hour and another at 140 hour = 2

so it will be = (Total up time) ÷ (number of breakdowns) .......2

= $\frac{2000}{2}$ = 1000

so here gap between occurrences is

gap between occurrences= 140 - 20

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and

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and

Failure rate (FR) will be

Failure rate (FR) = 1 ÷ MTBF ................3

Failure rate (FR) = R÷T ......................4

as here R is the number of failures and T is total time

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