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3 years ago

12

Two charges repel each other with a force. If the magnitude of both charges double, but the distance between them remains consta

nt, the force between them will
A. Double
B. Triple
C. Quadruple
D. Remain the same

1 answer:

-Dominant- [34]3 years ago

4 0

Answer:

C.

Explanation:

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4 years ago

Read 2 more answers

2. At noon, ship A was 12 nautical miles due north of ship B. Ship A was sailing south at 12 knots (nautical miles per hour; a n

AleksandrR [38]

Answer:

No

Explanation:

Let the reference origin be location of ship B in the beginning. We can then create the equation of motion for ship A and ship B in term of time t (hour):

A = 12 - 12t

B = 9t

Since the 2 ship motions are perpendicular with each other, we can calculate the distance between 2 ships in term of t

$d = \sqrt{A^2 + B^2} = \sqrt{(12 - 12t)^2 + (9t)^2}$

For the ships to sight each other, distance must be 5 or smaller

$d \leq 5$

$\sqrt{(12 - 12t)^2 + (9t)^2} \leq 5$

$(12 - 12t)^2 + (9t)^2 \leq 25$

$144t^2 - 288t + 144 + 81t^2 - 25 \leq 0$

$225t^2 - 288t + 119 \leq 0$

$(15t)^2 - (2*15*9.6)t + 9.6^2 + 26.84 \leq 0$

$(15t^2 - 9.6)^2 + 26.84 \leq 0$

Since $(15t^2 - 9.6)^2 \geq 0$ then

$(15t^2 - 9.6)^2 + 26.84 > 0$

So our equation has no solution, the answer is no, the 2 ships never sight each other.

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3 years ago