D. potential energy, because there is a bunch of water pent up, essentially stationary, waiting to roll down the steep mountain from the peak, so to say. if the dam were to be removed it would become kinetic.

4 0

3 years ago

Read 2 more answers

You are trying to overhear a juicy conversation, but from your distance of 23.0 m , it sounds like only an average whisper of 40

Len [333]

If its asking the distance for the 65 db then use a proportion, if otherwise pleas clarify. It sounds like a pretty juicy conversation.

4 0

3 years ago

Two electric charges A and B were placed facing each other at a distance of separation "r". The common electrostatic force betwe

lutik1710 [3]

Answer:

From the formula of force:

$F = \frac{kAB}{ {r}^{2} } \\$

since AB and k are constants:

$F \: \alpha \: \frac{1}{ {r}^{2} } \\ \\ F = \frac{x}{ {r}^{2} }$

x is a constant of proportionality

• when force is 4N, separation distance is 1

$4 = \frac{x}{1} \\ x = 4$

therefore, equation becomes

$F = \frac{4}{ {r}^{2} } \\$

when r is doubled, r becomes 2. find F:

$F = \frac{4}{ {2}^{2} } \\ \\ F = \frac{4}{4} \\ \\ { \underline{force \: is \: 1N}}$

5 0

2 years ago

The near point of a farsighted person's uncorrected eyes is 80 cm. what power contact lens should be used to move the near point

makvit [3.9K]

To solve this problem, we will get f and then we will use it to calculate the power.

So, for this farsighted person,
do = 25 cm and di = -80
Therefore:
1/f = (1/25) + (1/-80) = 0.00275 = 0.275 m

Power = 1/f = 1/0.275 = +3.6363 Diopeters.
This means that the lens is converging.

5 0

3 years ago