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bagirrra123 [75]
3 years ago
15

In physics, power is __________.

Physics
2 answers:
Sedaia [141]3 years ago
4 0

Answer:

The answer is A

Explanation:

erastovalidia [21]3 years ago
4 0

Answer:

A. How fast something works.

Explanation:

Power in physics is the rate at which work is done.

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If a cliff jumper leaps off the edge of a 100m cliff, how long does she fall before hitting the water? (assume zero air resistan
andrew-mc [135]
<h2>Answer:</h2>

<em>Hello, </em>

<h3><u>QUESTION)</u></h3>

Assuming that the initial velocity of the jumper is zero, on Earth any freely falling object has an acceleration of 9.8 m/s².  

<em>✔ We have : a = v/Δt = ⇔ Δt = v/a </em>

  • Δt = (√2xgxh)/9,8
  • Δt = (14√10)/9,8
  • Δt ≈ 4,5 s

4 0
3 years ago
What would be the indication if there is a high frequency and pitch?
il63 [147K]
Higher pitched sounds produce waves which are closer together than for lower pitched sounds. A smaller triangle or cymbal will make a relatively higher pitch note
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3 years ago
Energy has many different forms: ____________ energy is the energy of motion, for every moving object and living thing; ________
jok3333 [9.3K]

Answer:

1.kinetic

2.potential

3.directly

4.mass

6 0
3 years ago
A book has been lifted 1.5 meters into the air giving it 30J of potential energy. How much force was used to lift it
joja [24]

Answer:

Force on the object is 20 N

Explanation:

As we know that work done to raise the book from initial position to final position is known as potential energy stored in it

So here we know that

U = F.s

here we know that

U = 30 J

s = displacement = 1.5 m

so we have

30 = F(1.5)

F = 20 N

6 0
3 years ago
The near point of an eye is 48.5 cm. A corrective lens is to be used to allow this eye to focus clearly on objects at the distan
Masteriza [31]

Answer:

The focal length of the lens should be -51.5 cm (a concave lens).

Explanation:

The purpose of the lens is to make objects at 48.5 cm appear at the healthy near point. The healthy near point is 25.0 cm.

We use the lens formula

\dfrac{1}{f} = \dfrac{1}{u} + \dfrac{1}{v}

where <em>f</em> = focal length, <em>u</em> = object distance and <em>v</em> = image distance.

In this case, <em>u</em> = 48.5 cm and <em>v</em> = -25.0 cm.

<em>v</em> is negative because the image is virtual an not real. (Here, we are using the real-is-positive sign convention)

\dfrac{1}{f} = \dfrac{1}{48.5} + \dfrac{1}{-25.0} = -\dfrac{23.5}{1212.5}

f = -51.5

The negative sign indicates the lens is concave.

3 0
3 years ago
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