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hoa [83]
3 years ago
13

Light energy with very high amplitude would be perceived as ________.

Physics
2 answers:
s2008m [1.1K]3 years ago
6 0

Answer:

High amplitude is equivalent to loud sounds.

Explanation:

sashaice [31]3 years ago
5 0

Answer:

Brightness.

Explanation:

very bright.

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A marble is dropped from rest and falls for 2.3 seconds. Find its final velocity.
juin [17]

Answer:

23 m/s downward

__________________________________________________________

<em>Taking the downward direction as positive</em>

<u>We are given:</u>

Initial velocity of the marble (u) = 0 m/s

Time interval (t) = 2.3 seconds

Final velocity (v) = x m/s

<u>Solving for the Final velocity:</u>

<u>Acceleration of the Marble:</u>

We know that gravity will make the marble accelerate at a constant acceleration of 10 m/s

<u>Final velocity:</u>

v = u + at                                              [First equation of motion]

x = 0 + (10)(2.3)                                    [replacing the given values]

x = 23 m/s

Hence, after 2.3 seconds, the marble will move at a velocity of 23 m/s in the downward direction

4 0
3 years ago
Question in attachment​
zlopas [31]

3. A

4.B

3.Since the Earth rotates through two tidal “bulges” every lunar day, we experience two high and two low tides every 24 hours and 50 minutes

4.An eclipse of the Sun happens when the New Moon moves between the Sun and Earth, blocking out the Sun's rays and casting a shadow on parts of Earth

7 0
2 years ago
What is the special name for doctors who deal with cancer?
Sladkaya [172]
Oncology is the study of cancer, so the doctors are called oncologists
8 0
3 years ago
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In which situation does heating a substance generally change the temperature?
marta [7]
The correct answer is C
5 0
3 years ago
A real object is 10.0 cm to the left of a thin, diverging lens having a focal length of magnitude 16.0 cm. What is the location
amm1812

Answer:

A)6.15 cm to the left of the lens

Explanation:

We can solve the problem by using the lens equation:

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}

where

q is the distance of the image from the lens

f is the focal length

p is the distance of the object from the lens

In this problem, we have

f=-16.0 cm (the focal length is negative for a diverging lens)

p=10.0 cm is the distance of the object from the lens

Solvign the equation for q, we find

\frac{1}{q}=\frac{1}{-16.0 cm}-\frac{1}{10.0 cm}=-0.163 cm^{-1}

q=\frac{1}{-0.163 cm^{-1}}=-6.15 cm

And the sign (negative) means the image is on the left of the lens, because it is a virtual image, so the correct answer is

A)6.15 cm to the left of the lens

6 0
3 years ago
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