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3 years ago

What is it called when a surface takes light in without reflecting it

Physics

2 answers:

andrew-mc [135]3 years ago

7 0

Reflection from such a rough surface is called diffuse reflection and appears matte

Travka [436]3 years ago

5 0

Answer: Specular reflection

Explanation:

The law of reflection says that when a ray of light hits a surface, it bounces in a certain way, like a tennis ball thrown against a wall. ... Mirrors, however, don't scatter light in this way. With a smooth surface, light reflects without disturbing the incoming image. This is called specular reflection.

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Which requires more work, lifting a 10-kg load a vertical distance of 2 m or lifting a 5-kg load a vertical distance of 4 m?

fredd [130]

Both have the same work done.

5x4 = 20
10x2=20

3 0

4 years ago

What is the speed of a wave that has a frequency of 2,400 Hz and a wavelength of 0.75

likoan [24]

Answer:

1800 m/s

Explanation:

The equation is v = fλ

λ= 0.75

f = 2400 Hz

V = 2400 × 0.75

V = 1800 m/s

[ you did not give units for wavelength, I assumed it would be m/s]

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3 years ago

When aluminum foil is formed into a loose ball, it can float on water. But when the ball of foil is pounded flat with a hammer,

docker41 [41]

Air caught in the ball of foil makes the ball less dense than water

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3 years ago

There is a puncture in the inner tor of your bicycle. Give a method to detect the position of the puncture

Pani-rosa [81]

Answer:

Explanation:

The first method to engage is to listen to where the sound of air in the inner Tor escaping originated and look to see if u can find it. You can then feel the escape air with your hand.

You can Put it inside a container of water and see the bubble and rotate the inner tube to pass all of it through the water

7 0

3 years ago

if two point charges are separated by 1.5 cm and have charge values of 2.0 and -4.0, respectively, what is the value of the mutu

RUDIKE [14]

Complete question:

if two point charges are separated by 1.5 cm and have charge values of +2.0 and -4.0 μC, respectively, what is the value of the mutual force between them.

Answer:

The mutual force between the two point charges is 319.64 N

Explanation:

Given;

distance between the two point charges, r = 1.5 cm = 1.5 x 10⁻² m

value of the charges, q₁ and q₂ = 2 μC and - μ4 C

Apply Coulomb's law;

$F = \frac{k|q_1||q_2|}{r^2}$

where;

F is the force of attraction between the two charges

|q₁| and |q₂| are the magnitude of the two charges

r is the distance between the two charges

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

$F = \frac{k|q_1||q_2|}{r^2} \\\\F = \frac{8.99*10^9 *4*10^{-6}*2*10^{-6}}{(1.5*10^{-2})^2} \\\\F = 319.64 \ N$

Therefore, the mutual force between the two point charges is 319.64 N

4 0

3 years ago