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2 years ago

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When light waves hit ice, most of them bounce off and radiate back into space. Because of this, it can be said that ice is a(n)

2 answers:

Softa [21]2 years ago

8 0

Reflective~~~~~~~~~~~~~~~~~~~~~~

kirill [66]2 years ago

8 0

<u>Answer;</u>

Reflective surface

<u>Explanation;</u>

Reflection is one of the property of a wave that occurs when a wave hits a boundary which is a reflective surface. The wave bounces off upon hitting the reflective boundary, such that the angle of incidence is equal to the angle of reflection.
Therefore, ice is an example of a reflective surface as light waves bounces off and radiate back when they hit the surface of the ice.

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Does light travel faster through water or glass?

Len [333]

Answer: The refractive index or n shows the relationship between c and the speed v when light travels through a material. The equation is n=c/v. In water, the refractive index is 1.3 and in glass, it is 1.5. Therefore, light travels faster through water, than it travels through glass.

Explanation:

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3 years ago

What is the IMA of the following pulley system?

Furkat [3]

Answer: $IMA= \frac {F_r}{F_e}$

Explanation:

IMA stands for Ideal Mechanical advantage.

The IMA of pulley system can be defined as the ratio of output force to input force.

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The input force = $F_e$

The output force = $F_r$

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2 years ago

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VladimirAG [237]

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2 years ago

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Suzy drops a rock from the roof of her house. Mary sees the rock pass her 2.7 m tall window in 0.129 sec. From how high above th

vladimir1956 [14]

Answer:

h = 22.35 m

Explanation:

given,

initial speed of the rock,u = 0 m/s

length of the window,l = 2.7 m

time taken to cross the window,t = 0.129 s

Speed of the rock when it crosses the window

$v = \dfrac{l}{t}$

$v = \dfrac{2.7}{0.129}$

v = 20.93 m/s

height of the building above the window

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v² = u² + 2 g h

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h = 22.35 m

Hence, the height of the building above the top of window is equal to h = 22.35 m

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2 years ago

Help with 5 and the question below. BRAINLIEST FOR THE CORRECT ANSWER! Urgent Physics help!

krok68 [10]

Sound level at distance of 15 m is given as 20 dB

so intensity at this distance is given as

$L = 10 Log\frac{I}{I_0}$

$20 = 10 Log{I}{10^{-12}}$

$I_1 = 10^{-10}W/m^2$

now if we move closer to some some distance the sound level is now 50 dB

now the intensity is given as

$L = 10 Log\frac{I}{I_0}$

$50 = 10Log\frac{I}{10^{-12}}$

$I_2 = 10^{-7}W/m^2$

now we know that

$\frac{I_1}{I_2} = \frac{r_2^2}{r_1^2}$

$\frac{10^{-10}}{10^{-7}} = \frac{r_2^2}{15^2}$

$r_2 = 47 cm$

so now the distance from friend must be 47 cm

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2 years ago