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2 years ago

14

When light propagates from a material with a given index of refraction into a material with a smaller index of refraction, the s

peed of the light
A) incease
B) decrease
C) remain constant

1 answer:

SSSSS [86.1K]2 years ago

7 0

Answer: A) increase

Explanation:

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3. The expression 0.62 x10^3 is equivalent to...

Korolek [52]

$\\ \sf\longmapsto 0.62\times 10^3$

$\\ \sf\longmapsto 62\times 10^{-2}\times 10^3$

$\\ \sf\longmapsto 62\times 10^{-2+3}$

$\\ \sf\longmapsto 62\times 10^1$

$\\ \sf\longmapsto 62\times 10$

$\\ \sf\longmapsto 620$

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2 years ago

Read 2 more answers

A model rocket is shot directly upward, rises to its maximum height and then returns to its launch position in 10.0 s. Assuming

worty [1.4K]

d. 49.0 m/s

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3 0

3 years ago

She sights two sailboats going due east from the tower. The angles of depression to the two boats are 42o and 29o. If the observ

Reika [66]

Answer:

The boats are 934.65 feet apart

Explanation:

Given:

The angles of depression to the two boats are 42 degrees and 29 degrees

Height of the observation deck i = 1,353 feet

To Find:

How far apart are the boats (y )= ?

Solution:

<em><u>Step 1 : Finding the value of x(Refer the figure attached)</u></em>

We can use the tangent ratio to find the x value

$tan(42^{\circ}) = \frac{1353}{x}$

$x = \frac{1353}{tan(42^{\circ}) }$

x = 590.47 feet

<em><u>Step 2 : Finding the value of z (Refer the figure attached)</u></em>

$tan(29^{\circ}) = \frac{1353}{z }$

$z = \frac{1353}{tan(29^{\circ})}$

z = 1525.12 feet

<em><u>Step 3 : Finding the value of y (Refer the figure attached</u></em>)

y = z -x

y = 1525.12 - 590.47

y = 934.65 feet

Thus the two boats are 934.65 feet apart

3 0

3 years ago

Skater A travels with velocity of 3.2 m/s and has a momentum of 200 kg m/s

Ivenika [448]

Here's link $^{}$ to the answer:

bit. $^{}$ ly/3gVQKw3

5 0

2 years ago

a spring is compressed 15 centimeters when a 8.5 kilogram weight is set upon it. how much power would be needed to stretch out t

o-na [289]

Answer:

159.38 Watts

Explanation:

Initially;

Mass on the spring is 8.5 kg
Therefore, compression force is 85 N
Compression distance is 15 cm or 0.15 m

But;

F = kx

where F is the force of compression, k is the spring constant and x is the compression distance.

Thus;

k = F/x

= 85 N ÷0.15

= 566.67 N/m

We are required to determine the power needed to stretch the same spring for 1.5 m in 4 secs.

Power = Work done ÷ time

Work done is given by 0.5kx²

Therefore;

Power = 0.5kx²÷ t

= (0.5×566.67 N/m × 1.5² ) ÷ 4 seconds

= 159.38 Watts

Thus, the power needed is 159.38 watts

5 0

3 years ago