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Aloiza [94]
3 years ago
5

What does a refractory period prevent

Physics
1 answer:
Tamiku [17]3 years ago
5 0

Keister since the fence after I sent the van the refractory period prevents double counting the same event where is after I passed a van they prevent sensing the patient stimulus it’s after
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How much power is required to do 180 J of work in 2.4s?​
rjkz [21]
75
p = w / t
p = 180 / 2.4
p = 75
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During both the vernal equinox and _____________, there is an equal amount of night and day.
choli [55]
During both the Vernal Equinox and Autumnal equinox, there is an equal amount of night and day
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How is the direction of light changed when it travels from an optically denser medium to an optically rarer medium????? please a
ANTONII [103]

Answer:

The light bends away from the normal

Explanation:

We can solve the problem by using Snell's law:

n_1 sin \theta_1 = n_2 sin \theta_2

where:

n_1 is the index of refraction of the first medium

n_2 is the index of refraction of the second medium

\theta_1 is the angle of incidence (angle between the incoming ray and the normal to the interface)

\theta_2 is the angle of refraction (angle between the outcoming ray and the normal to the interface)

We can rearrange the equation as

sin \theta_2 = \frac{n_1}{n_2}sin \theta_1

In this problem, light travels from an optically denser medium to an optically rarer medium, so

n_1 > n_2

Therefore, the term \frac{n_1}{n_2} is greater than 1, so

sin \theta_2 > sin \theta_1\\\rightarrow \theta_2 > \theta_1

which means that the angle of refraction is greater than the angle of incidence, and so the light will bend away from the normal.

4 0
3 years ago
What does it mean to investigate science?
lapo4ka [179]

Answer:

Scientific investigation is a quest to find the answer to a question using the scientific method. In turn, the scientific method is a systematic process that involves using measurable observations to formulate, test or modify a hypothesis.

Explanation:

7 0
3 years ago
Two cars travel westward along a straight highway, one at a constant velocity of 85 km/h, and the other at a constant velocity o
spayn [35]

To solve letter a:

d1 = 85t1 = 16 km, 
85t1 = 16, 
t1 = 16 / 85 = 0.1882 h = 11.29 min. 

d2 = 115t2 = 16 km, 
115t2 = 16, 
t2 = 16 / 115 = 0.139 h = 8.35 min. 

t1 - t2 = 11.29 - 8.35 = 2.94 min. 
Car #2 arrives 2.94 minutes sooner.

To solve letter b:

15 min = 1/4 h = 0.25 h. 
d1 = d2, 
115t = 85(t + 0.25), 
115t = 85t + 21.25, 
115t - 85t = 21.25, 
30t = 21.25, 
t = 21.25 / 30 = 0.71 h, 

d = 115 * 0.71 = 81.65 km.

8 0
3 years ago
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