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3 years ago

Lisa sees the image below as she researches the behavior of light.

Physics

2 answers:

Elza [17]3 years ago

7 0

Refraction. When light is distorted when going through a substance like water, it is being refracted.

aalyn [17]3 years ago

6 0

The answer to the question is Refraction.

EXPLANATION:

The spoon half dipped inside the water is appeared to be bent. This is due to the optical phenomenon called refraction.

Refraction is the property of light by virtue of which light ray bends away or towards the normal at the point of incidence when it incidents obliquely at the refracting surface.

In the given figure, light rays are coming from air which is optically rarer medium as compared to water. So, the ray, after striking the air-water interface, gets bent towards the normal in the optically denser medium i.e water. It is so because the velocity of light in optically denser medium is less as compared to rarer medium.

We view the object due to the light. As light rays are bent, so the spoon is appeared to be bent.

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At what time after being ejected is the boulder moving at a speed 20.7 m/s upward?

Svetlanka [38]

The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.

<h3>What is the time after being ejected is the boulder moving at a speed 20.7 m/s upward?</h3>

The motion of the boulder is a uniformly accelerated motion, with constant acceleration

a = g = -9.8 $$$m / s^2$

downward (acceleration due to gravity).

By using Suvat equation:

v = u + at

where: v is the velocity at time t

u = 40.0 m/s is the initial velocity

a = g = -9.8 $$$m/s^2$ is the acceleration

To find the time t at which the velocity is v = 20.7 m/s

Therefore,

$$t=\frac{v-u}{a}=\frac{20.7-40}{-9.8}=2.0204 \mathrm{~s}$

The time after being ejected is the boulder moving at a speed 20.7 m/s upward is 2.0204 s.

The complete question is:

A large boulder is ejected vertically upward from a volcano with an initial speed of 40.0 m/s. Ignore air resistance. At what time after being ejected is the boulder moving at 20.7 m/s upward?

To learn more about uniformly accelerated motion refer to:

brainly.com/question/14669575

#SPJ4

4 0

1 year ago

Find the weight of an object with mass 80 kg on the moon ( g = 1.6 m/s^2)

SpyIntel [72]

Answer:

80kg = 133 Newtons I'm pretty sure this is right.

5 0

2 years ago

Why was the microscope and its improvement an important part of the development of the cell theory?

Tresset [83]

Answer:it’s is part of the cell theory because they where studying cells and to see it you need a microscope

Explanation:basically in the answer area

7 0

2 years ago

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Julio blows air across his hot bowl of soup. The tiny ripples he creates are similar to _____.

Allushta [10]

The tiny ripples on the soup are not only similar to wind-generated
waves ... they ARE wind-generated waves. This is a big part of the
reason why they bear such an uncanny resemblance.

8 0

3 years ago

Read 2 more answers

A particle makes 800 revolution in 4 minutes of a circle of 5cm. Find

vladimir1956 [14]

Answer:

i) The period of the particle is 0.3 seconds

ii) The angular velocity is approximately 20.94 rad/s

iii) The linear velocity is approximately 1.047 m/s

iv) The centripetal acceleration is approximately 6.98 m/s²

Explanation:

The given parameters are;

The number of revolution of the particle, n = 800 revolution

The time it takes the particle to make 800 revolutions = 4 minutes

The dimension of the circle = 5 cm = 0.05 m

Given that the dimension of the circle is the radius of the circle, we have;

i) The period of the particle, T = The time to complete one revolution

T = 1/(The number of revolutions per second)

∴ T = 1/(800 rev/(4 min × 60 s/min)) = 3/10 s

The period, T = 3/10 seconds = 0.3 seconds

ii) The angular velocity, ω = Angle covered/(Time)

800 revolutions in 4 minutes = Angle of (800 × 2·π) in 4 minutes

∴ ω = (800 × 2·π)/(4 × 60) = 20·π/3

The angular velocity, ω = 20·π/3 rad/s ≈ 20.94 rad/s

iii) The linear velocity, v = r × ω

∴ The linear velocity, v = 0.05 m × 20·π/3 rad/s = π/3 m/s ≈ 1.047 m/s

iv) The centripetal acceleration, $a_c$ = v²/r

∴ The centripetal acceleration, $a_c$ = (π/3)²/(0.05) = 20·π/9

The centripetal acceleration, $a_c$ = 20·π/9 m/s² ≈ 6.98 m/s²

4 0

2 years ago