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2 years ago

14

On a stormy night, a lighthouse emits a light that must travel through air, rain, and fog. As the light travels through these di

fferent media, what can be said about the light wave?
Select one:
a. Its wavelength remains constant.
b. Its speed remains constant.
c. Its speed is unaffected by the different mediums.
d. Its speed in air, rain, and fog is different.

2 answers:

klio [65]2 years ago

7 0

Answer:

CAN U GIVE ME A HOYAH

Explanation:

vlada-n [284]2 years ago

5 0

Answer:

HOOOOOOOYAAAAHHHHH

Explanation:

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A student uses the right-hand rule as shown. an illustration with a right hand with fingers curled and thumb pointed up. what is

Afina-wow [57]

The direction of the magnetic field in front of the wire closest to the student is on the left. The direction is found by the right-hand rule.

<h3>What is the right-hand rule?</h3>

The right-hand rule is a popular mnemonic for remembering how axes in three-dimensional space are oriented.

The fact that the three axes of three-dimensional space have two different orientations gives birth to the majority of the many left-hand and right-hand rules.

Using the right-hand rule, we can recall this diagram. Your thumb points in the direction of the magnetic force pushing on the moving charge

If you point your pointer finger in the direction of the positive charge and then your middle finger in the direction of the magnetic field.

To learn more about the right-hand rule refer to the link;

brainly.com/question/9750730

6 0

2 years ago

Which graph shows uniform motion.

bagirrra123 [75]

Answer:

1. shows uniform motion

Explanation:

5 0

3 years ago

Read 2 more answers

Where is the near point of an eye for which a contact lens with a power of +2.55 diopters is prescribed?Where is the far point o

tankabanditka [31]

Answer:

(a). The eye's near point is 68.98 cm from the eye.

(b). The eye's far point is 33.33 cm from the eye.

Explanation:

Given that,

Power = 2.55 D

Object distance = 25 cm for near point

Object distance = ∞ for far point

Suppose where is the far point of an eye for which a contact lens with a power of -3.00 D is prescribed for distant vision?

(a) We need to calculate the focal length

Using formula of power

$f =\dfrac{1}{P}$

Put the value into the formula

$f=\dfrac{100}{2.55}$

$f=39.21\ cm$

We need to calculate the image distance

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{39.21}=\dfrac{1}{v}-\dfrac{1}{-25}$

$\dfrac{1}{39.21}-\dfrac{1}{25}=\dfrac{1}{v}$

$\dfrac{1}{v}=-\dfrac{1421}{98025}$

$v=-68.98\ cm$

The eye's near point is 68.98 cm from the eye.

(b). We need to calculate the focal length

Using formula of power

$f =\dfrac{1}{P}$

Put the value into the formula

$f=-\dfrac{100}{3.00}$

$f=-33.33\ cm$

We need to calculate the image distance

Using formula of lens

$\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}$

Put the value into the formula

$\dfrac{1}{-33.33}=\dfrac{1}{v}-\dfrac{1}{\infty}$

$-\dfrac{1}{33.33}+\dfrac{1}{\infty}=\dfrac{1}{v}$

$\dfrac{1}{v}=-\dfrac{1}{33.33}$

$v=-33.33\ cm$

The eye's far point is 33.33 cm from the eye.

Hence, (a). The eye's near point is 68.98 cm from the eye.

(b). The eye's far point is 33.33 cm from the eye.

5 0

3 years ago

Which of the following is a contact force?

Nutka1998 [239]

The contact force is indeed caused by "Contact".

Air resistance is basically a type of friction, which is apparently present providing two object contact.

The rest selections are all interaction force, which is not necessarily caused by contact.

4 0

3 years ago

Read 2 more answers

A flywheel in a motor is spinning at 510 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75

8_murik_8 [283]

Complete Question

A flywheel in a motor is spinning at 510 rpm when a power failure suddenly occurs. The flywheel has mass 40.0 kg and diameter 75.0 cm . The power is off for 40.0 s , and during this time the flywheel slows down uniformly due to friction in its axle bearings. During the time the power is off, the flywheel makes 210 complete revolutions. At what rate is the flywheel spinning when the power comes back on(in rpm)? How long after the beginning of the power failure would it have taken the flywheel to stop if the power had not come back on, and how many revolutions would the wheel have made during this time?

Answer:

$\theta=274rev$

Explanation:

From the question we are told that:

Angular velocity $\omega=510rpm$

Mass $m=40.kg$

Diameter d $75=>0.75m$

Off Time $t=40.0s$

Oscillation at Power off $N=210$

Generally the equation for Angular displacement is mathematically given by

$\theta_{\infty}=\frac{w+w_0}{t}t$

$w=\frac{2*\theta_{\infty}}{t}-w_0$

$w=\frac{28210}{40*(\frac{1}{60})}-510$

$w=120rpm$

Generally the equation for Time to come to rest is mathematically given by

$t=(\frac{\omega_0}{\omega_0-\omega})t$

$t=(\frac{510}{510-120rpm})(40.0)(\frac{1}{60})$

$t=0.87min$

Therefore Angular displacement is

$\theta =(\frac{120+510}{2})0.87$

$\theta=274rev$

6 0

3 years ago