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Brut [27]
3 years ago
10

What can accurately be said about a resultant wave that displays both reinforcement and interference? A. Maximum interference oc

curs where the troughs of the two cmponent waves are in phase B. The crests of the two compnent waves are in phase C. The component waves have different frequencies D. Molecules remain in their normal positions at spots reinforcement
Physics
2 answers:
vitfil [10]3 years ago
5 0

The situation that can accurately be said about a resultant wave that displays both reinforcement and interference is that crests of the two component waves are in phase. The answer is letter B.

ziro4ka [17]3 years ago
5 0

What can accurately be said about a resultant wave that displays both reinforcement and interference?

A. The crests of the two component waves are in phase where interference occurs in the resultant wave.

B. The component waves have different frequencies.

C. Molecules remain in their normal positions at spots of reinforcement.

D. Maximum interference occurs where the troughs of the two component waves are in phase.

THE CORRECT ANSWER FOR THIS QUESTION IS B GOT 100% ON THE TEST

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Gravitational force exist between you and the building why are you not pulled towards the building?​
Nataly_w [17]

Answer:

You are pulled towards that building. At the same time, that building is pulled towards you. Neither object creates enough gravitational force to really do anything. That is why you never notice any affect by either body, (you and a building).

Explanation:

You will surely get attracted towards the building.But it takes a lot of time depending on their masses.

This happens only when you are away from earth with that building.

Both of you will get attracted to it

if a third party with mass more than you or building is with you.

If it is on the earth.. Then the gravity between you and the building is negligible compared to the earth.Hence you will not get attracted towards the building in this case.

3 0
2 years ago
A ticker-tape is moved through a ticker-timer for 5 seconds . if the timer is operating at 25 Hz. i.How many dots would have bee
svetlana [45]

Answer:

Explanation:

25 cycles/s (5 s) = 125 dots

Tape does Repetitious motion.

It might possibly be harmonic motion, but we do not have enough information to say for certain.

8 0
2 years ago
Column A is in the x-axis, and column B is on the y-axis. Which titles should replace A and B
pogonyaev

Column A: x-axis, input, domain

Column B: y-axis, output, range

Those are other ways to describe them

hope i helped:)

8 0
3 years ago
Red light of wavelength 651 nm produces photoelectrons from a certain photoemissive material. Green light of wavelength 521 nm p
Mnenie [13.5K]

Answer:

material work function is 0.956 eV

Explanation:

given data

red wavelength 651 nm

green wavelength 521 nm

photo electrons = 1.50 × maximum kinetic energy

to find out

material work function

solution

we know by Einstein photo electric equation  that is

for red light

h ( c / λr ) = Ф +  kinetic energy

for green light

h ( c / λg ) = Ф +  1.50 × kinetic energy

now from both equation put kinetic energy from red to green

h ( c / λg ) = Ф +  1.50 × (h ( c / λr ) - Ф)

Ф =( hc / 0.50) × ( 1.50/ λr  - 1/ λg)

put all value

Ф =( 6.63 ×10^{-34} (3 ×10^{8} )  / 0.50) × ( 1.50/ λr  - 1/ λg)

Ф =( 6.63 ×10^{-34} (3 ×10^{8} ) / 0.50 ) × ( 1.50/ 651×10^{-9}   - 1/ 521 ×10^{-9})

Ф = 1.5305  ×10^{-19} J  × ( 1ev / 1.6 ×10^{-19} J )

Ф = 0.956 eV

material work function is 0.956 eV

4 0
3 years ago
Read 2 more answers
A baseball hit just above the ground leaves the bat 27 m/s at 45° above the horizontal. A) How far away does the ball strike the
Sedbober [7]

Answer:

A) The ball hits the ground 74.45 m far from the hitting position.

B) Maximum height of the ball = 18.57 m

Explanation:

There are two types of motion in this horizontal and vertical motion.

We have velocity = 27 m/s at 45° above the horizontal

Horizontal velocity = 27cos45 = 19.09 m/s

Vertical velocity = 27sin45 = 19.09 m/s

Time to reach maximum height,

           v = u + at

           0 = 19.09 - 9.81 t

            t = 1.95 s

So total time of flight = 2 x 1.95 = 3.90 s

A) So the ball travels at 19.09 m/s for 3.90 seconds.

     Horizontal distance traveled = 19.09 x 3.90 = 74.45 m

     So the ball hits the ground 74.45 m far from the hitting position.

B) We have vertical displacement

              S = ut + 0.5 at²

              H = 19.09 x 1.95 - 0.5 x 9.81 x 1.95² = 18.57 m

    Maximum height of the ball = 18.57 m

6 0
2 years ago
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