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Andre45 [30]
4 years ago
11

Wave A has an amplitude of 2 and wave B has an amplitude of 2 as shown below. What will happen when the crest of wave A meets th

e trough of wave B? They will interfere to create a crest with an amplitude of 4. They will interfere to create a crest with an amplitude of 2. They will interfere to create a crest with an amplitude of 0. They will bounce off each another.
Physics
2 answers:
puteri [66]4 years ago
6 0
Since the two waves have equal amplitudes, if the crest of one wave
meets the trough of the other one, they'll add to produce a level of zero
at that location.
Ugo [173]4 years ago
3 0

The correct answer to the question is- They will interfere to create a crest with an amplitude of 0.

EXPLANATION:

Before going to answer this question, first we have to understand destructive interference.

The two super imposing waves are said to be doing destructive interference if the crest of one wave falls on the trough of another wave.

Let the amplitudes of  two waves are A and B.

The phases difference (\theta) will be 180 degree for destructive interference.

Hence,the amplitude of the resultant wave is calculated as -

                                          R =  \sqrt{A^2+B^2+2ABcos\theta}

                                             = \sqrt{A^2+B^2+2ABcos180}

                                             = \sqrt{A^2+B^2-2AB}

                                             = A-B

Here, the amplitude of first wave A = 2

  The amplitude of second wave B = 2

Hence, amplitude of resultant wave R = 2-2

                                                                = 0

Hence, the correct answer will be - They will interfere to create a crest with an amplitude of 0.

                               

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suggest an experiment to prove that the rate of evaporation of a liquid depends on its surface area vapour already present in su
gulaghasi [49]
That's two different things it depends on:

-- surface area exposed to the air
AND
-- vapor already present in the surrounding air.

Here's what I have in mind for an experiment to show those two dependencies:

-- a closed box with a wall down the middle, separating it into two closed sections;

-- a little round hole in the east outer wall, another one in the west outer wall,
and another one in the wall between the sections;
So that if you wanted to, you could carefully stick a soda straw straight into one side,
through one section, through the wall, through the other section, and out the other wall.

-- a tiny fan that blows air through a tube into the hole in one outer wall.

<u>Experiment A:</u>

-- Pour 1 ounce of water into a narrow dish, with a small surface area.
-- Set the dish in the second section of the box ... the one the air passes through
just before it leaves the box.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
=============================
-- Pour 1 ounce of water into a wide dish, with a large surface area.
-- Set the dish in the second section of the box ... the one the air passes through
just before it leaves the box.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
=============================
<span><em>Show that the 1 ounce of water evaporated faster </em>
<em>when it had more surface area.</em></span>
============================================
============================================

<u>Experiment B:</u>

-- Again, pour 1 ounce of water into the wide dish with the large surface area.
-- Again, set the dish in the second half of the box ... the one the air passes
through just before it leaves the box.
-- This time, place another wide dish full of water in the <em>first section </em>of the box,
so that the air has to pass over it before it gets through the wall to the wide dish
in the second section.  Now, the air that's evaporating water from the dish in the
second section already has vapor in it before it does the job.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
==========================================
<em>Show that it took longer to evaporate when the air </em>
<em>blowing over it was already loaded with vapor.</em>
==========================================
6 0
3 years ago
Ryan is driving home from work and notices a deer leaping onto the road about 25 m in front of his car. He immediately applies t
Anvisha [2.4K]

Answer:

mu = 0.56

Explanation:

The friction force is calculated by taking into account the deceleration of the car in 25m. This can be calculated by using the following formula:

v^2=v_0^2+2ax\\

v: final speed = 0m/s (the car stops)

v_o: initial speed in the interval of interest = 60km/h

    = 60(1000m)/(3600s) = 16.66m/s

x: distance = 25m

BY doing a the subject of the formula and replace the values of v, v_o and x you obtain:

a=\frac{v^2-v_o^2}{2x}=\frac{0m^2/s^2-(16.66m/s)^2}{2(25m)}=-5.55\frac{m}{s^2}

with this value of a you calculate the friction force that makes this deceleration over the car. By using the Newton second's Law you obtain:

F_f=ma=(1490kg)(5.55m/s^2)=8271.15N

Furthermore, you use the relation between the friction force and the friction coefficient:

F_f= \mu N=\mu mg\\\\\mu=\frac{F_f}{mg}=\frac{8271.15N}{(1490kg)(9.8m/s^2)}=0.56

hence, the friction coefficient is 0.56

6 0
3 years ago
Calculate the force needed to accelerate a car of a mass 1000 kg by 3 m/s2
beks73 [17]

Answer:

<h2>3000 N</h2>

Explanation:

The force acting on an object given it's mass and acceleration can be found by using the formula

force = mass × acceleration

From the question we have

force = 1000 × 3

We have the final answer as

<h3>3000 N</h3>

Hope this helps you

5 0
3 years ago
The average density of the body of a fish is 1080 kg/m3. To keep from sinking, the fish increases its volumeby inflating an inte
Dimas [21]

Answer:

f = 8 %

Explanation:

given,

density of body of fish = 1080 kg/m³

density of air = 1.2 Kg/m³

density of water = 1000 kg/m²

to protect the fish from sinking volume should increased by the factor f

density of fish + density of water x increase factor = volume changes in water                                                    

1080 +f x 1.2 =(1 + f ) x 1000                

1080 + f x 1.2 = 1000 + 1000 f      

998.8 f = 80                                  

f = 0.0800                            

f = 8 %                                        

the volume increase factor of fish will be equal to f = 8 %

7 0
3 years ago
A heavy solid disk rotating freely and slowed only by friction applied at its outer edge takes 120 seconds to come to a stop.
alisha [4.7K]

Answer:

The time is 16 min.

Explanation:

Given that,

Time = 120 sec

We need to calculate the moment of inertia

Using formula of moment of inertia

I=\dfrac{1}{2}MR^2

If the disk had twice the radius and twice the mass

The new moment of inertia

I'=\dfrac{1}{2}\times2M\times(2R)^2

I'=8I

We know,

The torque is

\tau=F\times R

We need to calculate the initial rotation acceleration

Using formula of acceleration

\alpha=\dfrac{\tau}{I}

Put the value in to the formula

\alpha=\dfrac{F\times R}{\dfrac{1}{2}MR^2}

\alpha=\dfrac{2F}{MR}

We need to calculate the new rotation acceleration

Using formula of acceleration

\alpha'=\dfrac{\tau}{I'}

Put the value in to the formula

\alpha=\dfrac{F\times R}{8\times\dfrac{1}{2}MR^2}

\alpha=\dfrac{2F}{8MR}

\alpha=\dfrac{\alpha}{8}

Rotation speed is same.

We need to calculate the time

Using formula angular velocity

\Omega=\omega'

\alpha\time t=\alpha'\times t'

Put the value into the formula

\alpha\times120=\dfrac{\alpha}{8}\times t'

t'=960\ sec

t'=16\ min

Hence, The time is 16 min.

5 0
3 years ago
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