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Andre45 [30]
3 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]3 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]3 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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The speed of an arrow fired from a compound
san4es73 [151]

Answer:

A.) The arrow`s range is 624,996 m

B.) The arrow`s range is 846.887 m, when the horse is galloping

Explanation:

We have a case of oblique movement. In these cases the movement in the X axis is a Uniform Rectelinear Movement (URM), and a Uniform Accelerated Movement (UAM) in the Y axis.

By the way, the equations that we use for the X axis will be from URM, and those for the Y axis wiil be from UAM.

<u>Equations</u>

X axis:

X=v_{ox}*t

v_{0x} =v_0cos(\alpha)

Y axis:

Y= Y_0 +v_{y0} t - \frac{g}{2} t^2

A.) First, it is necessary to know t, total time.

To figure out t value, we use UAM, since time is determined by this movement.

Now, at the end of the movement, Y=0, then

0= Y_0 +v_{y0} t - \frac{g}{2} t^2

0=2.4m+79m/s*sin(39)t-(1/2*9.81m/s^2)t^2

Caculate the segcond degree equation to obtain the two possible values for t:

t_1= 10.18 \\t_2= -0.04046

But, in physics, time it could not be negative, so we take t_1= 10.18

Caculate now:

X=79m/s*cos(\39)*10.18s= 624.996 m

B.) Now, the narrow has an additional speed, that could be sum to the speed due to the bow.

v_0= 79m/s+13m/s= 92m/s

Using the same procedure that item A, caculate X

First, we need to know the new time

0=2.4m+92m/s*sin(39)t-(1/2*9.81m/s^2)t^2

And we obtain:

t_1=11.845s\\t_2=-0.041s

One more time, we take the positive time: t_1=11.845s

Finally:

X=92m/s *cos(39)*11.845s=846.887 m

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nata0808 [166]
To calculate the specific heat capacity of an object or substance, we can use the formula

c = E / m△T

Where
c as the specific heat capacity,
E as the energy applied (assume no heat loss to surroundings),
m as mass and
△T as the energy change.

Now just substitute the numbers given into the equation.

c = 2000 / 2 x 5
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Therefore we can conclude that the specific heat capacity of the block is 200 Jkg^-1°C^-1
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Answer:

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T = is the applied torque

I_{p} = polar moment of inertia of the section

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Thus we can see that if a point is located at center i.e r = 0 there will be no shearing stresses at the center due to torque.

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