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3 years ago

Waves combine to produce a smaller or zero-amplitude wave in a process called

2 answers:

6 0

ANSWER: Destructive Interference

-This is the exact definition of the question you have provided, look this term up if you do not believe lol.

Hope this Helps!

meriva3 years ago

3 0

Destructive interference is what this process is called.

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What factor has the greatest impact on flexibility?

hjlf

That would be c. :) :) :)

4 0

3 years ago

250 W 12 km/h<br>4. Convert yoor answer from 3 to meters per second.

taurus [48]

Answer: 3.33 m/s

Explanation:

Assuming the questions is to convert 12 km/h to meter per second (m/s), let's begin:

In order to make the conversion, we have to know the following:

$1 km=1000 m$

And:

$1 h=3600 s$

Keeping this in mind, we can make the conversion:

$12 \frac{km}{h} \frac{1000 m}{1 km} \frac{1h}{3600 s}$

Then:

$12 \frac{km}{h}= 3.33 \frac{m}{s}$

5 0

3 years ago

Please help me asap

ddd [48]

Answer:

if I aint wrong it would 2nd one

8 0

3 years ago

A child slides down a hill on a toboggan with an acceleration of 1.8 m/s^2. If she starts at rest, how far has she traveled in :

DENIUS [597]

Explanation:

It is given that,

The acceleration of the toboggan, $a=1.8\ m/s^2$

Initial speed of the toboggan, u = 0

We need to find the distance covered by the toboggan. Using the second equation of motion as :

$s=ut+\dfrac{1}{2}at^2$

At t = 1 s

$s=\dfrac{1}{2}\times 1.8\times 1^2$

$s_1=0.9\ m$

At t = 2 s

$s=\dfrac{1}{2}\times 1.8\times 2^2$

$s_2=3.6\ m$

At t = 3 s

$s=\dfrac{1}{2}\times 1.8\times 3^2$

$s_3=8.1\ m$

Hence, this is the required solution.

4 0

3 years ago

We have an Atwood device, two blocks connect by a string strung over a pulley, but the twist this time is that both blocks are o

Zanzabum

The Acceleration of the system is 6.41 m/s².

Given,

α= 15°, m₁ = 7kg

β= 65°, m₂ = 11 kg

Let, a be the acceleration and T is the tensions at the end it's the cord.

Let, the mass m₂ be coming down along the inclined plane along the inclined surface towards downward m₂g sin β and the tension in the upward direction,

Resultant force, m₂a=m₂g sin β -T

11a=((11) ×g sin 65°) -T ...(i)

Now, considering the motion of m₁ which moves downwards, the forces are m₁g sinα, and T both are acting downwards.

Resultant force m₁a = m₁g sin α+T

7a =7g sin 15°+T ...(ii)

Solving both the equations by adding them,

18a=11gsin 65°+7g sin 15°-T+T

18a=11gsin 65°+7g sin 15°=115.45

a=115.45/18=6.41 m/s²

Hence, the Acceleration of the system is 6.41 m/s².

Learn more about the acceleration here:

brainly.com/question/22048837

#SPJ10

6 0

2 years ago