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lesya692 [45]
3 years ago
9

Which part of the wave is changed when there is interference?

Physics
2 answers:
slava [35]3 years ago
8 0

THE AMPLITUDE. (this is the first option on idgenuity)

Rasek [7]3 years ago
3 0
When 2 waves interefere (or collide with eachother), it usually affects the crest of the wave. If both waves collide with both crests, it will create an amplified crest, and the waves will pass through eachother afterwards. If a trough of a wave meets a crest, it will cause the crest to be lowered shortly before both continue on.
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Can anyone help? <br> Please and thank you
Juliette [100K]

Kinetic energy is calculated as such: KE=\frac{1}{2} mv^{2} | therefore, KE=0.913598 J

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What are facts about balloons?
Nataly_w [17]
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3 years ago
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Two pendulum bobs have equal masses and lengths (8.100 m). bob a is initially held horizontally while bob b hangs vertically at
Karo-lina-s [1.5K]
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4 years ago
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A 3.00 kg object is moving in the XY plane, with its x and y coordinates given by x = 5t³ !1 and y = 3t ² + 2, where x and y are
Hatshy [7]

Answer:

The net force acting on this object is 180.89 N.

Explanation:

Given that,

Mass = 3.00 kg

Coordinate of position of x= 5t^3+1

Coordinate of position of y=3t^2+2

Time = 2.00 s

We need to calculate the acceleration

a = \dfrac{d^2x}{dt^2}

For x coordinates

x=5t^3+1

On differentiate w.r.to t

\dfrac{dx}{dt}=15t^2+0

On differentiate again w.r.to t

\dfrac{d^2x}{dt^2}=30t

The acceleration in x axis at 2 sec

a = 60i

For y coordinates

y=3t^2+2

On differentiate w.r.to t

\dfrac{dy}{dt}=6t+0

On differentiate again w.r.to t

\dfrac{d^2y}{dt^2}=6

The acceleration in y axis at 2 sec

a = 6j

The acceleration is

a=60i+6j

We need to calculate the net force

F = ma

F = 3.00\times(60i+6j)

F=180i+18j

The magnitude of the force

|F|=\sqrt{(180)^2+(18)^2}

|F|=180.89\ N

Hence, The net force acting on this object is 180.89 N.

3 0
3 years ago
On a highway curve with a radius of 46 meters, the maximum force of static friction that can act on a 1,200 kg car going around
Mekhanik [1.2K]

Answer:

v\approx 16.956\,\frac{m}{s}

Explanation:

The motion of the vehicule on a highway curve can be modelled by the following equation of equilibrium:

\Sigma F = f = m\cdot \frac{v^{2}}{R}

The maximum speed is:

v = \sqrt{\frac{f\cdot R}{m} }

v = \sqrt{\frac{(7500\,N)\cdot (46\,m)}{1200\,kg} }

v\approx 16.956\,\frac{m}{s}

7 0
3 years ago
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