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zysi [14]
3 years ago
11

The of a wave is the density of the medium’s particles at the compression of the wave.

Physics
2 answers:
Westkost [7]3 years ago
7 0

'Amplitude' is the correct answer i just took the quiz

andrew11 [14]3 years ago
6 0
<span>Amplitude is the correct answer. I hope this helps.</span>
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Which scenario is in excample of a scientific way to think
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Can you explain this a bit more I don’t quite understand
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3 years ago
what was the temperature change in Celsius degree if it is changed from 44 degree fahrenheit to -56 degree fahrenheit​
wel

Answer:

-11.11 degree Celsius

Explanation:

The change was 44 degree fanhereit

To 56 degree fanhereit

Therefore the temperature range can be calculated as follows

56-44

= 12 degree fanhereit to Celsius

= 12-32×5/9

= -20×5/9

= 100/9

= -11.11 degree Celsius

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3 years ago
All contraceptives prevent STIs true or false
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7 0
4 years ago
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
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