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UNO [17]
3 years ago
15

The container is filled to yh 350 ml mark water is observation or inference?

Physics
1 answer:
Lorico [155]3 years ago
4 0
This is an observation.
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______ interference occurs when two waves are out of phase.
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Answer:

What interference occurs when two waves are out of phase?

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Destructive interference

Destructive interference occurs when the maxima of two waves are 180 degrees out of phase: a positive displacement of one wave is canceled exactly by a negative displacement of the other wave. The amplitude of the resulting wave is zero.

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Directions: Consider a 2-kg bowling ball sits on top of a building that is 40 meters tall. It falls to the ground. Think about t
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1. The bowling ball have more potential energy as it sit on top of the building

2.The bowling ball have same potential energy and kinetic energy as it is half way through its fall

3. The bowling ball have more kinetic energy just before it hits the ground

4. The potential energy of the bowling ball as it sits on top of the building is 784J

5. The potential energy of the ball as it is half way through the fall, 20 meters high is 392J

6. The kinetic energy of the ball as it is half way through the fall is 392J

7. The kinetic energy of the ball just before it hits the ground is 784J

Explanation:

Calculating potential energy and kinetic energy for all the instances,

1. ball on top of a 40 meters tall building

Potential energy at the top of building with a height of 40m = mgh

P.E = mgh =2*9.8*40= 784J

At the top pf the building since v=0 kinetic energy is zero

2. half way through a fall off a building that is 40 meters tall and travelling 19.8 meters per second

Potential energy when it is half way through fall = mgH

where H represents new height that is equal to 20m

hence P.E=mgH=2*9.8*20= 392J

Kinetic energy  of the ball is \frac{1}{2} mv^{2}  = \frac{1}{2} *2*19.8^{2}=392.04J

3.  Just about to hit the ground from a fall off a building that is 40 meters tall and travelling 28 meters per second.

The potential energy of ball just before it hits the ground = mgh= 2*9.8*0=0J

kinetic energy =\frac{1}{2} mv^{2}= 784J

3 0
3 years ago
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