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svet-max [94.6K]
3 years ago
12

What radiation lies at frequencies just below the frequencies of visible light?

Physics
1 answer:
vlabodo [156]3 years ago
3 0
<span>Infrared radiation also called as infrared light was discovered by Sir William Herschel in 1800.It is an electronic magnetic radiation with longer wavelengths than those of a visible light.It involves waves rather than particles. It lies at frequencies just below the frequencies of visible light.</span>
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A note is being played on a string on a guitar. The string is then pushed to the side while still in contact with the same fret.
Brut [27]

Answer:

See explanation

Explanation:

- When the string is pressed on a particular fret, the note is the same. That's because the string will sound from bridge to that fret-wire. It's always the same distance, so will always be the same note.

- Frets is that it splits the fingerboard into discrete diatonic parts, and well made fret-boards will mean the same note gets played on the same fret on the same string every time.

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3 years ago
A 4 kg bowling ball rolls at a speed of 5 m/s on the roof of a building that is 30 meters tall. What is its kinetic energy?
klasskru [66]

ans: B

Kinetic energy = \frac{1}{2} mv^{2}

= \frac{1}{2} (4)(5^{2} )

= 50J

ps. the height will only affect its potential energy, not kinetic.

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3 years ago
Human sense impressions are subjective and qualitative
Fittoniya [83]

Answer:trueee

Explanation:

7 0
3 years ago
A swimmer is capable of swimming at 1.4m/s in still water. a. How far downstream will he land if he swims directly across a 180m
ozzi

Answer:

t = 180 / 1.4 = 129 sec   (time to swim horizontally across river)

S = 129 sec * V     where V is speed of current and S is the distance he will be carried downstream

The problem does not specify V the speed of the river

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Read 2 more answers
A Tennis ball falls from a height 40m above the ground the ball rebounds
worty [1.4K]

If the ball is dropped with no initial velocity, then its velocity <em>v</em> at time <em>t</em> before it hits the ground is

<em>v</em> = -<em>g t</em>

where <em>g</em> = 9.80 m/s² is the magnitude of acceleration due to gravity.

Its height <em>y</em> is

<em>y</em> = 40 m - 1/2 <em>g</em> <em>t</em>²

The ball is dropped from a 40 m height, so that it takes

0 = 40 m - 1/2 <em>g</em> <em>t</em>²

==>  <em>t</em> = √(80/<em>g</em>) s ≈ 2.86 s

for it to reach the ground, after which time it attains a velocity of

<em>v</em> = -<em>g</em> (√(80/<em>g</em>) s)

==>  <em>v</em> = -√(80<em>g</em>) m/s ≈ -28.0 m/s

During the next bounce, the ball's speed is halved, so its height is given by

<em>y</em> = (14 m/s) <em>t</em> - 1/2 <em>g</em> <em>t</em>²

Solve <em>y</em> = 0 for <em>t</em> to see how long it's airborne during this bounce:

0 = (14 m/s) <em>t</em> - 1/2 <em>g</em> <em>t</em>²

0 = <em>t</em> (14 m/s - 1/2 <em>g</em> <em>t</em>)

==>  <em>t</em> = 28/<em>g</em> s ≈ 2.86 s

So the ball completes 2 bounces within approximately 5.72 s, which means that after 5 s the ball has a height of

<em>y</em> = (14 m/s) (5 s - 2.86 s) - 1/2 <em>g</em> (5 s - 2.86 s)²

==>  (i) <em>y</em> ≈ 7.5 m

(ii) The ball will technically keep bouncing forever, since the speed of the ball is only getting halved each time it bounces. But <em>y</em> will converge to 0 as <em>t</em> gets arbitrarily larger. We can't realistically answer this question without being given some threshold for deciding when the ball is perfectly still.

During the first bounce, the ball starts with velocity 14 m/s, so the second bounce begins with 7 m/s, and the third with 3.5 m/s. The ball's height during this bounce is

<em>y</em> = (3.5 m/s) <em>t</em> - 1/2 <em>g</em> <em>t</em>²

Solve <em>y</em> = 0 for <em>t</em> :

0 = (3.5 m/s) <em>t</em> - 1/2 <em>g t</em>²

0 = <em>t</em> (3.5 m/s - 1/2 <em>g</em> <em>t</em>)

==>  (iii) <em>t</em> = 7/<em>g</em> m/s ≈ 0.714 s

As we showed earlier, the ball is in the air for 2.86 s before hitting the ground for the first time, then in the air for another 2.86 s (total 5.72 s) before bouncing a second time. At the point, the ball starts with an initial velocity of 7 m/s, so its velocity at time <em>t</em> after 5.72 s (but before reaching the ground again) would be

<em>v</em> = 7 m/s - <em>g t</em>

At 6 s, the ball has velocity

(iv) <em>v</em> = 7 m/s - <em>g</em> (6 s - 5.72 s) ≈ 4.26 m/s

4 0
3 years ago
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