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Grace [21]
3 years ago
11

7) A ball is thrown upward at an initial velocity of 8.2 m/s, from a height of 1.8 meters above the ground. The height of the ba

ll h, in metres can be represented, after t seconds, is modelled by the equation h = –4.8t² + 8.2t + 1.8. (a) Determine the height of the ball after 1.7 seconds.

Physics
1 answer:
Dima020 [189]3 years ago
6 0

\\  \tt \longmapsto \: h =  - 4.8 {t}^{2}  + 8.2t + 1.8 \\ \\  \tt \longmapsto \: h =  - 4.8(1.7) {}^{2}  + 8.2 \times 1.7 + 1.8 \\ \\  \tt \longmapsto \:  - 4.8 \times 2.89 + 1.39 + 1.8 \\ \\  \tt \longmapsto \: 13.8 + 1.39 + 1.8 \\ \\  \tt \longmapsto \: 17.06

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Three noise sources produce volume (loudness) levels of 70, 73, and 80 dB when acting separately. When the sources act together,
Mashutka [201]

Sound at 70 dB is 70 dB louder than the human reference level.  That's 10⁷ times as much as the reference sound power.

Sound at 73 dB is 73 dB louder than the human reference level.  That's 10⁷.³  or  2 x 10⁷  times as much as the reference sound power.

Sound at 80 dB is 80 dB louder than the human reference level.  That's 10⁸  or 10 x 10⁷ times as much as the reference sound power.

Now we can adumup:

Intensity of all 3 sources = (10⁷) + (2 x 10⁷) + (10 x 10⁷)

Intensity = (13 x 10⁷) times the sound power reference intensity.

Intensity in dB = 10 log (13 x 10⁷) = 10 (7 + log(13)

Intensity = 70 + 10 log(13)

Intensity = 70 + 10 (1.114)

Intensity = 70 + 11.14

Intensity = <em>81.14 dB</em>

<em>______________________________________</em>

Looking at the questioner's profile, I seriously wonder whether I'll ever get a comment in return from this creature, and how I'll ever find out if my solution is correct.  For that matter, I'm also seriously questioning how and whether my solution will ever be used for anything.

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