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mezya [45]
2 years ago
7

18. A rock is thrown straight upward with an initial velocity of 12.2 m/s from a location where the acceleration due to gravity

is 9.8 m/s2. How high up does it go
Physics
1 answer:
motikmotik2 years ago
7 0

The height to which the rock can attain is 7.6 m

From the question given above, the following data were obtained:

Initial velocity (u) = 12.2 m/s

Acceleration due to gravity (g) = 9.8 m/s²

Final velocity (v) = 0 m/s (at maximum height)

<h3>Maximum height (h) =? </h3>

The maximum height to which the rock can attain can be obtained as follow:

v² = u² – 2gh (since the rock is going against gravity)

0² = 12.2² – (2 × 9.8 × h)

0 = 148.84 – 19.6h

Collect like terms

0 – 148.84 = – 19.6h

–148.84 = –19.6h

Divide both side by –19.6

h = –148.84 / –19.6

<h3>h = 7.6 m</h3>

Thus, the height to which the rock can attain is 7.6 m

Learn more: brainly.com/question/25711946

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Refer to the diagram shown below.

When the ball attains maximum height, it will have zero vertical velocity.
The maximum height, h. obeys the equation
0 = (22 m/s)² - 2*(9.8 m/s²)*(h m)
h = 22²/(2*9.8) = 24.694 m

Answer: The maximum height attained is 24.7 m (nearest tenth)

Part b.
The vertical height traveled by the ball obeys the equation
h = (22 m/s)t - (1/2)*(9.8 m/s²)*(t s)²
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4.9t² - 22t = 0
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4 years ago
Write a 150-word paragraph or two that describes at least three everyday things that exist or occur because of science.
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3 years ago
The sound level at a distance of 1.48 m from a source is 120 dB. At what distance will the sound level be 70.7 dB?
Tju [1.3M]

Answer:

The second distance of the sound from the source is 431.78 m..

Explanation:

Given;

first distance of the sound from the source, r₁ = 1.48 m

first sound intensity level, I₁ = 120 dB

second sound intensity level, I₂ = 70.7 dB

second distance of the sound from the source, r₂ = ?

The intensity of sound in W/m² is given as;

dB = 10 Log[\frac{I}{I_o} ]\\\\For \ 120 dB\\\\120 = 10Log[\frac{I}{1*10^{-12}}]\\\\12 =  Log[\frac{I}{1*10^{-12}}]\\\\10^{12} = \frac{I}{1*10^{-12}}\\\\I = 10^{12} \ \times \ 10^{-12}\\\\I = 1 \ W/m^2

For \ 70.7 dB\\\\70.7 = 10Log[\frac{I}{1*10^{-12}}]\\\\7.07 =  Log[\frac{I}{1*10^{-12}}]\\\\10^{7.07} = \frac{I}{1*10^{-12}}\\\\I = 10^{7.07} \ \times \ 10^{-12}\\\\I = 1 \times \ 10^{-4.93} \ W/m^2

The second distance, r₂, can be determined from sound intensity formula given as;

I = \frac{P}{A}\\\\I = \frac{P}{\pi r^2}\\\\Ir^2 =  \frac{P}{\pi }\\\\I_1r_1^2 = I_2r_2^2\\\\r_2^2 = \frac{I_1r_1^2}{I_2} \\\\r_2 = \sqrt{\frac{I_1r_1^2}{I_2}} \\\\r_2 =   \sqrt{\frac{(1)(1.48^2)}{(1 \times \ 10^{-4.93})}}\\\\r_2 = 431.78 \ m

Therefore, the second distance of the sound from the source is 431.78 m.

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