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2 years ago

The Steamboat Geyser in Yellowstone National Park shoots water into the air at 48.0 m/s. How

Physics

2 answers:

qwelly [4]2 years ago

7 0

Answer:

The maximum height reached by the water is 117.55 m.

Explanation:

Given;

initial velocity of the water, u = 48 m/s

at maximum height the final velocity will be zero, v = 0

the water is going upwards, i.e in the negative direction of gravity, g = -9.8 m/s².

The maximum height reached by the water is calculated as follows;

v² = u² + 2gh

where;

h is the maximum height reached by the water

0 = u² + 2gh

0 = (48)² + ( 2 x -9.8 x h)

0 = 2304 - 19.6h

19.6h = 2304

h = 2304 / 19.6

h = 117.55 m

Therefore, the maximum height reached by the water is 117.55 m.

Anon25 [30]2 years ago

4 0

Answer:

The maximum height reached by the water is 117.55 m.

Explanation:

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2 years ago

The Earth's biosphere is consists of

Firdavs [7]

Answer:

The part of the earth and its atmosphere in which living organisms exist or that is capable of supporting life.

Explanation:

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2 years ago

A projectile lands at the same height from which it was launched. which initial velocity will result

Serhud [2]

The required initial velocity that will result if a projectile lands at the same height from which it was launched is V₀ = V cosθ

First, we must understand that the component of the velocity along the vertical is due to maximum height achieved and expressed as usin θ.

The component of the velocity along the horizontal is due to the range of the object and is expressed as ucosθ.

If the <u>air resistance is ignored</u>, the velocity of the object will be constant throughout the flight and the initial velocity will be equal to the final velocity.

Hence the required initial velocity that will result if a projectile lands at the same height from which it was launched is V₀ = V cosθ

Learn more here; brainly.com/question/12870645

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3 years ago

Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit

Ber [7]

Solution :

Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

Acceleration due to gravity at 1000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

$= 822486 \times 10^{-8}$

$=0.822 \times 10^{-2} \ km/s$

= 8.23 m/s

Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

$= 673552 \times 10^{-8}$

$=0.673 \times 10^{-2} \ km/s$

= 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

$= 3371 \times 153.86 \times 10^{-8}$

= 5.18 m/s

Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

$= 153.84 \times 2371 \times 10^{-8}$

$=0.364 \times 10^{-2} \ km/s$

= 3.64 m/s

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3 years ago

Impact of electricity in the society

Travka [436]

Answer:

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2 years ago