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vagabundo [1.1K]
1 year ago
13

A firefighter is using a hose and the flow rate of the water leaving the hose is 0.032 m3/s. At the end of the hose, the nozzle

has a radius of 0.013 m. What is the speed of the water leaving the hose ?
Physics
1 answer:
Oksi-84 [34.3K]1 year ago
5 0

A firefighter is using a hose and the flow rate of the water leaving the hose. the speed of the water leaving the hose is mathematically given as

u = 9.947 m/s

<h3>What is the speed of the water leaving the hose?</h3>

Generally, the equation for Volume flow rate is  mathematically given as

(V)= (A) * (u)

Where

A= area

v=velocity

Therefore

V = A*u

V=0.032 m3/s

A=πR2

A= π*0.013

u=V/A

u=0.032/π*0.0322

u = 60.27m/s

In conclusion, the speed of the water leaving the hose

u = 60.27m/s

Read more about speed

brainly.com/question/7359669

#SPJ1

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R_3 < R_1 < R_2

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2) The second wire has length twice the first one: 2L, and same thickness, A. So its resistance is

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3) The third wire has length L (as the first one), but twice cross sectional area, 2A. So, its resistance is

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if a person can jump maximum along distance of 3m ,on the earth how far could be jump on the moon where acceleration due to grav
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Answer:

The person can jump 48 m on the Moon

Explanation:

The question parameters are;

The maximum long jump distance of a person on Earth, R_{max} = 3 m

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The distance the person can jump on the Moon is given as follows;

A person performing a jump across an horizontal distance on Earth (under gravitational force) follows the path of the motion of a projectile

The horizontal range, R_{max}, of a projectile motion is found by using the following formula

R_{max} = \dfrac{u^2}{g}

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Therefore, we have;

R_{max} = 3 \, m = \dfrac{u^2}{9.8 \, m/s^2 }

u² = 3 m × 9.8 m/s² = 29.4 m²/s²

Therefore, on the Moon, we have;

The acceleration due to gravity on the Moon, g_{Moon} = 1/16 × g

∴ g_{Moon} = 1/16 × g = 1/16 × 9.8 m/s² ≈ 0.6125 m/s²

R_{max \ Moon} = \dfrac{u^2}{g_{Moon}}   = \dfrac{29.4 \ m^2/s^2}{0.6125 \, m/s^2 } \approx 48 \, m

The maximum distance the person can jump on the Moon with the same velocity which was used on Earth is R_{max \ Moon} ≈ 48 m

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2 years ago
A scuba diver's tank contains 0.240 kg of o2 compressed into a volume of 3.10 l. what volume (in liters) would this oxygen occup
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Given:
m = 0.240 kg = 240 g, the mass of O₂
V = 3.10 L = 3.10 x 10⁻³ m³, the volume

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As an ideal gas,
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V = (nRT)/p
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When
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V = (0.2029 m³)*(10³ L/m³) = 202.9 L

Answer: 203 liters
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