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vagabundo [1.1K]
2 years 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]2 years 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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Answer:

Q_2 = +/- 295.75*Q

Explanation:

Given:

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- The second charge = Q_2

- The position of first charge x_1 = 2a

- The position of the second charge x_2 = 13a

- The net Electric Field produced at origin is E_net = 2kQ / a^2

Find:

Explain how many values are possible for the unknown charge and find the possible values.

Solution:

- The Electric Field due to a charge is given by:

                               E = k*Q / r^2

Where, k: Coulomb's Constant

            Q: The charge of particle

            r: The distance from source

- The Electric Field due to charge 1:

                               E_1 = k*Q_1 / r^2

                               E_1 = k*Q / (2*a)^2

                               E_1 = k*Q / 4*a^2

- The Electric Field due to charge 2:

                               E_2 = k*Q_2 / r^2

                               E_2 = k*Q_2 / (13*a)^2

                               E_2 = +/- k*Q_2 / 169*a^2

- The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:

                               E_net = E_1 + E_2

                               2kQ / a^2 = k*Q_1 / 4*a^2 +/- k*Q_2 / 169*a^2

- The two equations are as follows:

        1:                   2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2

                               2Q = Q / 4 + Q_2 / 169

                               Q_2 = 295.75*Q

        2:                    2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2

                               2Q = Q / 4 - Q_2 / 169

                               Q_2 = -295.75*Q

- The two possible values corresponds to positive and negative charge Q_2.

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Answer:

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