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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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Velocity increases as distance increase
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3 years ago
Two objects of equal mass collide on a horizontal frictionless surface. Before the collision, object A is at rest while object B
fenix001 [56]

Answer: 6m/s

Explanation:

Using the law of conservation of momentum, the change in momentum of the bodies before collision is equal to the change in momentum after collision.

After collision, the two objects will move at the same velocity (v).

Let mA and mB be the mass of the two objects

uA and uB be their velocities before collision.

v be their velocity after collision

Since the two objects has the same mass, mA= mB= m

Also since object A is at rest, its velocity = 0m/s

Velocity of object B = 12m/s

Mathematically,

mAuA + mBuB = (mA+mB )v

m(0) + m(12) = (m+m)v

0+12m = (2m)v

12m = 2mv

12 = 2v

v = 6m/s

Therefore the speed of the composite body (A B) after the collision is 6m/s

7 0
4 years ago
An object is moving east, and its velocity changes from 65 m/s to 25 m/s in 10 seconds Which describes the acceleration?
Rasek [7]

Answer:

4 m/s in negative acceleration

Explanation:

Acceleration = V- U/t

Where V is the final velocity

U is the initial velocity and t is the time given.

U = 65 m/s

V= 25 m/s

T= 10 seconds

Acceleration= (25m/s - 65m/s)÷10secs

= - 40/10

= -4m/s^2

Hence, it has a negative acceleration.

8 0
3 years ago
determine the pressure exerted on the surface of a submarine cruising 175 ft below the free surface of the sea. assume that the
Alenkasestr [34]

Answer:

92.81 psia.

Explanation:

The density of water by multiplying its specific gravity by the density of sea water.

SG = density of sea water/density of water

ρ = SG x ρw

1 kg/m3 = 62.4 lbm/ft^3

= 1.03 * 62.4

= 64.27lbm/ft^3.

The absolute pressure at 175 ft below sea level as this is the location of the submarine.

P = Patm +ρgh

= 14.7 + 64.27 * 32.2 * 175

Converting to pound force square inch,

= 14.7 + 64.27 * (32.2ft/s^2) * (175ft) * (1lbf/32.2lbm⋅ft/s^2) * (1ft^2/144in^2 )

= 14.7 + 78.11 psia

= 92.81 psia.

8 0
3 years ago
How do you solve 0.004 dm + 0.12508 dm?
Effectus [21]
0.004 of something added to 0.12508 of the same thing
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The thing could be a glass of water, a sheet of paper,
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In this problem, it just happens to be a dm. 
7 0
3 years ago
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