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almond37 [142]
3 years ago
5

What gauge pressure is required in the city water mains for a stream from a fire hose connected to the mains to reach a vertical

height of 15.0 m
Physics
2 answers:
Sliva [168]3 years ago
8 0

Answer:

The gauge pressure is 147000 Pa.

Explanation:

Height, h = 15 m

density of water, d= 1000 kg/m^3

gravity, g = 9.8 m/s^2

The gauge pressure is the pressure exerted by the fluid.

The pressure exerted by the fluid is given by

P  = h d g

P = 15 x 1000 x 9.8 =  147000 Pa

STALIN [3.7K]3 years ago
4 0

Answer:

The gauge pressure is equal to 147 kPa.

Explanation:

The pressure exerted by fluid is given by :

P=\rho gh

Where

\rho is density of water

h is height

So, put all the values,

P=1000\times 9.8\times 15\\\\P=147000\ Pa

or

P = 147 kPa

So, the gauge pressure is equal to 147 kPa.

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Which energy source is formed when organic matter is trapped underground without exposure to air or moisture?. . A. natural gas.
nikklg [1K]

Answer:

The correct answer is option B. coal

Explanation:

Coal is made of remains of organic material including trees and other vegetation which got trapped beneath the earth’s surface or at the bottom of the swamps. After burial below the ground the organic material was acted upon by the high temperature and pressure in the absence of air to form peat. Peat after further processing for a longer period of time converted into coal

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3 years ago
Consider a mechanical clutch that consists of two heavy disks that can engage or disengage. At the beginning disk 1 with mass m1
miv72 [106K]

Answer:

w = 4,786 rad / s ,  f = 0.76176 Hz

Explanation:

For this problem let's use the concept of angular momentum

       L = I w

The system is formed by the two discs, during the impact the system remains isolated, we have the forces are internal, this implies that the external torque is zero and the angular momentum is conserved

Initial Before sticking

      L₀ = 0 + I₂ w₂

Final after coupling

      L_{f} = (I₁ + I₂) w

The moments of inertia of a disk with an axis of rotation in its center are

      I = ½ M R²

How the moment is preserved

      L₀ = L_{f}

      I₂ w₂ = (I₁ + I₂) w

      w = w₂ I₂ / (I₁ + I₂)

Let's reduce the units to the SI System

      d₁ = 60 cm = 0.60 m

      d₂ = 40 cm = 0.40 m

      f₂ = 200 min-1 (1 min / 60 s) = 3.33 Hz

Angular velocity and frequency are related.

      w₂ = 2 π f₂

      w₂ = 2π 3.33

      w₂ = 20.94 rad / s

Let's replace

       w = w₂ (½ M₂ R₂²) / (½ M₁ R₁² + ½ M₂ R₂²)

       w = w₂ M₂ R₂² / (M₁ R₁² + M₂ R₂²)

Let's calculate

      w = 20.94 8 0.40² / (12 0.60² + 8 0.40²)

      w = 20.94 1.28 / 5.6

      w = 4,786 rad / s

Angular velocity and frequency are related.

      w = 2π f

      f = w / 2π

      f = 4.786 / 2π

      f = 0.76176 Hz

7 0
3 years ago
A car's bumper is designed to withstand a 5.04 km/h (1.4-m/s) collision with an immovable object without damage to the body of t
emmasim [6.3K]

Answer:

the magnitude of the average force on the bumper is 3189.8 N

Explanation:

Given the data in the question;

In terms of force and displacement, work done is;

W =F^> × x^>

W = Fxcos\theta    ------- let this be equation 1

where F is force applied, x is displacement and θ is angle between force and displacement.

Now, since the displacement of the bumper and force acting on it is in the same direction,

hence, θ = 0°

we substitute into equation 1

W = Fxcos( 0° )

W = Fx ------- let this be equation 2

Now, using work energy theorem,

total work done on the system is equal to the change in kinetic energy of the system.

W_{net = ΔKE

= \frac{1}{2}mv² -  \frac{1}{2}mu² --------- let this be equation 3

where m is mass of object, v is final velocity, u is initial velocity.

from equation 2 and 3

Fx = \frac{1}{2}mv² -  \frac{1}{2}mu²

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given that mass of car m = 830 kg, x = 0.255 m, v = 0 m/s, and u = 1.4 m/s

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F = \frac{830}{(2)(0.255)}( (0)² - (1.4)² )

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F = -3189.8 N

The negative sign indicates that the direction of the force was in opposite compare to the direction of the velocity of the car.

Therefore, the magnitude of the average force on the bumper is 3189.8 N

6 0
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