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3 years ago

Raw sewage is raised vertically by 5.49m in the amountof

Physics

1 answer:

uranmaximum [27]3 years ago

5 0

Answer:

P = 1235.7646 W

Explanation:

Given data:

height of raised sewage = 5.49 m

rate of sewage =1.89*10^6 lt/day

density of sewage = 1.050 kg/m^3

power is written as

$P = \frac{work}{time}$

work = m g h

$m - mass = ( \rho V)$

h -height

mass lifted per day $= ( 1.89*10^6)(1.050)$

= 1984500 kg

time = 24 hours* 3600 seconds per hour

power $P = \frac{( 1984500)(9.8)( 5.49)}{(24)(3600)}$

P = 1235.7646 W

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21. A 60 kg student on a scooter is pushed with a constant force of 40 N across a horizontal concrete driveway at a constant spe

DedPeter [7]

The magnitude of the force of friction is 40 N

Explanation:

To solve the problem, we just have to analyze the forces acting on the student and the scooter along the horizontal direction. We have:

- The constant pushing force forward, of magnitude F = 40 N

- The frictional force, acting backward, $F_f$

Since the two forces are in opposite direction, the equation of motion is

$F-F_f = ma$

where

m is the mass of the student+scooter

a is the acceleration

However, here the scooter is moving at constant speed: this means that its acceleration is zero, so

a = 0

And therefore,

$F - F_f = 0\\F_f = F$

which means that the magnitude of the force of friction is also equal to 40 N.

Learn more about force of friction here:

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3 years ago

What Do You Get When You Multiply An Object's mass times the acceleration?

emmasim [6.3K]

You get the net force acting on it ... the sum of the strengths and directions
of all the individual forces there may be.

7 0

3 years ago

You go to the hardware store to buy a new 50 ft garden hose. You find you can choose between hoses of ½ inch and 5/8 inch inner

omeli [17]

To solve this problem it is necessary to consider two concepts. The first of these is the flow rate that can be defined as the volumetric quantity that a channel travels in a given time. The flow rate can also be calculated from the Area and speed, that is,

Q = V*A

Where,

A= Cross-sectional Area

V = Velocity

The second concept related to the calculation of this problem is continuity, which is defined as the proportion that exists between the input channel and the output channel. It is understood as well as the geometric section of entry and exit, defined as,

$Q_1 = Q_2$

$V_1A_1=V_2A_2$

Our values are given as,

$A_1=\frac{1}{2}^2*\pi=0.785 in^2$

$A_2=\frac{5}{8}^2*\pi=1.227 in^2$

Re-arrange the equation to find the first ratio of rates we have:

$\frac{V_1}{V_2}=\frac{A_2}{A_1}$

$\frac{V_1}{V_2}=\frac{1.227}{0.785}$

$\frac{V_1}{V_2}=1.56$

The second ratio of rates is

$\frac{V2}{V1}=\frac{A_1}{A2}$

$\frac{V2}{V1}=\frac{0.785}{1.227}$

$\frac{V2}{V1}=0.640$

3 0

3 years ago

Identify the type of chemical reaction that is described.

Anastasy [175]

Water breaks down into hydrogen and oxygen: It is a decomposition reaction as a single substance decomposes to give two products.

$H_2O \rightarrow H_2 +\frac {1}{2}O_2$

Leaves make starch using chlorophyll and carbon dioxide: Synthesis reaction: as the synthesis reaction involves two or more than two reactants which join together to result into a single main product along with the formation of simple by products.

$6CO_2 + 6H_2O \rightarrow C_6H_{12}O_6 + 6O_2$

Food burns in oxygen gas and releases a lot of energy: Combustion: Combustion process involves the use of oxygen to give products along with release of energy.

$C_6H_{12}O_6 + 6O_2\rightarrow 6CO_2 + 6H_2O$

Adding vinegar (acid) to baking soda (alkali) gives a product that is neither acidic nor alkaline: Neutralization: acetic acid in vinegar reacts with soda (base) to give salt (neutral) and .

$CH_3COOH + NaHCO_3\rightarrow CH_3COONa +H_2O + CO_2$

6 0

3 years ago

A 1.5-kg mass attached to an ideal massless spring with a spring constant of 20.0 N/m oscillates on a horizontal, frictionless t

Molodets [167]

Answer:

Explanation:

a ) angular frequency ω = $\sqrt{\frac{k}{m} }$

k is spring constant and m is mass attached

ω = $\sqrt{\frac{20}{1.5} }$

= 3.6515 rad / s

frequency of oscillation n = 3.6515 / (2 x 3.14)

= .5814 s⁻¹

x = .1 mcos(ωt)

= .1 mcos(3.6515t)

b ) maximum speed = ωA , A is amplitude

= 3.6515 x .1

= .36515 m /s

36.515 cm /s

maximum acceleration = ω²A

= 3.6515² x .1

= 1.333 m / s²

c ) Kinetic energy at displacement x

= 1/2 m ω²( A²-x²)

potential energy =1/2 m ω²x²

so 1/2 m ω²( A²-x²) = 1/2 m ω²x²

A²-x² = x²

2x² = A²

x = A / √2

6 0

3 years ago