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scoundrel [369]
3 years ago
15

A car accelerates from O to 60. miles per hour in 5.2 seconds. Calculate acceleration in m/s seconds. Calculate acceleration in

m/s^2.
Engineering
1 answer:
DochEvi [55]3 years ago
7 0

Answer:

a=5.515\frac{m}{s^{2} }

Explanation:

The first thing we will do is convert the units. Miles per hour to meters per second.

1 mile=1609.34 mts.

1 hora=3600 segundos

Performing the operations

60\frac{mile}{h}=\frac{(60*1609.34)}{3600}\frac{m}{s}=26.822\frac{m}{s}

Now, we will use the acceleration formula

a=\frac{v}{t}

Where v = speed and t = time

Substituting the values ​​of t=5.2s

a=\frac{v}{t} =\frac{26.822\frac{m}{s} }{5.2s} =5.15\frac{m}{s^{2} }

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A 7-hp (shaft) pump is used to raise water to an elevation of 15 m. If the mechanical efficiency of the pump is 82 percent, dete
Natali [406]

The maximum volume flow rate of water is determined as 0.029 m³/s.

<h3>Power of the pump</h3>

The power of the pump is watt is calculated as follows;

1 hp = 745.69 W

7 hp = ?

= 7 x 745.69 W

= 5,219.83 W

<h3>Mass flow rate of water</h3>

η = mgh/P

mgh = ηP

m = ηP/gh

m = (0.82 x 5,219.83)/(9.8 x 15)

m = 29.12 kg/s

<h3>Maximum volume rate</h3>

V = m/ρ

where;

  • ρ is density of water = 1000 kg/m³

V = (29.12)/(1000)

V = 0.029 m³/s

Learn more about volume flow rate here: brainly.com/question/21630019

#SPJ12

5 0
2 years ago
A fluid of specific gravity 0.96 flows steadily in a long, vertical 0.71-in.-diameter pipe with an average velocity of 0.90 ft/s
KengaRu [80]

Answer:

0.00650 Ib s /ft^2

Explanation:

diameter ( D ) = 0.71 inches = 0.0591 ft

velocity = 0.90 ft/s ( V )

fluid specific gravity = 0.96 (62.4 )  ( x )

change in pressure ( P ) = 0 because pressure was constant

viscosity =  (change in p - X sin∅ ) D^{2} / 32 V

              = ( 0 - 0.96( 62.4) sin -90 ) * 0.0591 ^2  / 32 * 0.90

              = - 59.904 sin (-90) * 0.0035 / 28.8

              = 0.1874 / 28.8

  viscosity = 0.00650 Ib s /ft^2

8 0
3 years ago
Read 2 more answers
Water flows around a 6-ft diameter bridge pier with a velocity of 12 ft/s. Estimate the force (per unit length) that the water e
jolli1 [7]

Answer: hello the diagram related to your question is missing please the third image is the missing part of the question

Fx = 977.76 Ib/ft

Explanation:

<u>Estimate the force that water exerts on the pier </u>

V = 12 ft/s

D( diameter ) = 6 ft

first express the force  on the first half of the cylinder  as

Fx1 =  - -2\int\limits^\pi _\frac{\pi }{2}   {Ps*cos\beta *a} \, d\beta   ---------------- ( 1 )

where ; Fy = 0

Ps = Po + 1/2 Pv^2 ( 1 - 4 sin^2β )  ------------- ( 2 )

Input equation (2)  into equation ( 1 )         (note :  assuming Po = 0 )

attached below is the remaining part of the solution

3 0
3 years ago
Why is it reasonable to say that no system is 100% efficient?​
Virty [35]

Generally, frictional losses are more predominant for the machines being not 100% efficient. This friction leads to the loss of energy in the form of heat, into the surroundings. Some of the supplied energy may be utilised to change the entropy (measure of randomness of the particles) of the system.

5 0
3 years ago
An aquifer has three different formations. Formation A has a thickness of 8.0 m and hydraulic conductivity of 25.0 m/d. Formatio
saveliy_v [14]

Answer:

The horizontal conductivity is 41.9 m/d.

The vertical conductivity is 37.2 m/d.

Explanation:

Given that,

Thickness of A = 8.0 m

Conductivity = 25.0 m/d

Thickness of B = 2.0 m

Conductivity = 142 m/d

Thickness of C = 34 m

Conductivity = 40 m/d

We need to calculate the horizontal conductivity

Using formula of horizontal conductivity

K_{H}=\dfrac{H_{A}K_{A}+H_{A}K_{A}+H_{A}K_{A}}{H_{A}+H_{B}+H_{C}}

Put the value into the formula

K_{H}=\dfrac{8.0\times25+2,0\times142+34\times40}{8.0+2.0+34}

K_{H}=41.9\ m/d

We need to calculate the vertical conductivity

Using formula of vertical conductivity

K_{V}=\dfrac{H_{A}+H_{B}+H_{C}}{\dfrac{H_{A}}{K_{A}}+\dfrac{H_{B}}{K_{B}}+\dfrac{H_{C}}{K_{C}}}

Put the value into the formula

K_{V}=\dfrac{8.0+2.0+34}{\dfrac{8.0}{25}+\dfrac{2.0}{142}+\dfrac{34}{40}}

K_{V}=37.2\ m/d

Hence, The horizontal conductivity is 41.9 m/d.

The vertical conductivity is 37.2 m/d.

3 0
3 years ago
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