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Pavlova-9 [17]
3 years ago
6

A structure that extends from a bridge toform a firm foundation​

Engineering
2 answers:
igomit [66]3 years ago
8 0

Answer:

Explanation:

hi

Vlad [161]3 years ago
5 0
Jeanpaul because he was the first guy that was smart
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Define schema &amp; Instance...<br><br><br>don't spam..<br>​
zubka84 [21]

Answer:

<h3>Hey there ! </h3><h3>Here is your Answer Buddy | </h3>

<h3>• Define Instance :- </h3>

  • The data stored in database at a particular moment of time is called instance of database.

<h3>• Define Schema :- </h3>

  • Database schema defines the variable declarations in tables that belong to a particular database

<h3><u>The value of these variables at a moment of time is called the instance of that database.</u></h3><h3 />

Explanation:

<h3>Hope this helps !! </h3>
4 0
3 years ago
Read 2 more answers
A 1 250 kg car moving at a velocity of 30 km/hr along EDSA is accelerated by a force of 1 700 N. What will be its velocity after
Talja [164]
<h3><u>The velocity of the car after 10 s is 78.95 km/hr</u></h3>

Explanation:

<h2>Given:</h2>

m = 1,250 kg

v_i = 30 km/hr

F = 1,700 N

t = 10 s

<h2>Required:</h2>

Final velocity

<h2>Equation:</h2><h3>Force</h3>

F = ma

where: F - force

m - mass

a - acceleration

<h3>Acceleration</h3>

a = \frac{v_f \:-\:v_i}{t}

where: a - acceleration

v_i - initial velocity

v_f - final velocity

t - time elapsed

<h2>Solution:</h2><h3>Solve for acceleration using the formula for force</h3>

F = ma

Substitute the value of F and m

(1700 N) = (1250 kg)(a)

a = \frac{1700\:N}{1250\:N}

a = 1.36 m/s²

<h3>Solve for final velocity using the formula for acceleration</h3>
  • Convert 30 km/hr to m/s

= \frac{30\:km}{hr}\:×\:\frac{1000\:m}{1\:m}\:×\:\frac{1\:hr}{3600\:s}

= 8.33 m/s

  • Substitute the value of a, v_i and t

a = \frac{v_f \:-\:v_i}{t}

1.36\: m/s² \:= \:\frac{v_f \:-\:8.33\:m/s}{10\:s}

(10 \:s)1.36\: m/s² \:= \:v_f \:-\:8.33\:m/s

v_f\: =\: (10 \:s)1.36 \:m/s²\: + \:8.33\:m/s

v_f \: =\: 13.6 \:m/s \:+\: 8.33\:m/s

v_f\: =\:  21.93\: m/s

  • Convert to km/hr

= \frac{21.93\:m}{s}\:×\:\frac{1\:km}{1000\:m}\:×\:\frac{3600\:s}{1/:hr}

= 78.95\: km/hr

<h2>Final answer</h2><h3><u>The velocity of the car after 10 s is 78.95 km/hr</u></h3>
6 0
3 years ago
Water is flowing steadily through a 3-m long, 20 mm innerdiameter cast iron pipe. The water enters at a uniform velocity U and e
devlian [24]

Answer:

Hello your question is incomplete attached below is the complete question

answer :

U/r = R  = U_{max} [ 1 - (\frac{R}{R} )^2 ] = 0

∴ No slip condition is satisfied

Explanation:

Given that

Outlet velocity  U(r)  = U_{max} [ 1 - (\frac{r}{R} )^2 ]

<u>prove that the output velocity profile satisfies the no slip condition</u>

at no slip u = 0 ( i.e. at the pipe's inner surface )

at , r = R  ( inference  is at center )

hence  U/r = R  = U_{max} [ 1 - (\frac{R}{R} )^2 ] = 0

∴ No slip condition is satisfied

8 0
3 years ago
While discussing the MIL on OBD II systems: Technician A says that the MIL will flash if the PCM detects a fault that would dama
Damm [24]

Answer:

"C"

Explanation:

8 0
3 years ago
A plate in the shape of an isosceles triangle 3 feet high and 4 feet wide is submerged vertically in water, base doward, with th
Paul [167]

Answer:

The force exerted by the water on one side of the plate is F = 24*pg  

Explanation:

From the given question, the first step to take is to find  he force exerted by the water on one side of the plate.

Solution

Given that:

Let the pressure the  at a depth of y ft be = pgy lb/Pa

the area of the atrip is given as = f(y)*delta(y) = 4/3*(y-2)delta(y)

Then

we combine with the range for y as = y E [2 , 5]

Thus,

F = 4/3*pg * integral from (2 , 5) [y(y-2)] dy

Recall that,

p = water density

g= gravity of acceleration

so,

F = 4/3*pg * integral from (2 , 5) [y^2 - 2y]dy]

F = 4/3*pg * [y^3/3 - y^2] [2 , 5]

F = 4/3*pg * [18]

Finally, F = 24*pg  

7 0
3 years ago
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