In your own words, describe which engineering discipline (civil, mechanical, computer, or electrical) would be used to design a new sports arena.

Answers

Answer 1

Answer:

Civil engineering.

Explanation:

Civil engineering is a branch of engineering whose mission is the design and construction of structures and buildings for civil use, understood as the use of these structures by society in the context of their daily life. Thus, for example, this branch of engineering is responsible for the construction of bridges, dams, tunnels, etc., being in all cases structures that facilitate the normal development of life in society.

In this context, the design and construction of a new sports arena, as long as it meets these parameters, would be carried out by a civil engineer.


Related Questions

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 10 seconds? (Neglect friction)

Answers

The velocity of the car after 10 s is 78.95 km/hr

Explanation:

Given:

m = 1,250 kg

[tex]v_i[/tex] = 30 km/hr

F = 1,700 N

t = 10 s

Required:

Final velocity

Equation:Force

F = ma

where: F - force

m - mass

a - acceleration

Acceleration

a = [tex]\frac{v_f \:-\:v_i}{t}[/tex]

where: a - acceleration

[tex]v_i[/tex] - initial velocity

[tex]v_f[/tex] - final velocity

t - time elapsed

Solution:Solve for acceleration using the formula for force

F = ma

Substitute the value of F and m

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

a = [tex]\frac{1700\:N}{1250\:N}[/tex]

a = 1.36 m/s²

Solve for final velocity using the formula for accelerationConvert 30 km/hr to m/s

= [tex]\frac{30\:km}{hr}\:×\:\frac{1000\:m}{1\:m}\:×\:\frac{1\:hr}{3600\:s}[/tex]

= [tex] 8.33 m/s [/tex]

Substitute the value of a, [tex]v_i[/tex] and t

a = [tex]\frac{v_f \:-\:v_i}{t}[/tex]

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

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

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

[tex]v_f \: =\: 13.6 \:m/s \:+\: 8.33\:m/s[/tex]

[tex]v_f\: =\: 21.93\: m/s[/tex]

Convert to km/hr

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

= [tex]78.95\: km/hr[/tex]

Final answerThe velocity of the car after 10 s is 78.95 km/hr
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