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solmaris [256]
4 years ago
8

Fred wants to help his customer understand his vision for constructing the office interior. Which visual representation will bes

t explain Fred’s ideas?
A.
structural
B.
perspective projection
C.
auxiliary view
D.
cutting-plane view
E.
elevation drawing
Engineering
1 answer:
disa [49]4 years ago
8 0

Answer:

D. cutting-plane view

Explanation:

When using the cutting plane view, cutting plane lines are utilized to show a plane or planes in which a sectional view is shown. Section views give a clear indication of interior views of an object or structure including hidden parts  or features which are not directly visible through convectional ways such as direct observation.

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Water is flowing into the top of an open cylindrical tank (which has a diameter D) at a volume flow rate of Qi and the water flo
deff fn [24]

Answer:

Z = 3 + 0.23t

The water level is rising

Explanation:

Please see attachment for the equation

8 0
3 years ago
Read 2 more answers
E = 50V, R= 1000 ohms, then I =
Margarita [4]

Explanation:

<em>Current</em><em> </em><em>(</em><em>I)</em><em>=</em><em> </em><em><u>Voltage</u></em><em><u> </u></em><em><u>(</u></em><em><u>V)</u></em>

<em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>Resistance </em><em>(Ω)</em><em>. </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em>

<em>I=</em><em> </em><em><u>5</u></em><em><u>0</u></em><em><u>V</u></em>

<em> </em><em> </em><em> </em><em> </em><em>1000</em><em>(Ω)</em>

<em>I=</em><em> </em><em><u>5</u></em><em><u>V</u></em>

<em> </em><em> </em><em> </em><em> </em><em>1</em><em>0</em><em>0</em><em>(Ω)</em>

<em>I=</em><em> </em><em>0</em><em>.</em><em>0</em><em>5</em><em> </em><em>(</em><em>Amp)</em><em>. </em><em>Answer</em>

<em>Voltage=</em><em>V</em>

<em>Current</em><em>=</em><em>I</em>

<em>Resistance</em><em>=</em><em>(Ω)</em><em>. </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em> </em><em>(Ω)</em><em>=</em><em>Ohm</em>

<em>And </em><em>SI </em><em>unit</em><em> </em><em>of </em><em>current</em><em> </em><em>is</em><em>. </em><em>Ampere(</em><em>Amp)</em><em> </em>

<em><u>hope</u></em><em><u> this</u></em><em><u> helps</u></em><em><u> you</u></em>

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8 0
3 years ago
A 0.25in diameter steel rod BC is securely attached between two identical 1in diameter copper rods (AB and CD). Find the torque
Helen [10]

Answer:

Tmax= 46.0 lb-in

Explanation:

Given:

- The diameter of the steel rod BC d1 = 0.25 in

- The diameter of the copper rod AB and CD d2 = 1 in

- Allowable shear stress of steel τ_s = 15ksi

- Allowable shear stress of copper τ_c = 12ksi

Find:

Find the torque T_max

Solution:

- The relation of allowable shear stress is given by:

                             τ = 16*T / pi*d^3

                             T = τ*pi*d^3 / 16

- Design Torque T for Copper rod:

                             T_c = τ_c*pi*d_c^3 / 16

                             T_c = 12*1000*pi*1^3 / 16

                             T_c = 2356.2 lb.in

- Design Torque T for Steel rod:

                             T_s = τ_s*pi*d_s^3 / 16

                             T_s = 15*1000*pi*0.25^3 / 16

                             T_s = 46.02 lb.in

- The design torque must conform to the allowable shear stress for both copper and steel. The maximum allowable would be:

                             T = min ( 2356.2 , 46.02 )

                             T = 46.02 lb-in

6 0
3 years ago
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
Which of these is an example of a service job?
Sindrei [870]

Answer:

brainllest if right

A

Explanation:

7 0
3 years ago
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