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boyakko [2]
3 years ago
8

An artist is creating a gigantic mobile using old cars. On one end of the mobile, a 2100 lb Plymoth is suspended 30 feet from th

e fulcrum. In order for the mobile to be balanced, where should the artist suspend a 1400 lb Volkswagon?
a
20 feet from the fulcrum

b
7 feet from the fulcrum

c
45 feet from the fulcrum

d
on the fulcrum
Engineering
2 answers:
ozzi3 years ago
7 0

Answer:

  c  45 feet from the fulcrum

Explanation:

The moment at the fulcrum must be the same for each car. If the distance is d, then the artist must have ...

  (2100 lb)(30 ft) = (1400 lb)(d)

  2100·30/1400 ft = d = 45 ft

The Volkswagen should be 45 ft from the fulcrum.

Tresset [83]3 years ago
6 0

Answer:

<h2> C. <u>4</u><u>5</u><u> </u><u>feet</u><u> </u><u>from</u><u> </u><u>the</u><u> </u><u>fulcrum</u></h2>

  • <em>That</em><em> </em><em>artist</em><em> </em><em>should</em><em> </em><em>suspend</em><em> </em><em>a</em><em> </em><em>1</em><em>4</em><em>0</em><em>0</em><em> </em><em>lb</em><em>/</em><em>pounds</em><em> </em><em>Volkswagen</em><em>.</em>
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Igoryamba

Answer:

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3 years ago
The electron concentration in silicon at T = 300 K is given by
puteri [66]

Answer:

E=1.44*10^-7-2.6exp(\frac{-x}{18} )v/m

Explanation:

From the question we are told that:

Temperature of silicon T=300k

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7 0
3 years ago
An ideal vapor-compression refrigeration cycle operates at steady state with Refrigerant 134a as the working fluid. Saturated va
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Explanation:

Note: Refer the diagram below

Obtaining data from property tables

State 1:

\left.\begin{array}{l}P_{1}=1.25 \text { bar } \\\text { Sat - vapour }\end{array}\right\} \begin{array}{l}h_{1}=234.45 \mathrm{kJ} / \mathrm{kg} \\S_{1}=0.9346 \mathrm{kJ} / \mathrm{kgk}\end{array}

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State 3:

\left.\begin{array}{l}P_{3}=5 \text { bar } \\\text { Sat }-4 q\end{array}\right\} h_{3}=71-33 \mathrm{kJ} / \mathrm{kg}

State 4:

Throttling process  h_{4}=h_{3}=71.33 \mathrm{kJ} / \mathrm{kg}

(a)

Magnitude of compressor power input

\dot{w}_{c}=\dot{m}\left(h_{2}-h_{1}\right)=\left(8 \cdot 5 \frac{\mathrm{kg}}{\min } \times \frac{1 \mathrm{min}}{\csc }\right)(262.78-234 \cdot 45)\frac{kj}{kg}

w_{c}=4 \cdot 013 \mathrm{kw}

(b)

Refrigerator capacity

Q_{i n}=\dot{m}\left(h_{1}-h_{4}\right)=\left(\frac{g \cdot s}{60} k_{0} / s\right) \times(234 \cdot 45-71 \cdot 33) \frac{k J}{k_{8}}

Q_{i n}=23 \cdot 108 \mathrm{kW}\\1 ton of retregiration =3.51 k \omega

\ Q_{in} =6 \cdot 583 \text { tons }

(c)

Cop:

\beta=\frac{\left(h_{1}-h_{4}\right)}{\left(h_{2}-h_{1}\right)}=\frac{Q_{i n}}{\omega_{c}}=\frac{23 \cdot 108}{4 \cdot 013}

\beta=5 \cdot 758

3 0
3 years ago
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