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2 years ago

Two common types of glue

Engineering

2 answers:

dolphi86 [110]2 years ago

8 0

Answer:

white craft glue

yellow wood glue

Explanation:

the answer is this!

was it correct?sir!?

RSB [31]2 years ago

8 0

Answer:

epoxy

silicone

Explanation:

hope this helps

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The “Sun-Star” Company has purchased new ofﬁce furniture for their ofﬁces at a retail price of $100,000. An additional $12,000 h

Zinaida [17]

Based on the cost of the furniture, the following are true:

a. $10,400.
b. $93,600
c. $20,800

<h3>Depreciation in second year</h3>

Depreciation per year = (Cost of furniture - Salvage value) / Useful life

Cost will include both the purchase price and the charge for insurance and shipping.

= (100,000 + 12,000 - 8,000) / 10

= $10,400

<h3>BV at end of first year</h3>

= Cost - Depreciation

= 104,000 - 10,400

= $93,600

<h3>BV after 8 years </h3>

= Cost - (Depreciation x 8 years )

= 104,000 - (10,400 x 8)

= $20,800

In conclusion, depreciation is $10,400 per year.

Find out more about SL depreciation at brainly.com/question/13734742.

4 0

3 years ago

I need help with my autos

oksano4ka [1.4K]

Answer:

what is wrong with it and what is the question

Explanation:

6 0

3 years ago

A 1-ft rod with a diameter of 0.5 in. is subjected to a tensile force of 1,300 lb and has an elongation of 0.009 in. The modulus

iragen [17]

Answer:

E = 8.83 kips

Explanation:

First, we determine the stress on the rod:

$\sigma = \frac{F}{A}\\\\$

where,

σ = stress = ?

F = Force Applied = 1300 lb

A = Cross-sectional Area of rod = 0.5 $\pi \frac{d^2}{4} = \pi \frac{(0.5\ in)^2}{4} = 0.1963\ in^2$

Therefore,

$\sigma = \frac{1300\ lb}{0.1963\ in^2} \\\\\sigma = 6.62\ kips$

Now, we determine the strain:

$strain = \epsilon = \frac{elongation}{original\ length} \\\\\epsilon = \frac{0.009\ in}{12\ in}\\\\\epsilon = 7.5\ x\ 10^{-4}$

Now, the modulus of elasticity (E) is given as:

$E = \frac{\sigma}{\epsilon}\\\\E = \frac{6.62\ kips}{7.5\ x\ 10^{-4}}$

7 0

3 years ago

Determine the enthalpy, volume and density of 1.0 kg of steam at a pressure of 0.5 MN/m2 and with a dryness fraction of 0.96

Viktor [21]

Answer:

Enthalpy, hsteam = 2663.7 kJ/kg

Volume, Vsteam = 0.3598613 m^3 / kg

Density = 2.67 kg/ m^3

Explanation:

Mass of steam, m = 1 kg

Pressure of the steam, P = 0.5 MN/m^2

Dryness fraction, x = 0.96

At P = 0.5 MPa:

Tsat = 151.831°C

Vf = 0.00109255 m^3 / kg

Vg = 0.37481 m^3 / kg

hf = 640.09 kJ/kg

hg = 2748.1 kJ/kg

hfg = 2108 kJ/kg

The enthalpy can be given by the formula:

hsteam = hf + x * hfg

hsteam = 640.09 + ( 0.96 * 2108)

hsteam = 2663.7 kJ/kg

The volume of the steam can be given as:

Vsteam = Vf + x(Vg - Vf)

Vsteam = 0.00109255 + 0.96(0.37481 - 640.09)

Vsteam = 0.3598613 m^3 / kg

From the steam table, the density of the steam at a pressure of 0.5 MPa is 2.67 kg/ m^3

5 0

3 years ago

A certain part of the cast iron piping of a water distribution system involves a parallel section. Both parallel pipes have a di

Bezzdna [24]

Answer :

Explanation:

Given :

Length of pipe A $L_{A} = 1500$ m

Length of pipe B $L_{B} = 2500$ m

Flow rate through pipe A $Q_{A} = 0.4 \frac{m^{3} }{s}$

Diameter of pipe $D = 30 \times 10^{-2}$ m

Velocity from pipe A,

$V _{A} = \frac{Q_{A} }{A}$

$V _{A} = \frac{0.4 \times 4 }{\pi ( 30 \times 10^{-2} )^{2} }$

$V_{A} = 5.66$ $\frac{m}{s}$

Here, head loss is same because height is same.

$h_{a} = h_{b}$

$L_{A} V_{A} ^{2} = L_{B} V_{B} ^{2}$

$V_{B} = \sqrt{\frac{1500}{2500}} (5.66)$

$V_{B} = 4.38$ $\frac{m}{s}$

Now rate of flow from pipe B is,

$Q_{B} = V_{B} A$

$Q_{B} = \frac{\pi }{4} (0.3)^{2} \times 4.38$

$Q_{B} = 0.3094$ $\frac{m^{3} }{s}$

4 0

3 years ago