Market research is a good place to start the design process and usually involves asking questions about consumers.

8. Two 40 ft long wires made of differing materials are supported from the ceiling of a testing laboratory. Wire (1) is made of

san4es73 [151]

Answer:

Material K has a modulus of elasticity E=3.389× 10¹¹ Pa

Material H has a modulus of elasticity E=1.009 × 10⁹ Pa

Material K has higher value of modulus of elasticity than material H

Material K is stiffer.

Explanation:

Wire 1 material H

Length=L = 40 ft =12.192 m

Diameter= 3/8 in = 0.009525 m

Area= A= πr²,where r=0.009525/2 =0.004763

A=3.142*0.004763² =0.00007126 m²

Force, F= 225 lb= 225*4.45 =1001.25 N

Change in length =Δ L= 0.10 in = 0.00254

To find modulus of elasticity apply'

E=F*L/A*ΔL

E=1001.25*12.192/(0.004763*0.00254)

E= 1009027923.58 Pa

E=1.009 × 10⁹ Pa

For Wire 2 material K

Length=L= 40 ft =12.192 m

Diameter = 3/16 in = 0.1875 in = 0.004763 m

Area= πr² = 3.142 * (0.004763/2)² = 0.00000567154 m²

Force, F= 225 lb= 225*4.45 =1001.25 N

Change in length =Δ L= 0.25 in =0.00635 m

To find modulus of elasticity apply'

E=F*L/A*ΔL

E= (1001.25*12.192)/(0.00000567154 * 0.00635 )

E=338955422575 Pa

E=3.389× 10¹¹ Pa

Material K has a greater modulus of elasticity

The material with higher value of E is stiffer than that with low value of E.The stiffer material is K.

8 0

3 years ago

A container filled with a sample of an ideal gas at the pressure of 150 Kpa. The gas is compressed isothermally to one-third of

lyudmila [28]

Answer: c) 450 kPa

Explanation:

Boyle's Law: This law states that pressure is inversely proportional to the volume of the gas at constant temperature and number of moles.

$P\propto \frac{1}{V}$ (At constant temperature and number of moles)

$P_1V_1=P_2V_2$

where,

$P_1$ = initial pressure of gas = 150 kPa

$P_2$ = final pressure of gas = ?

$V_1$ = initial volume of gas = v L

$V_2$ = final volume of gas = $\frac{v}{3}L$

$150\times v=P_2\times \frac{v}{3}$

$P_2=450kPa$

Therefore, the new pressure of the gas will be 450 kPa.

7 0

4 years ago

Stream Piracy – Kaaterskill, NY. Check and double-click the Problem 15 folder. The dark blue and orange streams highlight the pr

baherus [9]

Answer:

b. The pirating streams are eroding headwardly to intersect more of the other streams’ drainage basins, causing water to be diverted down their steeper gradients.

Explanation:

From the Kaaterskill NY 15 minute map (1906), this shows two classic examples of stream capture.

The Kaaterskill Creek flow down the east relatively steep slopes into the Hudson River Valley. While, the Gooseberry Creek is a low gradient stream flowing down the west direction which in turn drains the higher parts of the Catskills in this area.

However, there is Headward erosion of Kaaterskill Creek which resulted to the capture of part of the headwaters of Gooseberry Creek.

The evidence for this is the presence of "barbed" (enters at obtuse rather than acute angle) tributary which enters Kaaterskill Creek from South Lake which was once a part of the Gooseberry Creek drainage system.

It should be noted again, that there is drainage divide between the Gooseberry and Kaaterskill drainage systems (just to the left of the word Twilight) which is located in the center of the valley.

As it progresses, this divide will then move westward as Kaaterskill captures more and more of the Gooseberry system.

5 0

3 years ago

Suppose within your web browser you click on a link to obtain a web page. The IP address for the associated URL is already cache

aleksandrvk [35]

Answer:

All the detailed steps are mentioned in pictures.

Explanation:

See attached pictures.

8 0

3 years ago

Come and look on my attachment

CaHeK987 [17]

Crazy Guy what do uh mean ?

4 0

3 years ago