A 3 m aluminum pole is kept at a residential site for construction

A rod that was originally 100-cm-long experiences a strain of 82%. What is the new length of the rod?

Ierofanga [76]

Answer: (b)

Explanation:

Given

Original length of the rod is $L=100\ cm$

Strain experienced is $\epsilon=82\%=0.82$

Strain is the ratio of the change in length to the original length

$\Rightarrow \epsilon =\dfrac{\Delta L}{L}\\\\\Rightarrow 0.82=\dfrac{\Delta L}{100}\\\\\Rightarrow \Delta L=82\ cm$

Therefore, new length is given by (Considering the load is tensile in nature)

$\Rightarrow L'=\Delta L+L\\\Rightarrow L'=82+100=182\ cm$

Thus, option (b) is correct.

8 0

3 years ago

A tungsten matrix with 20% porosity is infiltrated with silver. Assuming that the pores are interconnected, what is the density

daser333 [38]

Answer:

15.4 g/cm³, 17.4 g/cm³

Explanation:

The densities can be calculated using the formula below

ρ = (fraction of tungsten × ρt ( density of tungsten)) + (fraction of pores × ρp( density of pore)

fraction of tungsten = (100 - 20 ) % = 80 / 100 = 0.8

a) density of the before infiltration = ( 0.8 × 19.25) + (0.2 × 0) = 15.4 g/cm³

b) density after infiltration with silver

fraction occupied by silver = 20 / 100 = 0.2

density after infiltration with silver = ( 0.8 × 19.25) + (0.2 × 10) = 17.4 g/cm³

5 0

3 years ago

During a cold winter day, wind at 50 km/h is blowing parallel to a 4-m-high and 10-m-long wall of a house. If the air outside is

fenix001 [56]

Answer:

to determine the rate of heat loss from that wall by convection = 12780 watts

Explanation:

8 0

3 years ago

The alignment readings for the front of a vehicle are shown above. Camber and toe are within specification, caster is not. Techn

dlinn [17]

Answer:

B. B only

Given Information:

1. Camber and toe are within specification

2. Caster is not within specification

Technician A says that with the current settings, the left front tire tread may wear on the inside edge.

Technician B says that with the current settings, the vehicle may pull to the left

Explanation:

Lets discuss the effects of Camber, toe and caster misalignment

Effects of Camber and Toe misalignment:

Camber is the inward or outward tilt of the fron tires and is used to distribute load across the tread. Any misalignment causes uneven loading on the tires which results in tire wear on one edge.

The most common cause of tire wear on the inside edge is due to the camber misalignment which results in premature tire wear.

Another reason is of tire wear is vehicle’s toe. A slight misalignment of the toe reduces the life of the tire.

Since it is given that camber settings and toe settings are within specification therefore, tire tread wear on the inside edge cannot happen if camber and toe are within specification.

Technician A cannot be right.

Effects of Caster misalignment:

Whenever there is a misalignment of the castor then the vehicle will not be able to go in straight line rather it will pull to either left or right side. Caster misalignment also causes heavy or light steering depending upon the positive or negative misalignment of caster.

Since it is given that caster settings are not within specification therefore, the vehicle may pull to the left due to the caster misalignment.

Technician B must be right.

4 0

4 years ago

Liquid flows at steady state at a rate of 2 lb/s through a pump, which operates to raise the elevation of the liquid 100 ft from

Greeley [361]

Answer:

D) 1.04 Btu/s from the liquid to the surroundings.

Explanation:

Given that:

flow rate (m) = 2 lb/s

liquid specific enthalpy at the inlet ( $h_{1}=40.09$ Btu/lb)

liquid specific enthalpy at the exit ( $h_{2}=40.94$ Btu/lb)

initial elevation ( $z_1=0ft$ )

final elevation ( $z_2=100ft$ )

acceleration due to gravity (g) = 32.174 ft/s²

$W_{cv}$ = 3 Btu/s

The energy balance equation is given as:

$Q_{cv}-W{cv}+m[(h_1-h_2)+(\frac{V_1^2-V_2^2}{2})+g(z_1-z_2)]=0$

Since kinetic energy effects are negligible, the equation becomes:

$Q_{cv}-W{cv}+m[(h_1-h_2)+g(z_1-z_2)]=0$

Substituting values:

$Q_{cv}-(-3)+2[(40.09-40.94)+\frac{32.174(0-100)}{778*32.174} ]=0\\Q_{cv}+3+2[-0.85-0.1285 ]=0\\Q_{cv}+3+2(-0.9785)=0\\Q_{cv}+3-1.957=0\\Q_{cv}+1.04=0\\Q_{cv}=-1.04\\$

The heat transfer rate is 1.04 Btu/s from the liquid to the surroundings.

8 0

3 years ago