A short framing member that fills the space between the rough sill and the soleplate is a

. A roadway is being designed capable of allowing 70 mph vehicle speed. The superelevation around one curve is 0.05 inches per i

adelina 88 [10]

The minimum radius of the curve to the vehicle to provide safe travel is 2339.2 ft

Explanation:

Given -

Speed allowed, v = 70mph

1 mph = 1.467 fps (feet per second)

Therefore, v = 70 X 1.467 fps

Superelevation, e = 0.05 inches/inch

Coefficient of side friction, fs = 0.09

Minimum radius of the curve, r = ?

We know,

r = v² / g ( fs + e/100 )

where,

g = acceleration due to gravity in feet

g = 32.2 lb.ft / lbf.s²

r = ( 70 X 1.467 )² / 32.2 ( 0.09 + 0.05 )

r = 2339.2 ft

Therefore, the minimum radius of the curve to the vehicle to provide safe travel is 2339.2 ft

6 0

3 years ago

Were women treated as equals to men in early aviation history?

Dmitry_Shevchenko [17]

Answer:

No

Explanation:

In the early Aviation history men have always dominated the world of aviation and women are notoriously under-represented, particularly in technical and leadership roles. Therefore, women was treated unequal.

8 0

3 years ago

Read 2 more answers

An 80-L vessel contains 4 kg of refrigerant-134a at a pressure of 160kPa. Determine (a) the temperature, (b) the quality, (c) th

makvit [3.9K]

Answer:

temperature -15.6 C, quality x=0.646, enthalpy h=667.20 KJ, volume of vapor phase Vg= 79.8 L

Explanation:

property table for R-134a

https://www.ohio.edu/mechanical/thermo/property_tables/R134a/R134a_PresSat.html

at 160 KPa , temperature = -15.66 C

quality x=mass of vapour/ total mass of liq-vap mixture

alternaternately: x=(v-vf)/(vg-vf)

v=total volume i.e. volume of container"80L" 80L=0.08 cubic meter

vf=vol of liquid phase vg=vol of vapor phase vf, vg values at 160Kpa

x=(0.08-0.0007437)/(0.1235-0.0007437)=0.646

enthalpy

h=hf+xhfg hf, hfg values at 160Kpa

h=hf+xhfg=31.2+0.646(209.9)=166.80 KJ/Kg

for 4Kg R-134a h=m(166.80 KJ/Kg )=667.20 KJ

volume of vapor phase

vg at 160Kpa=0.1235 cubic meter for quality=1.

in this case quality=0.646 , so it will occupy 64.6% space of the vapor phase at quality=1.

vol. of vapor phase=0.646*0.1235=0.0798 cubic meter=79.8 L

7 0

3 years ago

Under conditions for which the same roojm temperature is mainteined bt a heating or cooling system, it is not uncommon for a per

bonufazy [111]

Answer:

Net heat transfers: during summer = 52.253 W/m^2 during winter = 119.375 w/m^2

Explanation:

Given data :

Room temperature throughout the year = 20⁰c = 293 k

Room temperature during summer = 27⁰c = 300 k

Room temperature during winter = 14⁰c = 287 k

surface temperature of a person throughout the year = 32⁰c = 305 k

coefficient of heat transfer by natural convection (h)= 2 w /m^2 k

emissivity = 0.9

An explanation to the condition of feeling chilled during the winter and comfortable during summer can be explained with the calculation below

The heat transfer from the surface body of a person the the room is carried out by convection and this can be calculated as

q = hΔt = 2 * ( 305 - 293) = 2 * 12 = 24 w /m^2

also calculate heat transfer through radiation using this formula

$q_{rad} =$ εσ [ (Temperature of body)^4- (temperature of room at each season)^4 ]

ε = 0.90,(emissivity) σ = 5.67 *10^-8 w/^2 . k ( Boltzmann's constant)

during summer :

$q_{rad}$ = (0.90)*(5.67*10^-8)* ( 305^4 - 300^4 ) = 28.253 W/m^2

therefore the net heat transfer during the summer = 24 w/m^3 + 28.253 W/m^3 = 52.253 W/m^2

During winter :

$q_{rad}$ = (0.90)*(5.67*10^-8) * ( 305^4 - 287^4 ) = 95.375 W/m^3

therefore the net heat transfer during the winter

= 24 w/m^3 + 95.375w/m^3 = 119.375 w/m^2

7 0

3 years ago

The Environmental Protection Agency (EPA) has standards and regulations that says that the lead level in soil cannot exceed the

DENIUS [597]

Answer:

See below

Explanation:

Check One-Sample T-Interval Conditions

Random Sample? √

Sample Size ≥30? √

Independent? √

Population Standard Deviation Unknown? √

One-Sample T-Interval Information

Formula --> $CI=\bar{x}\pm t^*(\frac{S_x}{\sqrt{n}})$
Sample Mean --> $\bar{x}=390.25$
Critical Value --> $t^*=2.0096$ (given $df=n-1=50-1=49$ degrees of freedom at a 95% confidence level)
Sample Size --> $n=50$
Sample Standard Deviation --> $S_x=30.5$

Problem 1

The critical t-value, as mentioned previously, would be $t^*=2.0096$ , making the 95% confidence interval equal to $CI=\bar{x}\pm t^*(\frac{S_x}{\sqrt{n}})=390.25\pm2.0096(\frac{30.5}{\sqrt{50}})\approx\{381.5819,398.9181\}$

This interval suggests that we are 95% confident that the true mean levels of lead in soil are between 381.5819 and 398.9181 parts per million (ppm), which satisfies the EPA's regulated maximum of 400 ppm.

3 0

2 years ago