A picture of a ______ is the label pictogram on a chemical that warns of a product's acute toxicity.

13–27. The conveyor belt is moving downward at 4 m>s. If the coefficient of static friction between the conveyor and the 15-k

Feliz [49]

Answer:

See explanation for step by step procedure to get answer.

Explanation:

Given that:

The conveyor belt is moving downward at 4 m>s. If the coefficient of static friction between the conveyor and the 15-kg package B is ms = 0.8, determine the shortest time the belt can stop so that the package does not slide on the belt.

See the attachments for complete steps to get answer.

4 0

2 years ago

Six years ago, an 80-kW diesel electric set cost $160,000. The cost index for this class of equipment six years ago was 187 and

exis [7]

Answer:

new boiler total cost = $229706.825

new boiler total cost = $127512

Explanation:

given data

power p1 = 80 kW

cost C = $160000

cost index CI 1 = 187

cost index CI 2= 194

cost capacity factor f = 0.6

power p2 = 120 kW

current cost = $18000

to find out

total cost and cost of 40 kW

solution

we consider here CN cost for new boiler and CO cost for old boiler

and x is capacity of new boiler

first we find old boiler current cost that is

current cost CO = C × $\frac{CI 1 }{CI 2 }$ .............1

put here value

current cost = 160000 × $\frac{194 }{187 }$

new current cost = $165989.304

and

use here power sizing technique for 124 kW

CN/CO = $(\frac{p2}{p1} )^{f}$ ...............2

put here value and find CN

CN/CO = $(\frac{p2}{p1} )^{f}$

CN / 165989.304 = $(\frac{120}{80} )^{0.6}$

CN = 211706.825

so new cost = $211706.825

so

total cost for new boiler is

total cost = new cost + current cost

total cost = 211706.825 + 18000

new boiler total cost = $229706.825

and

for 40 kW new cost will be

use equation 2

CN/CO = $(\frac{p2}{p1} )^{f}$

CN / 165989.304 = $(\frac{40}{80} )^{0.6}$

CN = 109512

so new cost is $109512

so

total cost for new boiler is

total cost = new cost + current cost

total cost = 109512 + 18000

new boiler total cost = $127512

7 0

2 years ago

A platinum resistance temperature sensor has a resistance of 120 Ω at 0℃ and forms one arm of a Wheatstone bridge. At this tempe

oksian1 [2.3K]

Answer : 9.36ohms/ temperature

Explanation:

Expression for the variation of resistance of platinum with temperature

Rt= Ro(1+*t)

Rt= resistance @ t°C

Ro= resistance @ 0°C

*= temperature coefficient of resistance

Calculate the change in resistance by putting 120ohms for Ro,

0.0039/K for *

20°C for t

Using this formula:

Rt = Ro(1+*t)

Rt- Ro = Ro*t

= (120ohms)(0.0039/K)(20°C)

= 9.36ohms/K

8 0

2 years ago

An electrical utility delivers 6.25E10 kWh of power to its customers in a year. What is the average power required during the ye

Sindrei [870]

Answer:

The overall Utility delivered to customers in a year 'U' = 6.25 X 10¹⁰Kwh

However, the average power P, required for a year, t = ? Kw

Expressing their relationship, we will have

U = P x t

Given t = 1 year = 24 x 365 hours (assume a year operation is 365 days)

t = 8760 hours

P = $\frac{62500000000}{8760}$

P = 7134.7Kw

Hence, the average power required during the year is 7,135Kw

Now to calculate the energy used by the power plant in a year (in quads)?

Recall, Efficiency, η = Power Output/Power Input (100)

so, we have

η = P₀/P₁, given

0.45 = $\frac{7134.7Kw}{P₁}$

P₁ = 15,855Kw

the total energy E₁ used in a year = 15,855x24x365 = 138.89MJoules

So to convert this to quads, Note;

1 quads of energy = 10¹⁵ Joules

The total energy used is 0.000000139 quads

Now to find the cubic feet of natural gas required to generate this power?

Note: 0.29Kwh of Power generated = 1 cubic feet of natural gas used

Since, the power plant generated = 62500000000Kwh

The cubic feet of natural gas used = $\frac{62500000000}{0.29}$

Hence, 2.155x10²⁰cubic feet of N.gas was used to generate this much power.

8 0

2 years ago

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

2 years ago