Ask question

Ask question

All categories

kolbaska11 [484]

3 years ago

12

An angle is observed repeatedly using the same equipment and procedures producing the data below:35 ∘ 40'00",35 ∘ 40'10",35 ∘ 40

'10", and 35 ∘ 39'55"A) Calculate the standard deviation.B)Calculate the standard deviation of the mean.

1 answer:

Helen [10]3 years ago

6 0

Answer: (a) +/- 7.5° (b) +/- 3.75°

Explanation:

See attachment

You might be interested in

The chart describes four people’s credit histories.

icang [17]

Answer:

D). Eesha pays more than the minimum payment each month.

Explanation:

Eesha would be considered most creditworthy among the given persons as she not only pays on time but also repays more than the minimum amount assigned to pay each month. In order to test the creditworthiness of an individual, his ontime debt paying capability is tested at first followed by the past credit repayment history and the credit score. Except Eesha, all the given candidates have failed to make timely repayment of their debts and hence, they cannot be considered creditworthy.

7 0

3 years ago

Read 2 more answers

A pressure-temperature curve shows that pressure and temperature are ____ in a refrigeration system.

erica [24]

A pressure-temperature curve demonstrates that pressure and temperature are directly related in a refrigeration system.

<h3>What is refrigeration system?</h3>

In both commercial and residential settings, refrigeration systems are crucial because they provide cooling or maintain a specific room temperature. With a refrigerant flowing through the units, a refrigeration cycle entails heat exchange, compression, and expansion. The cold room's refrigeration system keeps it running.

It is merely a method of moving heat from one location to another. The method of refrigeration that is most frequently utilized is the vapor-compression system. It is widely employed in sizable cold spaces, such as commercial chillers. A physiological technique for monitoring or diagnosing acute respiratory distress syndrome during mechanical ventilation is the pressure-volume (PV) curve.

Hence, A pressure-temperature curve demonstrates that pressure and temperature are directly related in a refrigeration system.

To learn more about refrigeration system refer to:

brainly.com/question/26395073

#SPJ4

7 0

2 years ago

1. A six lane freeway (three lanes in each direction) currently operates at maximum LOS C conditions. The lanes are 11 ft wide,

kiruha [24]

Answer:

The answer is " $4,071 \ \frac{veh}{h}$ ".

Explanation:

Compute the Free-Flow Speed estimation (FFS)

The formulation:

$FFS=75.4-f_{LW}-f_{LC}-3.22 TRD^{0.84}$

The lane width change is 11 ft. The width of the lane.

$F_{LW} \ is \ 19 \ \frac{m}{h}$

Adaptation for lateral clearing for both the night shoulder

Right side shoulder equivalent 4 ft

In one way clearance or 3 lanes.

$f_{LC} \ is \ 0.8 \ \frac{m}{h}$

$FFS = 75.4 -1.9 -0.8 -3.22(\frac{3}{6})^{0.84}$

$= 75.4 -1.9 -0.8 - 1.799\\\\= 70.901\ \frac{mi}{h}\\\\= 70 \ \frac{mi}{h}$

Adjustment factor (f_{HV})

The eqlivdents for the passenger vehicles (PCEs). Any corresponding segment of its moving field. Take the $E_T \ value \ 2.5$ and the $E_R \ value \ 2.0$

$It \ replace \ 10\% \ with \ p_r, \ 2.5 \ with\ E_r, \ 0 \ with \ P_R, \ E_R's \ and \ 2.0:$

$f_{(HV)}= \frac{1}{1 + \frac{10}{100} (2.5-1)+(0)(2.0-1)}$

$= 0.869$

Compute volume(V) hourly:

Please take 15 Minute passenger car eqvialent flow rate for the LOS Parameters for the LOS C and FFS of $70 \frac{mi}{h}$ So, $v_p$ is worth $1,735 \frac{pc}{\frac{h}{In}}$

Consider that $f_p$ is 1.00 for commuters.

Replace $v_p$ with $1,735 \frac{pc}{\frac{h}{In}}$

Thus, the value of $v_p$ , is $1,735 \frac{pc}{\frac{h}{In}}$

Consider for commuters the value of f_p, is 1.00.

Substitute for v_p,

$0.90 \ for \ PHF\ 3 for \ N,\ 0.869 \ for\\F_{HV} \ and \ \ 1.00 \ for\ f_v \\\\1,735 = \frac{V}{0.9 \times 3 \times 0.869 \times 1} \\\\V= 1,735 \times 0.9 \times 0.869 \\= 4,070.83 = 4,071 \ \ \frac{veh}{h}$

6 0

3 years ago

What is your understanding and opinion on the validity of the stand-by equipment maintenance requirement?

Fed [463]

Answer:

The stand-by equipment is technically required in case where the we need some urgent equipment for the purpose of maintenance in emergency and if the other equipment system get fails. the term stand by means backup equipment and component.

In the reliability engineering, we always need to provide an extra equipment or component in case of emergency. It is basically used so that it does not affect any type of productivity in the organization. It is also increase the redundancy of the equipment.

6 0

3 years ago

Magnesium sulfate has a number of uses, some of which are related to the ability of the anhydrate form to remove water from air

lara31 [8.8K]

Answer:

Magnesium sulphate is in two forms

Magnesium anhydrate and magnesium hepta hydrate

Density of MgSO4= 2.66 g/ml

Density of MgSO4. 7H2O= 1.68 g/ml

M. W of MgSO4= 120 kg/kmol

In order to form 1000 Kg 20wt% solution

A) the Kg of water required before addition of crystals for MgSO4 =(1000) (0.80)= 800 kg

Average density= (0.20) (2.66) +(0.80) (1) = 1.332 g/ml = 1332 Kg/m3

Volume of water required= (800/1332) = 600.60 litres

For hepta hydrate

Average density= (0.20) (1.68) +(0.80) (1) = 1.136 g/ml=

1136 kg/m 3

MgSO4. 7 H2O, mole fraction of water present =

7×18/(120+(7×18) )=0.512 = 51.2 mol%= 13.6 wt%

1000 kg of (20 wt%) solution required

Amount of pure MgSO4 needed= 1000(.20) = 200 Kg

But it contains 13.6% water hence total weight of

MgSO4. 7H2O= 200/(1-0.136) = 231.48 kg

Amount of water required= (1000-231.48) = 768.51 Kg

Volume of water required= 768.51/1136= 676.51 Litres

B) diameter of tank= 0.30 m

V= 3.142(d2)(H) /4

For pure MgSO4 , volume of water required=600.60 litres

H = 8.49 m

For hepta hydrate volume of water required=676.51 litres

H= 9.57 m

C) height of tank after addition of crystals

Total volume in case of pure MgSO4=

1000/(1332) = 0.7507= 750.7 litres

H= 10.62 m

Total volume of hepta hydrate =

1000/(1136) = 0.8802= 880.2 litres

H= 12.45 m

D) after dissolution of crystals

The volume after dissolution in case of pure MgSO4

= volume of water= 600.67 litres

H=8.49 m

in case of hepta hydrate

The volume after dissolution= ( volume of water + volume of water in MgSO4. 7H2O)

= (676.51+(231.48×0.136) = 707.99 Litres

H= 10.01 m

Part B question

A) Weight of house= 1×105 lb= 2.20 ×105

Force applied by house= (2.20×105) (9.81) = 2.15×106 N

Pressure of ballon= 1.05 atm= 1.063×105 Pa

Diameter= 9.5 inch= 0.2413 m

Area= (3.142) (0.2413) 2/4= 0.0457 m2

F= P×A= 4857.91 N

To float the house

Both forces should be equal

So force applied by 1 ballon= 4857.91 N

Total force required=2.15×106 N

Number of ballons required= (2.15×106) /4857.91=442.57=443 ballons

B) the ballon pressure must be greater than atmospheric pressure

If both pressures are equal then there will be no air flow

If outside pressure is higher, then air would flow from outside to inside of ballon causing deflation, hence pressure inside the ballon must be higher than outside

6 0

4 years ago