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2 years ago

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Type the correct answer in each box. Use numerals instead of words. If necessary, use / for the fraction bar. In a survey conduc

ted in an office, 60% of the sample respondents agreed that employees should have flexible working hours. If the margin of error is ±2%, the expected population proportion that wants flexible working hours is between __% and __%

1 answer:

jarptica [38.1K]2 years ago

3 0

Answer: The expected population that wants flexible working hours in ranging from 58% to 62 %.

Explanation:

Percentage of sample respondents agreed for flexible working hours = 60%

The margin of the error = ± 2%

The expected population that wants flexible working hours in ranging from:

$\text{percentage of agreed}-2\%$ lower range

$\text{percentage of agreed}+\2%$ upper range

$60\%-2\% to 60\%+2\%$ that is 58% to 62 %.

The expected population that wants flexible working hours in ranging from 58% to 62 %.

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A force of 250 N is applied to a hydraulic jack piston that is 0.02 m in diameter. If the piston that supports the load has a di

Vikentia [17]

Answer:

option B

Explanation:

given,

Force exerted by the hydraulic jack piston = F₁ = 250 N

diameter of piston, d₁ = 0.02 m

r₁ = 0.01 m

diameter of second piston, d₂ = 0.15 m

r₂ = 0.075 m

mass of the jack to lift = ?

now,

$\dfrac{F_1}{A_1} =\dfrac{F_2}{A_2}$

$\dfrac{250}{\pi r_1^2} =\dfrac{F_2}{\pi r_2^2}$

$\dfrac{250}{0.01^2} =\dfrac{F_2}{0.075^2}$

$F_2= \dfrac{250}{0.01^2}\times {0.075^2}$

F₂ = 14062.5 N

F = m g

$m = \dfrac{F}{g}$

$m = \dfrac{14062.5}{9.8}$

m = 1435 Kg

hence, the correct answer is option B

5 0

2 years ago

You travel in a circle, whose circumference is 8 kilometers, at an average speed of 8 kilometers/hour. If you stop at the same p

Schach [20]

Velocity = (displacement) / (time)

Displacement = straight-line distance between start-point and end-point

If you stop at the same point you started from, then
your displacement for the trip is zero, and your average
velocity is also zero.

5 0

2 years ago

Read 2 more answers

Biologists think that some spiders "tune" strands of their web to give enhanced response at frequencies corresponding to those a

garik1379 [7]

Answer:

T=2.94*10^-10 N/m.

Explanation:

Biologists think that some spiders "tune" strands of their web to give enhanced response at frequencies corresponding to those at which desirable prey might struggle. Orb spider web silk has a typical diameter of 20μm, and spider silk has a density of 1300 kg/m³.

To have a fundamental frequency at 150Hz , to what tension must a spider adjust a 14cm -long strand of silk?

l=length of the spider silk, 14cm

velocity of wave = √(T/μ)

where T = tension and

μ = mass per unit length)

λ/2=l

for fundamental frequency λ/2 =14cm

(λ= wavelength of standing wave; as there will be no node

except the endpoints of silk strand)

λ = 28 cm = 0.28 m

and since frequency * wavelength = speed of wave. we have,

150 * 0.28 = √(T/μ) ..................(#)

now μ = mass/length = [volume * density]/length = [(length*area) * density] / length = area * density

= [π * (10 * 10^(-6))²] * 1300 = 13π * 10^(-8).

now putting this in equation (#) we get

150 * 0.28 = √(T/[13π * 10^(-8)]).

thus T = [13π * 10^(-8)] * (42)² =

2.94*10^-10 N/m.

6 0

2 years ago

A ball with a mass of 0.5 kilograms is lifted to a height of 2.0 meters and dropped. It bounces back to a height of 1.8 meters.

Degger [83]

Hi, thank you for posting your question here at Brainly.

To compute for the change in potential energy, the equation would be:

delta PE = mg*delta h
delta PE = 0.5*9.81*(2-1.8)
delta Pe = 0.98 J

The potential energy is converted to kinetic energy.

3 0

2 years ago

Read 2 more answers

Air at 3 104 kg/s and 27 C enters a rectangular duct that is 1m long and 4mm 16 mm on a side. A uniform heat flux of 600 W/m2 is

ad-work [718]

Answer:

$T_{out}=27.0000077$ ºC

Explanation:

First, let's write the energy balance over the duct:

$H_{out}=H_{in}+Q$

It says that the energy that goes out from the duct (which is in enthalpy of the mass flow) must be equals to the energy that enters in the same way plus the heat that is added to the air. Decompose the enthalpies to the mass flow and specific enthalpies:

$m*h_{out}=m*h_{in}+Q\\m*(h_{out}-h_{in})=Q$

The enthalpy change can be calculated as Cp multiplied by the difference of temperature because it is supposed that the pressure drop is not significant.

$m*Cp(T_{out}-T_{in})=Q$

So, let's isolate $T_{out}$ :

$T_{out}-T_{in}=\frac{Q}{m*Cp}\\T_{out}=T_{in}+\frac{Q}{m*Cp}$

The Cp of the air at 27ºC is 1007 $\frac{J}{kgK}$ (Taken from Keenan, Chao, Keyes, “Gas Tables”, Wiley, 1985.); and the only two unknown are $T_{out}$ and Q.

Q can be found knowing that the heat flux is 600W/m2, which is a rate of heat to transfer area; so if we know the transfer area, we could know the heat added.

The heat transfer area is the inner surface area of the duct, which can be found as the perimeter of the cross section multiplied by the length of the duct:

Perimeter:

$P=2*H+2*A=2*0.004m+2*0.016m=0.04m$

Surface area:

$A=P*L=0.04m*1m=0.04m^2$

Then, the heat Q is:

$600\frac{W}{m^2} *0.04m^2=24W$

Finally, find the exit temperature:

$T_{out}=T_{in}+\frac{Q}{m*Cp}\\T_{out}=27+\frac{24W}{3104\frac{kg}{s} *1007\frac{J}{kgK} }\\T_{out}=27.0000077$

$T_{out}$ =27.0000077 ºC

The temperature change so little because:

The mass flow is so big compared to the heat flux.
The transfer area is so little, a bigger length would be required.

3 0

2 years ago