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3 years ago

Kellie jogs 6.0 km in 54 minutes, then 1.0 km in 16 minutes. What is her average speed in km/min please show your work

Physics

1 answer:

satela [25.4K]3 years ago

3 0

Answer:

Vprom = 0.00347[km/min]

Explanation:

We can calculate each of the average speeds and then perform the overall average between the two speeds.

V1 = 6/54

V1 = 0.111[km/min]

V2 = 1/16

V2 = 0.0625[km/min]

$V_{prom} = \frac{V_{1} + V_{2}}{2} \\V_{prom} = \frac{0.1111 + 0.0625}{2}\\V_{prom} = 0.00347 [km/min]$

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Holding onto a tow rope moving parallel to a frictionless ski slope, a 68.7 kg skier is pulled up the slope, which is at an angl

Furkat [3]

Answer:

a) $F = 78.606\,N$ , b) $F = 88.911\,N$

Explanation:

a) Let consider two equations of equilibrium, the first parallel to ski slope and the second perpendicular to that. The equations are, respectively:

$\Sigma F_{x'} = F - m\cdot g \cdot \sin \theta = 0\\\Sigma F_{y'} = N - m\cdot g \cdot \cos \theta = 0$

The force on the skier is:

$F = m \cdot g \cdot \sin \theta$

$F = (68.7\,kg)\cdot (9.807\,\frac{m}{s^{2}} )\cdot \sin 6.7^{\textdegree}$

$F = 78.606\,N$

b) The equations of equilibrium are the following:

$\Sigma F_{x'} = F - m\cdot g \cdot \sin \theta = m\cdot a\\\Sigma F_{y'} = N - m\cdot g \cdot \cos \theta = 0$

The force on the skier is:

$F = m\cdot (a + g \cdot \sin \theta)$

$F = (68.7\,kg)\cdot (0.150\,\frac{m}{s^{2}}+9.807\,\frac{m}{s^{2}}\cdot \sin 6.7^{\textdegree})$

$F = 88.911\,N$

3 0

3 years ago

One of the most efficient engines built so far has the following characteristics: combustion chamber temperature = 1900°C, exhau

suter [353]

Answer:

actual efficiency is 47.78 %

Carnot efficiency is 67.65 %

power output is 5.20 × 10^3 hp

Explanation:

given data

temperature = 1900°C = 1900+ 273 K = 2173 K

exhaust temperature = 430°C = 430 + 273 K = 703 K

fuel = 7.0 × 10^9 cal

work = 1.4 × 10^10 J

to find out

actual efficiency and Carnot efficiency and power output of engine

solution

first we find actual efficiency that is = work / heat input

put the value and

input energy = 7.0 × 10^9 cal (4.184 J/1 cal) = 29.29 × 10^9 J

actual efficiency = 1.4 × 10^10 / ( 29.29 × 10^9 )

actual efficiency = 0.4778

actual efficiency is 47.78 %

and

Carnot efficiency is = 1 - ( 703 / 2173 )

so Carnot efficiency is = 0.67648

Carnot efficiency is 67.65 %

and

power output = work / time

power output = 1.4 × 10^10 / 3600 sec

power output = 3.88 × 10^6 W

power output = 3.88 × 10^6 W / 746 hp

so power output is 5.20 × 10^3 hp

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3 years ago

Which describes how to generate a real image of an object using a concave mirror?The object must be closer to the mirror than th

iren2701 [21]

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2 years ago

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Lunna [17]

Answer:

Question 1 is C.

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When connecting a step down transformer what does the h1 symbol denote?

densk [106]

Answer: Primary coil

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