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svet-max [94.6K]
3 years ago
8

The coefficient of kinetic friction between a suitcase and the floor is 0.272. You may want to review (Pages 196 - 203) . Part A

If the suitcase has a mass of 80.0 kg , how far can it be pushed across the level floor with 660 J of work
Physics
1 answer:
vazorg [7]3 years ago
8 0

Answer:

Explanation:

The work required to push will be equal to work done by friction . Let  d be the displacement required .

force of friction = mg x μ where m is mass of the suitcase , μ be the coefficient of friction

work done by force of friction

mg x μ x d   = 660

80 x 9.8 x .272 x d = 660

d = 3 .1 m .

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3 years ago
How much work would have to be done to bring a 1150kg automobile traveling at 86km/h to a stop?​
krok68 [10]

Explanation:

We have,

Mass of an automobile is 1150 kg

The automobile traveling at 86 km/h and then it comes to stop.

86 km/h = 23.88 m/s

It is required to find work done by the automobile.

Concept used : Work energy theorem

Th change in kinetic energy of an object is equal to the work done by it. The work done is then given by :

W=\dfrac{1}{2}m(v^2-u^2)

Here, v = 0

W=-\dfrac{1}{2}mu^2\\\\W=-\dfrac{1}{2}\times 1150\times (23.88)^2\\\\W=-327896.28\ J

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3 years ago
We can reasonably model a 90-W incandescent lightbulb as a sphere 7.0cm in diameter. Typically, only about 5% of the energy goes
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Answer:

292.3254055 W/m²

469.26267 V/m

1.56421\times 10^{-6}\ T

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P = Power of bulb = 90 W

d = Diameter of bulb = 7 cm

r = Radius = \frac{d}{2}=\frac{7}{2}=3.5\ cm

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c = Speed of light = 3\times 10^8\ m/s

The intensity is given by

I=\frac{P}{A}\\\Rightarrow I=\frac{90}{4\pi 0.035^2}\\\Rightarrow I=5846.50811\ W/m^2

5% of this energy goes to the visible light so the intensity is

I=0.05\times 5846.50811\\\Rightarrow I=292.3254055\ W/m^2

The visible light intensity at the surface of the bulb is 292.3254055 W/m²

Energy density of the wave is

u=\frac{1}{2}\epsilon_0E^2

Energy density is also given by

\frac{I}{c}=\frac{1}{2}\epsilon_0E^2\\\Rightarrow E=\sqrt{\frac{2I}{c\epsilon_0}}\\\Rightarrow E=\sqrt{\frac{2\times 292.3254055}{3\times 10^8\times 8.85\times 10^{-12}}}\\\Rightarrow E=469.26267\ V/m

The amplitude of the electric field at this surface is 469.26267 V/m

Amplitude of a magnetic field is given by

B=\frac{E}{c}\\\Rightarrow B=\frac{469.26267}{3\times 10^8}\\\Rightarrow B=1.56421\times 10^{-6}\ T

The amplitude of the magnetic field at this surface is 1.56421\times 10^{-6}\ T

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