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3 years ago

Determine the work that is being done by tension in pulling the box 178.0 cm along the table.

Physics

1 answer:

garri49 [273]3 years ago

3 0

1. A force of 25.0 Newtons is applied so as to move a 5.0 kg mass a distance of 20.0 meters. How much work was done?
Ans. W = F × d = 25 N × 20.0 m = 50 J
2. A force of 120 N is applied to the front of a sled at an angle of 28.0 above the horizontal so as to pull the sled a distance of 165 meters. How much work was done by the applied force?
Ans. W = F •d cos  = 120 N • 165 m cos 28.0 = 17,482.36 J
3. A sled, which has a mass of 45.0 kg., is sitting on a horizontal surface. A force of 120 N is applied to a rope attached to the front of the sled such that the angle between the front of the sled and the horizontal is 35.0o. As a result of the application of this force the sled is pulled a distance of 500 meters at a relatively constant speed. How much work was done to this sled by the applied force?
Ans. W = F• d cos  = 120 N • 500 m • cos 35.0 = 49,149.12 N
4. A rubber stopper, which has a mass of 38.0 grams, is being swung in a horizontal circle which has a radius of R = 1.35 meters. The rubber stopper is measured to complete 10 revolutions in 8.25 seconds.
a. What is the speed of the rubber stopper?
Ans. v =  v = •  v = .•(. ) v = 10.29 m/s .
b. How much force must be applied to the string in order to keep this stopper moving in this circular path at a constant speed?
Ans. The Centripetal Force, Fc = • = . • (. /)= 29.76 N .
c. How far will the stopper move during a period of 25.0 seconds?
Ans. d = vt = 10.29 m/s25 s = 257.25 m
d. How much work is done on the stopper by the force applied by the string during 25.0 seconds?
Ans. 0 J. The force is towards the center and the distance is around the perimeter of the circle. They are at right angles to each other. For work to be done, the force has to be in the same direction as the displacement.
5. How much work would be required to lift a 12.0 kg mass up onto a table 1.15 meters high?
Ans. W = F• d = mg d = 12 kg  9.8 m/s2  1.15 m = 135.24 J

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<span>Resistance of automobile light is equal to 12 ohms.
</span>Resistance = Voltage / Current
where
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Current = 1.0 Amperes
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Discuss how a photon (aka the light particle) can be affected by gravity despite being massless.

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Explanation:

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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

$1.56421\times 10^{-6}\ T$

Explanation:

P = Power of bulb = 90 W

d = Diameter of bulb = 7 cm

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

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

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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