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Natasha_Volkova [10]
3 years ago
13

A driver is traveling along a straight road at the speed limit of 60 mph. After two minutes, she slows at a constant rate to a s

top at a stop light. Two minutes later, the light turns green and she accelerates at a constant rate back up to the speed limit. Three and a half minutes later, she again slows at a constant rate to a stop. After three minutes, she performs a U-turn, then accelerates at a constant rate back up to 60 mph. Two minutes later, she reaches her destination and slows at a constant rate to a stop. Assume that each period of slowing down and speeding up lasts 30 s and that the driver is initially moving in the +x- direction. Create a graph of the driver's velocity versus time that represents her trip.

Physics
1 answer:
Yuri [45]3 years ago
7 0

Explanation :

From the given information, the graph is plotted.

It is given that, a driver is traveling along a straight road at the speed limit of 60 mph. Initially, he was at point A.

After two minutes, she slows at a constant rate to a stop at a stop light. BC shows this part.

Two minutes later, the light turns green and she accelerates at a constant rate back up to the speed limit. CD shows this part.

Three and a half minutes later, she again slows at a constant rate to a stop. DE shows this part.

Hence, this is the required solution.

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A horizontal uniform meter stick supported at the 50-cm mark has a mass of 0.50 kg hanging from it at the 20-cm mark and a 0.30
ElenaW [278]

Answer:

70 cm

Explanation:

0.5 kg at 20 cm

0.3 kg at 60 cm

x = Distance of the third 0.6 kg mass

Meter stick hanging at 50 cm

Torque about the support point is given by (torque is conserved)

0.5(50-20)=0.3(60-50)+0.6x\\\Rightarrow x=\dfrac{0.5(50-20)-0.3(60-50)}{0.6}\\\Rightarrow x=20\ cm

The position of the third mass of 0.6 kg is at 20+50 = 70 cm

7 0
4 years ago
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A 31 kg block is initially at rest on a horizontal surface. A horizontal force of 83 N is required to set the block in motion. A
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Answer:

The static and kinetic coefficients of friction are 0.273 and 0.181, respectively.

Explanation:

By Newton's Laws of Motion and definition of maximum friction force, we derive the following two formulas for the static and kinetic coefficients of friction:

\mu_{s} = \frac{f_{s}}{m\cdot g} (1)

\mu_{k} = \frac{f_{k}}{m\cdot g} (2)

Where:

\mu_{s} - Static coefficient of friction, no unit.

\mu_{k} - Kinetic coefficient of friction, no unit.

f_{s} - Static friction force, in newtons.

f_{k} - Kinetic friction force, in newtons.

m - Mass, in kilograms.

g - Gravitational constant, in meters per square second.

If we know that f_{s} = 83\,N, f_{k} = 55\,N, m = 31\,kg and g = 9.807\,\frac{m}{s^{2}}, then the coefficients of friction are, respectively:

\mu_{s} = \frac{83\,N}{(31\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

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\mu_{k} = \frac{55\,N}{(31\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)}

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Answer:

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Explanation:

72 hours in 3 days

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round to 0.22 km/h

For future reference, Distance/Time= Speed

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