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Rudiy27
2 years ago
6

When a bicycle tire gets mud on it, the tire becomes slippery. What does the mud do to the tire? A. Increases the force of gravi

ty B. Increases the force of friction C. It decreases the force of friction D. It decreases the force of gravity
Physics
2 answers:
larisa86 [58]2 years ago
8 0
Question: <span>When a bicycle tire gets mud on it, the tire becomes slippery. What does the mud do to the tire?

C. It decreases the force of friction.

First mud is not a solid. It is more of a slow liquid like corn syrup. And when you are caught in mud or trying to ride past the mud you loose friction with the ground.
You could also resemble friction with traction. In a car. If there is water & oil mixed on the road and you go past it at a high mph range. You will loose traction and start skipping against the mixed substance. This reduces your traction (friction) with the ground.

Final Answer:

C. It decreases the force of friction
</span>
Flura [38]2 years ago
7 0
The answer to this is C.
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A car with a mass of 1600 kg is towing a trailer with a mass of 420 kg. The car
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3 years ago
Suppose the rocket in the Example was initially on a circular orbit around Earth with a period of 1.6 days. Hint (a) What is its
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Answer:

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The orbital speed is v= 2.6*10^{3} m/s

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Explanation:

Generally angular velocity is mathematically represented as

            w = \frac{2 \pi}{T}

Where T is the period which is given as 1.6 days = 1.6 *24 *60*60 = 138240 sec

       Substituting the value

         w = \frac{2 \pi}{138240}

             = 4.54*10^ {-5} rad /sec

At the point when the rocket is on a circular orbit  

   The gravitational force =  centripetal force and this can be mathematically represented as

              \frac{GMm}{r^2} = mr w^2

Where  G is the universal gravitational constant with a value  G = 6.67*10^{-11}

            M is the mass of the earth with a constant value of M = 5.98*10^{24}kg

            r is the distance between earth and circular orbit where the rocke is found

               Making r the subject

                     r = \sqrt[3]{\frac{GM}{w^2} }

                        = \sqrt[3]{\frac{6.67*10^{-11} * 5.98*10^{24}}{(4.45*10^{-5})^2} }

                        = 5.78 *10^7 m

The orbital speed is represented mathematically as

                   v=wr

Substituting value

                  v= (5.78*10^7)(4.54*10^{-5})

                     v= 2.6*10^{3} m/s    

The escape velocity is mathematically represented as

                            v_e = \sqrt{\frac{2GM}{r} }

Substituting values

                             = \sqrt{\frac{2(6.67*10^{-11})(5.98*10^{24})}{5.78*10^7} }

                             v_e= 3.72 *10^3 m/s

7 0
3 years ago
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