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balandron [24]
3 years ago
7

A motorcyclist is traveling at 58.3 mph on a flat stretch of highway during a sudden rainstorm. The rain has reduced the coeffic

ient of static friction between the motorcycle tires and the road to 0.070 when the motorcyclist slams on the brakes.
(a) What is the minimum distance required to bring the motorcycle to a complete stop?
(b) What would be the stopping distance if it were not raining and the coefficient of static friction were 0.580?

Physics
1 answer:
Novosadov [1.4K]3 years ago
8 0

Explanation:

Below is an attachment containing the solution.

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Zero according to ohm's law R= V / i

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1. A 25.35-g piece of iron absorbs 1562.75 Joules of heat energy, and its temperature
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Answer:

Q = C M T        where C is the specific, M the mass, T the temperature change

Note 1 cal = 4.19 Joules

1562.75 J / (4.19 J/cal) = 378 cal

C = Q / (M * T) = 378 cal / (25.35 g * 155 deg C)

C = .096 cal / g deg C

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A bicyclist of mass 112 kg rides in a circle at a speed of 8.9 m/s. If the radius of the circle is 15.5 m, what is the centripet
kogti [31]
The centripetal force, Fc, is calculated through the equation, 
                                    Fc = mv²/r
where m is the mass,v is the velocity, and r is the radius. 
Substituting the known values,
                                     Fc = (112 kg)(8.9 m/s)² / (15.5 m)
                                         = 572.36 N
Therefore, the centripetal force of the bicyclist is approximately 572.36 N. 
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A rocket of mass 1000kg uses 5kg of fuel and oxygen to produce exhaust gases ejected at 500m/s calculate the increase its veloci
Vlad [161]

Answer:

Approximately \rm 2.5\; m \cdot s^{-1}.

Explanation:

Let the increase in the rocket's velocity be \Delta v. Let v_0 represent the initial velocity of the rocket. Note that for this question, the exact value of  v_0 doesn't really matter.

The momentum of an object is equal to its mass times its velocity.

  • Mass of the rocket with the 5 kg of fuel: 1000.
  • Initial velocity of the rocket and the fuel: v_0.
  • Hence the initial momentum of the rocket: 1000\,v_0.
  • Mass of the rocket without that 5 kg of fuel: 1000 - 5 = 995.
  • Final velocity of the rocket: v_0 + \Delta v.
  • Hence the final momentum of the rocket: 995\,(v_0 + \Delta v).
  • Mass of the 5 kg of fuel: 5.
  • Final velocity of the fuel: v_0 - 500 (assuming that the the 500 m/s in the question takes the rocket as its reference.)
  • Hence the final momentum of the fuel: 5\,(v_0 - 500).

Momentum is conserved in an isolated system like the rocket and its fuel. That is:

Sum of initial momentum = Sum of final momentum.

1000\,v_0 = 995\,(v_0 + \Delta v) + 5\,(v_0 - 500).

Note that 1000\, v_0 appears on both sides of the equation. These two terms could hence be eliminated.

0 = 995\, \Delta v - 5\times 500.

\displaystyle \Delta v = \frac{5}{995}\times 500 \approx \rm 2.5\; m \cdot s^{-1}.

Hence, the velocity of the rocket increased by around 2.5 m/s.

5 0
3 years ago
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