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butalik [34]
3 years ago
6

Driving on asphalt roads entails very little rolling resistance, so most of the energy of the engine goes to overcoming air resi

stance. But driving slowly in dry sand is another story. If a 1200 kg car is driven in sand at 5.2 m/s, the coefficient of rolling friction is 0.06. In this case, nearly all of the energy that the car uses to move goes to overcoming rolling friction, so you can ignore air drag in this problem.
a. What propulsion force is needed to keep the car moving forward at a constant speed?

b. What power is required for propulsion at 5.2 m/s?
If the car gets 15 mpg when driving on sand, what is the car's efficiency? Assume the density of gasoline is 719.7 kg/m^3.
Help please!
Physics
1 answer:
Bad White [126]3 years ago
8 0
A) Agreed. 
<span>b) Value agreed but units should be W (watts). </span>

<span>c) Here's one method... </span>

<span>15 miles = 24140 m </span>

<span>1 gallon of gasoline contains 1.4×10⁸ J. </span>

<span>So moving a distance of 24140m requires gasoline containing 1.4×10⁸ J </span>

<span>Therefore moving a distance of 1m requires gasoline containing 1.4×10⁸/24140 = 5800 J </span>

<span>Overcoming rolling resitance for 1m requires (useful) work = force x distance = 1000x1 = 1000J </span>

<span>So 5800J (in the gasoline) provides 1000J (overcoming rolling resistance) of useful work for each metre moved. </span>

<span>Efficiency = useful work/total energy supplied </span>
<span>= 1000/5800 </span>
<span>= 0.17 (=17%) </span>
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Answer:

a) velocity v = 322.5m/s

b) time t = 19.27s

Explanation:

Note that;

ads = vdv

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a is acceleration

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a = 6 + 0.02s

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\int\limits^s_0 {a} \, ds  = \int\limits^v_0 {v} \, dv\\ \int\limits^s_0 {6+0.02s} \, ds  = \int\limits^v_0 {v} \, dv\\ 6s + \frac{0.02s^{2} }{2} = \frac{1}{2} v^{2} \\v = \sqrt{12s + 0.02s^{2} } .....................1 \\\\\\

Remember that

v = \frac{ds}{dt} \\\frac{ds}{v} = dt\\\int\limits^s_0 {\frac{ds}{\sqrt{12s+0.02s^{2} } } } \, ds = \int\limits^t_0 {} \, dt \\t=  (5\sqrt{2} ) ln  \frac{| [s + 300 + \sqrt{(s^{2}  + 600s)} ] |}{300} .......2

substituting s = 2km =2000m, into equation 1

v = 322.5m/s

substituting s = 2000m into equation 2

t = 19.27s

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3 years ago
If some raps 3words per second then how many words would he say in 9,999,999,990 seconds ​
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nikitadnepr [17]

Answer:

C. The change of internal energy of a system is the sum of work and heat spent on it.

Explanation:

The law of conservation of Energy states that energy cannot be destroyed but can only be converted or transformed from one form to another. Therefore, the sum of the initial kinetic energy and potential energy is equal to the sum of the final kinetic energy and potential energy.

Mathematically, it is given by the formula;

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Ki and Kf are the initial and final kinetic energy respectively.

Ui and Uf are the initial and final potential energy respectively.

The law of conservation of Energy is another way to describe the law of Thermodynamics. It states that the change of internal energy of a system is the sum of work and heat spent on it.

Mathematically, it is given by the formula;

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Answer

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