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mash [69]
3 years ago
11

A car of the future requires 15kW of power to travel along a level road at 65km/h. A physics student wishes to use the car to dr

ive from Bowser to Qualicum, a distance of about 25 km. The engine is driven by heat energy recovered from a special material which has a specific heat capacity of 9800 J/kg • degree Celsius, If the car has 350 kg of this material, the outside air temperature is 15 °c and the engine is 88% efficient, to what temperature must the material be heated?
Physics
1 answer:
Luda [366]3 years ago
7 0
Power = Energy / time, The time is the one it takes to cover 25 km at 65 km/h, so:  time = 25/65*60*60 = 1384.6 seconds.

Now the total energy needed for that is: Energy = Power * time = 15,000 * 1384.6 = 20769239.8 J.

The heat provides those J. The heat for a difference in temperature is:

Heat = mass * c_h * Difference_of_temperature,

Heat = 350 * 9,800 * (T-15)*0.88,

where the 0.88 is because some of the heat is lost, efficiency 88%. Now, you can equal both results and get T:

350*9,800*0.88*(T-15) = 20769239.8

T-15 = 6.88, and T = 21.88 C

The answer is T= 21.88 C

NB: this specific heat, 9,800, is more than 2,000 time sthat of water, it is really a substance that does not exist,. It seems He has the highest (5,300) but it's a gas!
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The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 2.0 rev/s in 10.0 s. At th
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Answer:

22 revolutions

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2 rev/s = 2*(2π rad/rev) = 12.57 rad/s

The angular acceleration when it starting

\alpha_a = \frac{\Delta \omega}{\Delta t} = \frac{12.57}{10} = 1.257 rad/s^2

The angular acceleration when it stopping:

\alpha_o = \frac{\Delta \omega}{\Delta t} = \frac{-12.57}{12} = -1.05 rad/s^2

The angular distance it covers when starting from rest:

\omega^2 - 0^2 = 2\alpha_a\theta_a

\theta_a = \frac{\omega^2}{2\alpha_a} = \frac{12.57^2}{2*1.257} = 62.8 rad

The angular distance it covers when coming to complete stop:

0 - \omega^2 = 2\alpha_o\theta_o

\theta_o = \frac{-\omega^2}{2\alpha_o} = \frac{-12.57^2}{2*(-1.05)} = 75.4 rad

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centripetal force "F" experienced by the penny of "m" at distance "r" from axis of rotation is given as

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in the above equation , mass of penny "m"  and angular speed "w" of the turntable is same at all places. hence the centripetal force directly depends on the radius .

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