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algol13
3 years ago
15

a 1430 kg car speeds up from 7.50 m/s to 11.0 m/s in 9.30 s. Ignoring friction, how much power did that require?

Physics
1 answer:
iogann1982 [59]3 years ago
4 0
Kinetic energy = (1/2) (mass) (speed²)

Original KE = (1/2) (1430 kg) (7.5 m/s)²  =  40,218.75 joules

Final KE  =  (1/2) (1430 kg) (11.0 m/s)²  =   86,515 joules

Work done during the acceleration = (40218.75 - 86515) = 46,296.25 joules

Power = work/time = 46,296.25 joules / 9.3 sec  =  4,978.1 watts .
_______________________________

One interesting thing to notice while you're (I'm) solving this problem
is the fact that although the car's speed only increased by about 47%,
its kinetic energy more than doubled ... increased by about 115%. 
That's because kinetic energy is proportional to the SQUARE of the
speed.

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In which step of the scientific method would a scientist make use of graphs
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Which of the following statements is true about earth's magnetic poles
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Explanation:

According to classical physics, a magnetic field always has two associated magnetic poles (north and south), the same happens with magnets. This is because for <em>classical physics</em>, naturally, magnetic monopoles can not exist.  

In this context, Earth is similar to a magnetic bar with a north pole and a south pole. This means, the axis that crosses the Earth from pole to pole is like a big magnet.  

Now, by convention, on all magnets the north pole is where the magnetic lines of force leave the magnet and the south pole is where the magnetic lines of force enter the magnet.  Then, for the case of the Earth, the north pole of the magnet is located towards the geographic south pole and the south pole of the magnet is near the geographic north pole.  

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4 0
4 years ago
Refrigerant 134a enters a well-insulated nozzle at 200 lbf/in.2, 200°F, with a velocity of 120 ft/s and exits at 50 lbf/in.2 wit
riadik2000 [5.3K]

Answer:

x = 0.75801 = 75.801%

T_2 = 72..78 degree F

Explanation:

From superheated R 134 a properties table

At 200 lb/in^2 and 200 degree F

h_1 = 138.99 Btu/lbm

steady flow energy equation is givena s

h_1 + \frac{v_1^2}{2}  = h_2 + \frac{v_2^2}{2}

138.99 + \frac{120^2}{2\times 25037} = h_2 + \frac{1500^2}{2 \times 25037}

h_2 = 94.344 Btu/lbm

At 90 lb/in2 Tsat = 72.78 degree F

h_f = 35.715 Btu/lbm

hfg  = 77.345 Btu/lbm

h = hf + x hfg

94.344 = 35.715+ x \times 77.345

solving for x we get

x = 0.75801 = 75.801%

T_2 = 72..78 degree F

5 0
3 years ago
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