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denis23 [38]
3 years ago
14

TECHNOLOGY A Lamborghini Aventador accelerates from rest to 100 km / h in 2.9 s. The mass of the car is 1575 kg Calculate the po

wer supplied by the car's engine in HP without considering drag and friction (I HP 745.7 W)
Physics
1 answer:
DerKrebs [107]3 years ago
6 0

At 100 km/hr, the car's kinetic energy is

KE = (1/2) (mass) (speed)²

KE = (1/2) (1575 kg) ( [100 km/hr] x [1000 m/km] x [1 hr/3600 sec] )²

KE = (787.5 kg) (27.78 m/s)²

KE = 607,639 Joules

In order to deliver this energy in 2.9 seconds, the engine must supply

(607,639 J / 2.9 sec) = 209,531 watts

<em>Power = 281 HP</em>

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Charging a balloon and rubbing it on wool is an example of static electricity. :)
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true or false? destructive interference occurs when a trough meets up with another trough given location along the medium
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Answer:Poopy-di scoop

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Does the force of kinetic friction depend on the weight of the block? explain.
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7 0
2 years ago
What is the weight of a 8-kg substance in N, kN, kg·m/s2, kgf, lbm·ft/s2, and lbf?
kvv77 [185]

Answer:

W = 78.48N\\W =0.0784kN\\W = 8kgf\\

W= 3.6lbf\\W= 115.2 lbm*ft/s2

Explanation:

if

m=8kg=3.6lb\\

and g=9.81 m/s2=32.16 ft/s2

and

W=m*g

we can just replace de mass and gravity and we have

W = 8kg * 9.81 \frac{m}{s^{2} } =78.48N\\W = \frac{78.48N}{1000} =0.0784kN\\W = 8kgf\\

W= 3.6lbf\\W= 3.6lbm *32.16 ft/s2 =115.2 lbm*ft/s2

5 0
3 years ago
A particle with a charge of 3.00 elementary charges moves through a potential difference of 4.50 volts. What is the change in el
GuDViN [60]

Answer:

7.2\cdot 10^{-19} J

Explanation:

The change in electrical potential energy of a charged particle moving through a potential difference is given by

\Delta U = q \Delta V

where

q is the magnitude of the charge of the particle

\Delta V is the potential difference

In this problem:

- the charge of the particle is 3.00 elementary charges, so

q=3e=3\cdot 1.6\cdot 10^{-19} J=4.8\cdot 10^{-19}J

- the potential difference is

\Delta V=4.50 V

So, the change in electrical potential energy is

\Delta U=(1.6\cdot 10^{-19}C)(4.50 V)=7.2\cdot 10^{-19} J

7 0
3 years ago
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