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2 years ago

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a car's engine is only 12 % efficient at converting chemical energy in gasoline to mechanical energy. If it takes 18000 N of for

ce to keep the car moving at a constant speed of 21 m/s, how much chemical energy would be needed to move the car a distance of 450m at this speed?

1 answer:

Georgia [21]2 years ago

6 0

I wish i could help but idk the answer

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algol13

the answer should be C.

4 0

3 years ago

Read 2 more answers

What is the fundamental cause of air circulation in earths atmosphere

Digiron [165]

<span>A differences in the warmth, or moisture level as well as neighbouring areas of pressure in </span>air cause air<span> to circulate. in the earth's atmosphere.</span>

4 0

3 years ago

Suppose you are myopic (nearsighted). You can clearly focus on objects that are as far away as 52.5 cm away. You can clearly foc

Lilit [14]

Answer:

Explanation:

Image of distant object will be made at far point or at 52.5 so

object distance u = infinity

image distance v = - 52.5 cm

focal length required = f

Lens formula

1 / v - 1 / u = 1 / f

1 / - 52.5 - 0 = 1 / f

f = -52.5 cm

= -.525 m

Power P = 1 / f = - 1 / .525

= - 1.90

now , for eye with glass we shall find new near point .

v = ?

u = - 17.2 cm

f = - 52.5 cm

1 / v - 1 / u = 1 / f

1 / v + 1 / 17.2 = - 1 / 52.5

1 / v = - 1 / 17.2 - 1 / 52.5

= - .05813 - .019

= - .07713

u = - 12.96 cm

so new near point will be 12.96 cm

5 0

3 years ago

What would happen if you took a really strong metal box and you started teleporting in hundreds and hundreds of small cotton bal

mojhsa [17]

It will not explode since the mass of the cotton balls is so low but rather will most likely break the lock and hinges and come out.

4 0

3 years ago

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Earth orbits once around the sun every 365.25 days at an average radius of 1.5X10^11 m. Earth's mass is 6X10^24 kg. What distanc

Katyanochek1 [597]

(a) The distance the Earth travels every year is the perimeter of the orbit; since we have the average radius of the orbit, we can calculate the perimeter:
$d=2\pi r = 2 \pi (1.5 \cdot 10^{11}m)=9.4 \cdot 10^{11}m$

(b) The average centripetal acceleration is given by
$a_c = \frac{v^2}{r}$
where v is the average speed of the Earth, that we can find by dividing the distance covered by the Earth in one year (=the perimeter of the orbit) by the the time corresponding to 1 year:
t=1 year=365.25 days=3.2 \cdot 10^7 s
So the velocity is:
$v= \frac{d}{t}= \frac{9.4 \cdot 10^{11}m}{3.2 \cdot 10^7 s} =2.94 \cdot 10^4 m/s$
And so the centripetal acceleration is:
$a_c = \frac{v^2}{r}= \frac{(2.94 \cdot 10^4 m/s)^2}{9.4 \cdot 10^{11}m} =9.2 \cdot 10^{-4} m/s^2$

6 0

3 years ago