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Nimfa-mama [501]
3 years ago
10

What speed would a fly with a mass of 0.72 g need in order to have the same kinetic energy as a 1250 kg automobile traveling at

a speed of 11 m/s?
Physics
1 answer:
Elena L [17]3 years ago
5 0
Kinetic of automobile

Mass m = 1,250 Kg;         V = 11 m/s

Formula: K.E = 1/2 mV²

               K.E = 1/2(1,250 Kg)(11 m/s)²

               K.E = 75,625 J

Speed required for insect to have the same kinetic energy as automobile

Mass of insect = 0.72 g convert to Kg   m = 7.2 x 10⁻⁴ Kg

K.E = 1/2 mV²  Derive V =?

 V = 2 K.E/m

 V = √2(75,625 J)/7.2 x 10⁻4 Kg

 V = √2.1 x 10⁸ m²/s²

 V = 14,491.34 m/s  (velocity of insect)









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Where is the near point of an eye for which a contact lens with a power of +2.55 diopters is prescribed?Where is the far point o
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\dfrac{1}{39.21}=\dfrac{1}{v}-\dfrac{1}{-25}

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The eye's near point is 68.98 cm from the eye.

(b). We need to calculate the focal length

Using formula of power

f =\dfrac{1}{P}

Put the value into the formula

f=-\dfrac{100}{3.00}

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We need to calculate the image distance

Using formula of lens

\dfrac{1}{f}=\dfrac{1}{v}-\dfrac{1}{u}

Put the value into the formula

\dfrac{1}{-33.33}=\dfrac{1}{v}-\dfrac{1}{\infty}

-\dfrac{1}{33.33}+\dfrac{1}{\infty}=\dfrac{1}{v}

\dfrac{1}{v}=-\dfrac{1}{33.33}

v=-33.33\ cm

The eye's far point is 33.33 cm from the eye.

Hence, (a). The eye's near point is 68.98 cm from the eye.

(b). The eye's far point is 33.33 cm from the eye.

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