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jolli1 [7]
3 years ago
5

The speed of the train is 72 kilometers per hour find its speed in meters per second

Physics
1 answer:
pshichka [43]3 years ago
5 0

<em>20 meters per second</em>

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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
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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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What is the best thermal insulator? aluminum copper steel or glass
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Answer:

Copper is a better heat conductor than glass. In general, we think of metals as good conductors. In fact, metals vary widely in their conductivity but, overall, they are better conductors of heat than most liquids and gasses. Other solids also vary in their capability to conduct heat.

Explanation:

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Describe how the properties of sound waves change as they spread out in a spherical pattern.
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They go up and down
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What is the information gathered in an investigation is called
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Clues or evidence
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Find electric field at point p which is a distance l away from the both +q and -q
denis-greek [22]

Answer:

\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

Explanation:

As given point p is equidistant from both the charges

It must be in the middle of both the charges

Assuming all 3 points lie on the same line

Electric Field due a charge q at a point ,distance r away

=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{r^{2} }

Where

  • q is the charge
  • r is the distance
  • E is the permittivity of medium

Let electric field due to charge q be F1 and -q be F2

I is the distance of P from q and also from charge -q

⇒

F1=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }

F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

⇒

F1+F2=\frac{1}{4\times(pie)\times\text{E}} \times\frac{q}{I^{2} }+\frac{1}{4\times(pie)\times\text{E}} \times\frac{-q}{I^{2} }

8 0
4 years ago
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