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GrogVix [38]
3 years ago
7

Suppose someone caused a book to fall from the shelf to the ground. Compared to the total sum of kinetic and potential energy th

e book has while on the shelf, about how much would it have when it is halfway from the shelf to the ground? no energy, half as much,twice as much, or the same amount? Pretty easy question.​
Physics
1 answer:
gavmur [86]3 years ago
6 0

Answer:

book resting on a shelf has no potential energy relative to the shelf since it has a height of zero meters relative to the shelf. However, if the book is elevated to some height above the shelf, then it has potential energy proportional to the height at which it resides above the shelf.

An object can have both kinetic and potential energy at the same time. For example, an object which is falling, but has not yet reached the ground has kinetic energy because it is moving downwards, and potential energy because it is able to move downwards even further than it already has. The sum of an object's potential and kinetic energies is called the object's mechanical energy.

As an object falls its potential energy decreases, while its kinetic energy increases. The decrease in potential energy is exactly equal to the increase in kinetic energy.

Another important concept is work.

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Suppose a NASCAR race car rounds one end of the Martinsville Speedway. This end of the track is a turn with a radius of approxim
sergiy2304 [10]

Answer:

10.53m/s²

Explanation:

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r is the radius of the track = 57.0m

Substitute the given values into the formula:

a = \frac{24.5^2}{57} \\\\a = \frac{600.25}{57}\\ \\a = 10.53m/s^{2}

Hence the centripetal acceleration of the race car is 10.53m/s²

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Kinetic energy is: which one is the correct answer A,B,C,or D A.inversely proportional to mass and velocity B.directly proportio
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Answer: A

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Read 2 more answers
A hollow metal sphere has a 5 cm inner radius and a 12 cm outer radius. There is a particle at the center of the sphere. The sur
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Answer:

Explanation:

 The surface charge density on the inside surface of the sphere is -150nC/m2

Charge on the inner surface = - 4 π x (.05)² x 150 x 10⁻⁹

= - 4.71 x 10⁻⁹ C

This is the bound charge . This charge will be bound by charge at the centre

equal to the same charge but of opposite nature .

So charge at the centre

Q = + 4.71 x 10⁻⁹ C

B) Imagine a Gaussian surface with radius equal to 4 cm

If E be electric field on the surface

Flux through the surface

4 π x (.04)² x E = Q / ε₀

= 4.71 x10⁻⁹ / 8.85 x 10⁻¹²

= .532 x 10³

E = .532 x 10³ / 4 π x (.04)²

= 26473 N/C

C )Imagine a Gaussian surface with radius equal to 8 cm

If E be electric field on the surface

Flux through the surface

4 π x (.08)² x E = Q / ε₀

= 0  / 8.85 x 10⁻¹² ( Total charge inside will be -Q + Q = 0 )

= 0

D)

Total charge on the outer surface

= 4 π x (.12)² x 150 x 10⁻⁹

= 27.13 x 10⁻⁹ C

Imagine a Gaussian surface with radius equal to 20 cm

If E be electric field on the surface

Flux through the surface

4 π x (.12)² x E = Q / ε₀

= 27.13 x10⁻⁹ / 8.85 x 10⁻¹² ( - Q + Q will cancel each other )

= 3.065 x 10³

E = 3.065 x 10³ / 4 π x (.12)²

= 16946.43  N/C

3 0
3 years ago
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