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Setler79 [48]
2 years ago
15

Give an example of a physical entity that is quantized. State specifically what the entity is and what the limits are on its val

ues.
Physics
2 answers:
konstantin123 [22]2 years ago
7 0

One example of a physical entity that is quantized is:

The amount of money in your pocket.

The amount can't have any fraction of 1 cent.  

Its value must be an integer-multiple of cents, or 0.01 dollar.    

When it increases or decreases, it jumps from one integer number of cents to the next integer number.  It doesn't "slide" from one to the next.  It can never have a value between two integer numbers of cents.

Bogdan [553]2 years ago
3 0

Answer:

A charge is a physical entity that has been quantized. The limits on its values are the value of a charged particle quantized in the state where 'n'...

Explanation:

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Answer:

\gamma  \: dynamo \: converts \: the \: mechanical \: energy \: into \: electrical \: energy \: .

Explanation:

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6 0
3 years ago
Read 2 more answers
You hold a ruler that has a charge on its tip 6 cm above a small piece of tissue paper to see if it can be picked up. The ruler
Fed [463]

Answer:

9.81 × 10⁻¹⁰ C

Explanation:

Given:

Distance between the tissue and the tip of the scale, r = 6 cm = 0.06 m

Charge on the ruler, Q = - 12 μC = - 12 × 10⁻⁶ C

Mass of the tissue = 3 g = 0.003 Kg

Now,

The force required to pick the tissue, F = mg

where, g is the acceleration due to gravity

also,

The force between (F) the charges is given as:

F=\frac{kQq}{r^2}

where,

q is the charge on the tissue

k is the Coulomb's constant = 9 × 10⁹ Nm²/C²

thus,

mg=\frac{kQq}{r^2}

on substituting the respective values, we get

0.003\times9.81=\frac{9\times10^9\times(-12\times10^{-6})\times q}{0.06^2}

or

q = 9.81 × 10⁻¹⁰ C

Minimum charge required to pick the tissue paper is 9.81 × 10⁻¹⁰ C

3 0
3 years ago
24. Every magnet, regardless of its shape, has twoO magnetic charges.O magnetic poles.magnetic fields.O magnetic domains.
Irina18 [472]
I would have to say 2 Magnetic Poles.
7 0
3 years ago
A student pushed a box 32.0 m across a smooth, horizontal floor using a constant force of 124 N. If the force was applied for 8.
Brilliant_brown [7]

The power developed is 500 W ( to the nearest Watt)

Power(P) is the rate at which work is done. Work done (W) is the product of the force applied on the object and the displacement (s) made by the point of application of the force.

P = \frac{W}{t}

W= F*s

Therefore,

P=\frac{F*s}{t}

Substitute the given values of force , displacement and time

F = 124 N,s = 32.0 m,t = 8.0 s

P =\frac{W*s}{t} =\frac{124N*22.0s}{8.0s} =496 W

Thus the Power can be rounded off to the nearest value of 500 W

3 0
3 years ago
Read 2 more answers
If you know the amount of air pressure exerted on a tabletop, how can you calculate the force exerted on the tabletop? Multiply
babymother [125]

Answer:

Multiply the air pressure by the area of the tabletop.

Explanation:

The relationship between pressure, force and area is given by:

p=\frac{F}{A}

where in this case, p is the air pressure, F is the force exerted and A the area of the tabletop. By re-arranging the equation, we can solve for F, the force exerted:

p=\frac{F}{A}\\p\cdot A=\frac{F}{A}\cdot A\\pA=F

So, the correct answer is:

The force exerted on the tabletop can be found by multiplying the air pressure by the area of the tabletop.

7 0
3 years ago
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