1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Setler79 [48]
3 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]3 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]3 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:

You might be interested in
A technological device that can be used to see sound waves is an
Bond [772]
The answer is oscilloscope
4 0
3 years ago
A + 8.42 nC point charge and a - 3.75 nC point charge are 2.73 cm apart. What is the electric field strength at the midpoint bet
aev [14]

Given:

The charge q1 = 8.42 nC

The charge q2 = -3.75 nC

The distance between the charges is d = 2.73 cm

To find the electric field strength at the midpoint between the two charges.

Explanation:

The distance between the charges and the midpoint is d' =d/2

The electric field strength can be calculated by the formula

E\text{ = }\frac{k|q|}{d^{\prime2}}

The electric field strength at the midpoint due to the charge q1 will be

\begin{gathered} E_1=\frac{9\times10^9\times8.42\times10^{-9}\text{ C}}{(\frac{2.73}{2}\times10^{-2})^2} \\ =\text{ 4.07}\times10^5\text{ V/m} \end{gathered}

The electric field strength at the midpoint due to the charge q2 will be

\begin{gathered} E_2=\frac{9\times10^9\times3.75\times10^{-9}}{(\frac{2.73}{2}\times10^{-2})^2} \\ =1.8\times10^5\text{ V/m} \end{gathered}

The electric field strength at the midpoint between the charges will be

\begin{gathered} E=E_1+E_2 \\ =4.07\times10^5+1.8\times10^5 \\ =\text{ 5.87}\times10^5\text{ V/m} \end{gathered}

The electric field strength at the midpoint between the two charges is 5.87 x 10^(5) m.

4 0
1 year ago
You are sitting in your car at rest at a traffic light with a bicyclist at rest next to you in the adjoining bicycle lane. As so
grigory [225]

Answer:

Explanation:

Time duration during which acceleration exists in  bicycle =

23 / 12 = 1.91 s

Time duration during which acceleration exists in car

= 47 / 8 = 5.875 s

Distance covered by bicycle during acceleration ( t = 1.91 s )

= 1/2 x 12 x (1.91)²

= 21.88 mi

Distance covered by car during this time ( t = 1.91 s )

= 1/2 x 8 x (1.91)²

7.64 mi ,

velocity of car after 1.91 s

= 8 x 1.91 = 15.28 mi/h

Let after time 1.91 , time taken by them to meet each other be t

Total distance covered by cycle = total distance covered by car

21.88 + 23 t = 7.64 + 15.28t + 4 t²

21.88 = 7.64 - 7.72t +4 t²

4 t² -7.72 t -14.24 = 0

t = 2.83 s

Total time taken

= 2.83 + 1.91

= 4.74 s

So after 4.74 s they will meet each other.

b ) Maximum distance occurs when velocity of both of them becomes equal .

Velocity after 1.91 s of bicycle

12 x 1.91 = 23 mi/h

Velocity after 1.91 s of car

8 x 1.91 = 15.28 mi/h . Let after time t , the velocity of car becomes 23

15.28 + 8t = 23

t = .965 s

So after time .965 s , car has velocity equal to that of bicycle.

The bicycle will travel a distance of

= 21.88 + .965 x 23 = 44.075 mi

car will travel a distance of

7.64 + 15.28 x .965 + .5 x 8 x .965²

= 7.64 + 14.75 + 3.72

= 26.11 mi

Distance between car and bicycle

= 44.075 - 26.11 = 17.965 mi

= 17.965 x 1760

= 31618.4 ft.

5 0
3 years ago
What is the radius of a circle thats the diameter of 480?
Luda [366]
The answer is 12.36. hoped this helped!
4 0
3 years ago
A fox runs for 12 seconds at a speed of 9.65 m/s. How much distance does it cover?
serg [7]

Answer:

115.8

Explanation:

6 0
3 years ago
Read 2 more answers
Other questions:
  • A car travels 400 km in 5 hours. what is the average speed of the car (in km/hr) ?
    15·1 answer
  • Which of the following radioactive emissions is the most penetrating?
    12·2 answers
  • A 139-turn circular coil of radius 2.13 cm is immersed in a uniform magnetic field that is perpendicular to the plane of the coi
    5·1 answer
  • In a lightning bolt, a large amount of charge flows during a time of 1.2 x 10-3 s. Assume that the bolt can be represented as a
    5·1 answer
  • Using Newton's First Law, predict what will happen when a car traveling on an icy road approaches a sharp turn
    14·1 answer
  • Find the final velocity of two balls colliding head on if the ball with velocity v2i = −21.4 cm/s has a mass equal to half that
    15·1 answer
  • The strong nuclear force felt by a single proton in a large nucleus
    6·1 answer
  • A girl whose mass is 40kg walk up a flight of 20steps each 15mm hight in 10seconds.find power developed by the girl showing the
    6·1 answer
  • Table 1
    9·1 answer
  • Your English teacher runs 0.80 km north and 1.20 km west. What is the magnitude of the resultant vector of this position?
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!