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Schach [20]
3 years ago
6

According to quantum physics, measuring velocity of a tiny particle with an electromagnet

Physics
1 answer:
lidiya [134]3 years ago
4 0

Answer:

Option A.

Explanation:

In quantum physics <u>there is a law to relate the position and the momentum of the particle</u>, it says that if we know with precision where is a quantum particle, we can not know the momentum of this particle, in other words, the velocity of the particle. So, when we measure the velocity of the particle we find the correct value of the particle, but we can not determine with accuracy where is the particle. This law is known as the Heisenberg's uncertainty principle and, its expressed as follows:    

\Delta x \Delta p \geq \frac{h}{4 \pi}

<em>where Δx: is the position's uncertainty, Δp: is the momentum's uncertainty and h: is the Planck constant.</em>  

Therefore, the correct answer is A: measuring the velocity of a tiny particle with an electromagnet has no effect on the velocity of the particle. It only affects the determination of the particle's position.      

I hope it helps you!

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To tugboats pull a disabled supertanker. Each tug exerts a constant force of 1.50•10^6 N, one at an angle 19.0° west of north, a
irina1246 [14]

Answer:

Work done by a tug boat, W = 1.735 x 10⁸ J

Explanation:

Given,

The of each tugboat, F = 1.5 x 10⁶ N

The angle of each tugboat forms with the resultant force, θ = 19°

The displacement of the supertanker, s = 710 m

The individual tugboat will be responsible for the displacement, d = 710/2

                                                                                                               = 355 m

The displacement component in each tugboat direction = 355 · sin θ meter

Therefore, the work done by each tugboat is

                                           W = F x S    joules

Substituting the values in the above equation

                                            W = 1.5 x 10⁶  x  355 · sin θ

                                                = 1.735 x 10⁸ J

Hence, the work done by each tugboat is, W = 1.735 x 10⁸ J                        

6 0
3 years ago
Which group was guaranteed voting rights by the Fifteenth Amendment?
castortr0y [4]

Explanation:

The 15th Amendment to the U.S. Constitution granted African American men the right to vote by declaring that the "right of citizens of the United States to vote shall not be denied or abridged by the United States or by any state on account of race, color, or previous condition of servitude." Although ratified on February 3, 1870, the promise of the 15th Amendment would not be fully realized for almost a century. Through the use of poll taxes, literacy tests and other means, Southern states were able to effectively disenfranchise African Americans. It would take the passage of the Voting Rights Act of 1965 before the majority of African Americans in the South were registered to vote.

3 0
4 years ago
Read 2 more answers
A bicycle wheel with radius 0.3 m rotates from rest to 3 rev/s in 5 s. What is the magnitude and direction of the total accelera
AlekseyPX

Answer:

Explanation:

Given

Radius of bicycle wheel r=0.3\ m

Initial angular velocity \omega _0=0

It rotates 3 revolution in 5 s therefore

\omega =2\pi 3=\6\pi =18.85\ rad/s

using \omega =\omega _0+\alpha t

where \alpha =angular\ acceleration

\omega =Final\ angular\ velocity

t=time

\alpha =\frac{18.85}{5}=3.77 rad/s^2

Total acceleration of any point will be a vector sum of tangential acceleration and centripetal acceleration

\omega at t=1

\omega =0+3.77\times 1=3.77 rad/s

a_c=\omega ^2\cdot r

a_c=(3.77)^2\cdot 0.3=4.26 m/s^2

Tangential acceleration a_t=\alpha \times r

a_t=3.77\times 0.3=1.13 m/s^2

a_{net}=\sqrt{a_t^2+a_c^2}

a_{net}=\sqrt{(1.13)^2+(4.26)^2}

a_{net}=4.41 m/s^2

                       

7 0
4 years ago
To analyze the experiment used to determine the properties of an electron. In 1909, Robert Millikan performed an experiment invo
sweet-ann [11.9K]

Answer:

C has 5 electrons

Explanation:

Given:

The data acquired from the experiment performed by Millikan:

Q_a = 3.20 x10^{-19}  C

Q_b = 4.80 x10^{-19}  C

Q_c = 8.00 x 10^{-19}  C

Q_d = 9.60 x 10^{-19}  C

Find:

How many Electrons were present in drop C

Solution:

It is known that the charge of an electron e = 1.602 *10^-19 C / electron.

Hence the number of electrons n in drop C will be:

      n = Q_c / e

      n = 8.00 x 10^{-19}  / 1.602*10^-19

      n = 4.99 = 5 electrons  

Answer: The drop C contains 5 electrons.

5 0
3 years ago
8x^3/27y^8×9y^3/12x^2
konstantin123 [22]
2x/ 9y^5  this is your answer

4 0
4 years ago
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