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azamat
3 years ago
10

At which SQR3 step should a student highlight important terms?

Physics
1 answer:
slega [8]3 years ago
7 0
I think its c revising
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An electron is moving east in a uniform electric field of 1.47 {\rm N/C} directed to the west. At point A, the velocity of the e
zimovet [89]

Answer:

a) v = 6.25\times 10^{5} m/s

b) v = 1.73\times 10^{4} m/s

Explanation:

Given data:

Electric field = 1.47 N/C

velocity of electron is 4.55\times 10^5 m/s

distance of point b from point A is 0.55 m

we know that acceleration of particle is given as

a) for electron

a =\frac{q E}{m}

a = \frac{1.6\times 10^{-19} \times 1.47}{9.1\times 10^{-31}}

a = 2.58\times 10^{11} m/s^2

from equation of motion we have

v^2 = u^2 + 2aS

      = 20.7025 \times 10^{10} + 2\times 2.58\times 10^{11} \times 0.355

v = 6.25\times 10^{5} m/s

b) for proton

a = \frac{1.6\times 10^{-19} \times -1.47}{1.6\times 10^{-27}}

a = -1.41\times 10^{8} m/s^2

from equation of motion we have

v^2 = u^2 + 2aS

       = 3.8025 \times 10^{8} - 2\times 1.41\times 10^{8} \times 0.355

v = 1.73\times 10^{4} m/s

3 0
3 years ago
A uniform magnetic field is perpendicular to the plane of a circular loop of diameter 13 cm formed from wire of diameter 2.6 mm
I am Lyosha [343]

Answer:

Rate of change of magnetic field is 3.466\times 10^3T/sec        

Explanation:

We have given diameter of the circular loop is 13 cm = 0.13 m

So radius of the circular loop r=\frac{0.13}{2}=0.065m

Length of the circular loop L=2\pi r=2\times 3.14\times 0.065=0.4082m

Wire is made up of diameter of 2.6 mm

So radius r=\frac{2.6}{2}=1.3mm=0.0013m

Cross sectional area of wire A=\pi r^2=3.14\times0.0013^2=5.30\times 10^{-6}m^2

Resistivity of wire \rho =2.18\times 10^{-8}m

Resistance of wire R=\frac{\rho L}{A}=\frac{2.18\times 10^{-8}\times 0.4082}{5.30\times 10^{-6}}=1.67\times 10^{-3}ohm

Current is given i = 11 A

So emf  e=11\times 1.67\times 10^{-3}=0.0183volt

Emf induced in the coil is e=-\frac{d\Phi }{dt}=-A\frac{dB}{dt}

0.0183=5.30\times 10^{-6}\times \frac{dB}{dt}

\frac{dB}{dt}=3.466\times 10^3=T/sec

8 0
3 years ago
What are the three rules for writing the electron configuration of elements
astraxan [27]

<em></em>

Answer:

<u><em>The aufbau principle</em></u>

<u />

<u><em>The Pauli exclusion principle</em></u>

<u><em></em></u>

<u><em>Hund's rule of maximum multiplicity</em></u>

Explanation:

<u><em>The aufbau principle:</em></u>

<em></em>

The fundamental electronic configuration is achieved by placing the electrons one by one in the different orbitals available for the atom, which are arranged in increasing order of energy.

<u><em>The Pauli exclusion principle:</em></u>

<em></em>

Two electrons of the same atom cannot have their four equal quantum numbers.  Because each orbital is defined by the quantum numbers n, l, and m, there are only two possibilities ms = -1/2 and ms = +1/2, which physically reflects that each orbital can contain a maximum of two electrons, having opposite spins

<u><em>Hund's rule of maximum multiplicity:</em></u>

This rule says that when there are several electrons occupying degenerate orbitals, of equal energy, they will do so in different orbitals and with parallel spins, whenever this is possible.  Because electrons repel each other, the minimum energy configuration is one that has electrons as far away as possible from each other, and that is why they are distributed separately before two electrons occupy  the same orbital.

4 0
3 years ago
Please someone help!!!
antoniya [11.8K]
I know it’s the Coulomb’s law and that I’m pretty sure the answer would be C.Inverse Square.
4 0
3 years ago
A mother and her 35.0 -kg child are riding an escalator to the third level of a shopping mall. If the child's gravitational pote
notka56 [123]

The increase in potential energy of his mother if her mass is 56.0 kg will be 6031.97 J.

<h3>What is gravitational potential energy?</h3>

The energy that an item has due to its location in a gravitational field is known as gravitational potential energy.

The potential energy increases by 3773 J

PE₂-PE₁=mg(h₂-h₁)

3773 J = 35.0 × 9.81 × (h₂-h₁)

(h₂-h₁) = 10.98

Case 2 ;

ΔPE =?

ΔPE=mg(h₂-h₁)

ΔPE=56.0 × 9.81 ×10.98

ΔPE=6031.97 J.

Hence, the increase in potential energy of his mother if her mass is 56.0 kg will be 6031.97 J.

To learn more about the gravitational potential energy, refer;

brainly.com/question/3884855#SPJ1

#SPJ1

8 0
1 year ago
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