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cestrela7 [59]
3 years ago
11

What are the difference between conductor and insulator? ​

Physics
1 answer:
Anvisha [2.4K]3 years ago
5 0

Answer:

  • A conductor allows energy e.g. electric charge or heat to pass through it easily. While an insulator does not let electric current or heat to travel through it.
  • Insulators have strong molecular bonds. While molecular bonds are very weak in conductors.
  • Insulators have very low conductivity. While in conductors, it is very high.
  • Insulators have a very high resistance and therefore the electrons are held together very firmly. The conductors, on the other hand, have a very low resistance.
  • Insulators do not have any electric field, neither inside nor on the surface. While in conductors, it is found on the surface and continues to be zero in the inner part of the conductor.
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A gaseous sample of this element is bombarded by photons of various energies (in these same units). Match each photon to the res
solniwko [45]
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A charged particle moves in a circular path in a uniform magnetic field.Which of the following would increase the period of the
Bond [772]

Answer:

Increasing its charge

Increasing the field strength

Explanation:

For a charged particle moving in a circular path in a uniform magnetic field, the centripetal force is provided by the magnetic force, so we can write:

qvB = m\frac{v^2}{r}

where

q is the charge

v is the velocity

B is the magnetic field

m is the mass

r is the radius of the orbit

The period of the motion is

T=\frac{2\pi r}{v}

Re-arranging for r

r=\frac{Tv}{2\pi}

And substituting into the previous equation

qvB = m \frac{Tv^3}{2\pi}

Solving for T,

T=\frac{2\pi q B}{m v^2}

So we see that the period is:

- proportional to the charge and the magnetic field

- inversely proportional to the mass and the square of the speed

So the following will increase the period of the particle's motion:

Increasing its charge

Increasing the field strength

4 0
3 years ago
Use the Bohr model to address this question. When a hydrogen atom makes a transition from the 5 th energy level to the 2nd, coun
iris [78.8K]

Answer:

A. 2.82 eV

B. 439nm

C. 59.5 angstroms

Explanation:

A. To calculate the energy of the photon emitted you use the following formula:

E_{n1,n2}=-13.4(\frac{1}{n_2^2}-\frac{1}{n_1^2})     (1)

n1: final state = 5

n2: initial state = 2

Where the energy is electron volts. You replace the values of n1 and n2 in the equation (1):

E_{5,2}=-13.6(\frac{1}{5^2}-\frac{1}{2^2})=2.82eV

B. The energy of the emitted photon is given by the following formula:

E=h\frac{c}{\lambda}   (2)

h: Planck's constant = 6.62*10^{-34} kgm^2/s

c: speed of light = 3*10^8 m/s

λ: wavelength of the photon

You first convert the energy from eV to J:

2.82eV*\frac{1J}{6.242*10^{18}eV}=4.517*10^{-19}J

Next, you use the equation (2) and solve for λ:

\lambda=\frac{hc}{E}=\frac{(6.62*10^{-34} kg m^2/s)(3*10^8m/s)}{4.517*10^{-19}J}=4.39*10^{-7}m=439*10^{-9}m=439nm

C. The radius of the orbit is given by:

r_n=n^2a_o   (3)

where ao is the Bohr's radius = 2.380 Angstroms

You use the equation (3) with n=5:

r_5=5^2(2.380)=59.5

hence, the radius of the atom in its 5-th state is 59.5 anstrongs

8 0
3 years ago
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