All categories

3 years ago

Which of the following is a noncontact force? A. electrical forces B. gravitational forces C. magnetic forces D. all of these

Physics

2 answers:

Tamiku [17]3 years ago

7 0

Answer:

D. all of these

Explanation:

Electric forces, gravitational forces and magnetic forces are non contact forces. These forces are also the fundamental forces. Non contact forces are the forces which applied on a body without any real touch. Generally these non contact forces are in equilibrium.

goblinko [34]3 years ago

6 0

D. all of these

hope this helps.

You might be interested in

In a physics laboratory experiment, a coil with 200 turns enclosing an area of 11.8 cm^2 is rotated during the time interval 4.9

Zanzabum

Answer:

Explanation:

Flux through the coil = nBA , n is no of turns , B is magnetic flux and A a is area of the coli

= 200 x 5.6 x 10⁻⁵ x 11.8 x 10⁻⁴

= 13216 x 10⁻⁹ weber .

b ) When the coil becomes parallel to magnetic field , flux through it will become zero.

c ) e m f induced = change in flux / time

= 13216 x 10⁻⁹ / 4.9 x 10⁻²

= 2697.14 x 10⁻⁷ V

= 269.7 x10⁻⁶

269.7 μV.

6 0

3 years ago

A medieval instrument used to determine the position of the sun:

Stels [109]

Answer: astrolabe

Explanation:

4 0

2 years ago

What is the name of the chart that contain elements?

prisoha [69]

Answer:

Element Chart

Explanation:

It is a chart that provides many different elements.

7 0

3 years ago

An electron and a proton each have a thermal kinetic energy of 3kBT/2. Calculate the de Broglie wavelength of each particle at a

S_A_V [24]

Answer:

Given:

Thermal Kinetic Energy of an electron, $KE_{t} = \frac{3}{2}k_{b}T$

$k_{b} = 1.38\times 10^{- 23} J/k$ = Boltzmann's constant

Temperature, T = 1800 K

Solution:

Now, to calculate the de-Broglie wavelength of the electron, $\lambda_{e}$ :

$\lambda_{e} = \frac{h}{p_{e}}$

$\lambda_{e} = \frac{h}{m_{e}{v_{e}}$ (1)

where

h = Planck's constant = $6.626\times 10^{- 34}m^{2}kg/s$

$p_{e}$ = momentum of an electron

$v_{e}$ = velocity of an electron

$m_{e} = 9.1\times 10_{- 31} kg$ = mass of electon

Now,

Kinetic energy of an electron = thermal kinetic energy

$\frac{1}{2}m_{e}v_{e}^{2} = \frac{3}{2}k_{b}T$

$}v_{e} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{e}}}$

$}v_{e} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{9.1\times 10_{- 31}}}$

$v_{e} = 2.86\times 10^{5} m/s$ (2)

Using eqn (2) in (1):

$\lambda_{e} = \frac{6.626\times 10^{- 34}}{9.1\times 10_{- 31}\times 2.86\times 10^{5}} = 2.55 nm$

Now, to calculate the de-Broglie wavelength of proton, $\lambda_{e}$ :

$\lambda_{p} = \frac{h}{p_{p}}$

$\lambda_{p} = \frac{h}{m_{p}{v_{p}}$ (3)

where

$m_{p} = 1.6726\times 10_{- 27} kg$ = mass of proton

$v_{p}$ = velocity of an proton

Now,

Kinetic energy of a proton = thermal kinetic energy

$\frac{1}{2}m_{p}v_{p}^{2} = \frac{3}{2}k_{b}T$

$}v_{p} = \sqrt{2\frac{\frac{3}{2}k_{b}T}{m_{p}}}$

$}v_{p} = \sqrt{\frac{3\times 1.38\times 10^{- 23}\times 1800}{1.6726\times 10_{- 27}}}$

$v_{p} = 6.674\times 10^{3} m/s$ (4)

Using eqn (4) in (3):

$\lambda_{p} = \frac{6.626\times 10^{- 34}}{1.6726\times 10_{- 27}\times 6.674\times 10^{3}} = 5.94\times 10^{- 11} m = 0.0594 nm$

7 0

3 years ago

How many atoms are in SO<br> 3

kkurt [141]

Answer:

There are 6.87 x 1023 atoms in 1.14 mol SO3, or sulfur trioxide :

3 0

3 years ago