PHYSICS COLLEGE

What is the correct formula for barium nitride? A. Ba3N2,
в. BaN
с. Ва2N3
D. Ba2N

Answers

Answer 1
Answer: The formula for barium nitride is Ba(NO3)2

Related Questions

A proton (charge e), traveling perpendicular to a magnetic field, experiences the same force as an alpha particle (charge 2e) which is also traveling perpendicular to the same field. The ratio of their speeds, vproton/valpha is:
What is required for a sound wave to be reflected?
A remote controlled car is moving in a vacant parking lot. The velocity of the car as a function of time is given by v= [5.00 m/s – (0.0180 m/s3)t^2 ]i+[2.00 m/s + (0.550 m/s2)t ]j .a) What are ax(t) and ay(t), the x- and y- components of cars acceleration as a function of time?b) What are the magnitude and direction of the velocity of the car at t= 8 sec?c) What is the magnitude and direction of cars acceleration at t=8 sec
A uniform-density 8 kg disk of radius 0.25 m is mounted on a nearly frictionless axle. Initially it is not spinning. A string is wrapped tightly around the disk, and you pull on the string with a constant force of 41 N through a distance of 0.9 m. Now what is the angular speed
What is the angular width of a person's thumb viewed at arm's length? Assume that the width of the thumb is 17.3 mm and that the distance between the eyes and the thumb is 71.9 cm. Use the small-angle approximation and then convert the answer to degrees.

Two parallel wires I and II that are near each other carry currents i and 3i both in the same direction. Compare the forces that the two wires exert on each other. A. The wires exert equal magnitude attractive forces on each other. B. Wire I exerts a stronger force on wire II than II exerts on I.C. Wire II exerts a stronger force on wire I than I exerts on II. D. The wires exert equal magnitude repulsive forces on each other. E. The wires exert no forces on each other.

Answers

Answer:

A. The wires exert equal magnitude attractive forces on each other.

Explanation:

Magnetic field due to current i on current 2i

B₁ = 10⁻⁷ x 2 i / r where r is distance between the two wires

Force on wire II due to wire I per unit length

= magnetic field x current in wire II

= B₁ x 2 i

= [ 10⁻⁷ x 2 i / r ]  x 2i

= 4  x 10⁻⁷ i² / r

Magnetic field due to current 2i on current i

B₂ = 10⁻⁷ x 4 i / r where r is distance between the two wires

Force on wire I due to wire II per unit length

= magnetic field x current in wire I

= B₂ x  i

= [ 10⁻⁷ x 4 i / r ]  x i

= 4  x 10⁻⁷ i² / r

So final forces on each wire are same .

This force will be attractive in nature . The direction of force can be known from fleming's right  hand rule .

A smart phone charger delivers charge to the phone, in the form of electrons, at a rate of -0.75. How many electrons are delivered to the phone during 27 min of charging?

Answers

Answer:

The no. of electrons is 7.59* 10^(21)

Solution:

According to the question:

The rate at which the charge is delivered is given by:

(dQ)/(dt) = - 0.75

Now,

\int_(0)^(Q)dQ = - 0.75\int_(0)^(27 min) dt

Q = -0.75t|_(0)^(27 min)

Q= -0.75* 27* 60 = - 1215 C

No. of electrons, n can be calculated from the following relation:

Q = ne

where

e = electronic charge =1.6* 10^(- 19) C

Thus

n = (Q)/(e)

n= (1215)/(1.6* 10^(- 19))

n = 7.59* 10^(21)

Ultraviolet light is typically divided into three categories. UV-A, with wavelengths between 400 nm and 320 nm, has been linked with malignant melanomas. UV-B radiation, which is the primary cause of sunburn and other skin cancers, has wavelengths between 320 nm and 280 nm. Finally, the region known as UV-C extends to wavelengths of 100 nm. (a) Find the range of frequencies for UV-B radiation. (b) In which of these three categories does radiation with a frequency of 7.9 * 1014 Hz belong

Answers

Answer:

a) The UV-B has frequencies between 9.375x10^(14)Hz and 1.071x10^(15)Hz

b) The radiation with a frequency of 7.9x10^(14)Hz belong to the UV-A category.

Explanation:

(a) Find the range of frequencies for UV-B radiation.

Ultraviolet light belongs to the electromagnetic spectrum, which distributes radiation along it in order of different frequencies or wavelengths.

Higher frequencies:  

  • Gamma ray
  • X ray
  • Ultraviolet rays
  • Visible region

Lower frequencies:

  • Infrared
  • Microwave
  • Radio waves

That radiation is formed by electromagnetic waves, which are transverse waves formed by an electric field and a magnetic field perpendicular to it. Any of those radiations will have a speed of 3x10^{8]m/s in vacuum.

The velocity of a wave can be determined by means of the following equation:

c = \nu \cdot \lambda(1)  

Where c is the speed of light, \nu is the frequency and \lambda is the wavelength.

Then, from equation 1 the frequency can be isolated.

\nu = (c)/(\lambda)(2)

Before using equation 2 to determine the range of UV-B it is necessary to express \lambda in units of meters in order to match with the units from c.

   

\lambda = 320nm . (1m)/(1x10^(9)nm) ⇒ 3.2x10^(-7)m

\lambda = 280nm . (1m)/(1x10^(9)nm) ⇒ 2.8x10^(-7)m

\nu = (3x10^(8)m/s)/(3.2x10^(-7)m)

\nu = 9.375x10^(14)s^(-1)

\nu = 9.375x10^(14)Hz

\nu = (3x10^(8)m/s)/(2.8x10^(-7)m)

\nu = 1.071x10^(15)Hz

Hence, the UV-B has frequencies between 9.375x10^(14)Hz and 1.071x10^(15)Hz

(b) In which of these three categories does radiation with a frequency of 7.9x10^(14)Hz belong.

The same approach followed in part A will be used to answer part B.

Case for UV-A:

\lambda = 400nm . (1m)/(1x10^(9)nm) ⇒ 4x10^(-7)m

\nu = (3x10^(8)m/s)/(4x10^(-7)m)

\nu = 7.5x10^(14)s^(-1)

\nu = 7.5x10^(14)Hz

Hence, the UV-A has frequencies between 7.5x10^(14)Hz and 9.375x10^(14)Hz.

Therefore, the radiation with a frequency of 7.9x10^(14)Hz belongs to UV-A category.

Why aren’t there more craters on earth than the moon.

Answers

Answer:

Earths atmosphere causes most small to medium meteors to burn up disintergrate before hitting the earth the moon does not have a protective atmosphere

Jolene travels north 5 miles and then goes west 3 miles before coming straight back south 2 miles. What is her distance

Answers

Answer:

mnbhngbfcvdxc

Explanation:

"For the lowest harmonic of pipe with two open ends, how much of a wavelength fits into the pipe’s length?"

Answers

Answer:

0.5 lambda(wavelength)

Explanation:

We know that

The first harmonic for both side open ended pipe is

L= 1/2lambda

So L = 0.5*wavelength
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