State Schrodinger equation.

A mass spectrometer is being used to separate common oxygen-16 from the much rarer oxygen-18, taken from a sample of old glacial

Nataly_w [17]

Answer:

0.092 m

Explanation:

A charged moving particle immersed in a region with magnetic field follows a circular trajectory at constant speed (uniform circular motion), since the magnetic forces acts perpendicular to the direction of motion of the particle.

Since the magnetic force acts as centripetal force, we can write:

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

where

q is the charge of the particle

v is its velocity

B is the strength of the magnetic field

m is the mass of the particle

r is the radius of the orbit

Solving the equation for r,

$r=\frac{mv}{qB}$

For the ion of oxygen-16, we have:

$m_A=2.66\cdot 10^{-26}kg$

$q_A = 1.6\cdot 10^{-19}C$ (it is singly charged)

$v_A=2.90\cdot 10^6 m/s$

$B_A=1.30 T$

So the radius of its orbit is

$r_A=\frac{m_A v_A}{q_A B_A}=\frac{(2.66\cdot 10^{-26})(2.90\cdot 10^6)}{(1.6\cdot 10^{-19})(1.30)}=0.371 m$

For the ion of oxygen-18, we have:

$m_B = \frac{18}{16}m_A = 2.99\cdot 10^{-26}kg$

$q_B = 1.6\cdot 10^{-19}C$ (it is singly charged)

$v_B=2.90\cdot 10^6 m/s$

$B_B=1.30 T$

So the radius of its orbit is

$r_B=\frac{m_B v_B}{q_B B_B}=\frac{(2.99\cdot 10^{-26})(2.90\cdot 10^6)}{(1.6\cdot 10^{-19})(1.30)}=0.417 m$

After each ion has travelled a semicircle, the separation between the two ions will be twice the difference in their radius, so:

$d=2(r_B-r_A)=2(0.417-0.371)=0.092 m$

3 0

3 years ago

Given that coal used by electric power plants has a heating value of 27.5 million btus metric ton (25 million btus per ton), det

fredd [130]

Answer:

• 36.4 kg of coal.

• 80 pounds of coal.

Explanation:

Using proportionality constant,

Mass of coal = 1,000,000/27,500,000 btus/metric ton

= 0.0364 metric tons of coal

Mass of coal = 1,000,000/25,000,000 btus/ton

= 0.04 tons of coal.

Converting metric tons to kilogram,

1 metric ton = 1000kg,

0.0364 metric ton;

= 36.4 kg of coal.

Converting tons to pounds,

1 ton = 2000 pounds,

0.04 metric ton;

= 80 pounds of coal.

3 0

3 years ago

The aeroplane lands at a speed of 80 m/s

strojnjashka [21]

The mass of the aeroplane is 300,000 kg.

<h3>What is Newton's second law of motion?</h3>

It states that the force F is directly proportional to the acceleration a of the body and its mass.

The law is represented as

F =ma

where acceleration a = velocity change v / time interval t

Given is the aeroplane lands at a speed of 80 m/s. After landing, the aeroplane takes 28 s to decelerate to a speed of 10 m/s. The mean resultant force on the aeroplane as it decelerates is 750 000 N.

The force expression will be

F = mv/t

Substitute the values and we have

750000 = m x (80 -10)/ 28

750,000 = m x 2.5

m = 300,000 kg

Thus, the mass of the aeroplane is 300,000 kg.

Learn more about Newton's second law of motion.

brainly.com/question/13447525

#SPJ1

8 0

2 years ago

A small bag of sand is released from an ascending hot-air balloon whose upward constant velocity is v0 = 2.45 m/s. knowing that

lakkis [162]

When you draw an illustration for this problem, you would come up with the same drawing as shown in the picture. As the hot-air balloon travels upwards, there is a slight time when the bag of sand rises up until it reaches the maximum height. Then, it goes back down to the ground. The total time would be t₁ + t₂. The solution is as follows:

H = v₀²/2g = (2.45)²/2(9.81) = 0.306 m
t₁ = H/v₀ = 0.306 m/2.45 m/s = 0.125 s

t₂ = √2(H + 98.8)/g = √2(0.306+ 98.8)/9.81
t₂ = 4.495 s

Total time = 0.125 s + 4.495 s = 4.62 seconds

8 0

3 years ago

At some distance from a point charge, the electric potential is 635.0 V and the magnitude of the electric field is 189.0 N/C. Fi

Nataliya [291]

Answer:

The distance from the charge is 3.35 m.

Explanation:

Given that,

Electric potential, V = 635 V

Magnitude of electric field, E = 189 N/C

We need to find the distance from the charge. We know that the relation between electric field and electric potential is given by :

$E=\dfrac{V}{d}$

d is the distance from charge

$d=\dfrac{V}{E}\\\\d=\dfrac{635}{189}\\\\d=3.35\ m$

So, the distance from the charge is 3.35 m. Hence, this is the required solution.

8 0

3 years ago

State Schrodinger equation.​

State Schrodinger equation.