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marta [7]
3 years ago
14

How many orbitals are allowed in a subshell if the angular momentum quantum number for electrons in that subshell is 3?

Physics
1 answer:
svp [43]3 years ago
4 0

Answer:

7 orbitals are allowed in a sub shell if the angular momentum quantum number for electrons in that sub shell is 3.

Explanation:

We that different values of m for a given l provide the total number of ways in which a given s, p,d and f sub shells in presence of magnetic field can be arranged in space along x, y ,z- axis or total number of orbitals into which a given subshell can be divided.

    Range for given l lies between -l to +l .

The possible values of m are -3 , -2 , -1 , 0 , 1 ,2 , 3 .

    Total number of orbitals are 7.

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zhannawk [14.2K]

Hacks in real life are tricks to help solve a problem with a simple solution.

Say for example, add a pinch of salt to whip egg white faster.

5 0
2 years ago
Read 2 more answers
An airplane flies in a loop (a circular path in a vertical plane) of radius 160 m . The pilot's head always points toward the ce
notka56 [123]

Answer:

a) 39.6 m/s b) 4123 N

Explanation:

a) At the top of the loop, all of the forces point downwards (force of gravity and normal force).

Fnet=ma

ma=m(v^2/R) (centripetal acceleration)

mg=m(v^2/R)

m cancels out (this is why pilot feels weightless) so,

g=(v^2/R)

9.8 m/s^2 = v^2/160 m

v^2=1568 m^2/s^2

v=39.6 m/s

b) At the bottom of the loop, the normal force and the force of gravity point in opposite directions. The normal force is the weight felt.

Convert 300 km/hr to m/s

300 km/hr=83.3 m/s

Convert pilot's weight into mass:

760 N = 77.55 kg

Fnet=ma

n-mg=m(v^2/R)

n=(77.55 kg)(((83.3 m/s)^2)/160 m)+(77.55 kg)(9.8 m/s^2)

n=3363.2 N+760 N=4123 N

5 0
2 years ago
What is the average speed of an object with a total distance traveled of 45 meters in 50 seconds
Anettt [7]

Answer: 0.9m/s

Explanation:

Use equation for speed:

V=S/t

S-distance; S=45m

t-time; t=50 s

V=S/t

V=45m/50s

V=0.9 m/s

7 0
3 years ago
Can someone help me with 1-2
ira [324]
Can't really plot a graph here for question 1.

2a) The car speeds up from A to B. The car travels at a constant speed from B to C. The car slows down to a stop from C to D.

b) From the graph, at 10 seconds, the car is moving at 20 m/s.
4 0
3 years ago
Calculate the distance traveled by a projectile as a function of launch angle. Compare the distances for two projectiles launche
DaniilM [7]

Answer:

R = x_{max} = \frac{v^2\sin(2\theta)}{g}\\\frac{R_1}{R_2} = \frac{\sin(2\theta_1}{\sin(2\theta_2}

Explanation:

Using kinematics equations:

\Delta x = v_{0x}t\\\Delta y = -\frac{1}{2}gt^2+v_{0y}t

Use \Delta y = 0 due to condition of distance traveled.

Solving second equation for time, there are two solutions. t=0 and

t=\frac{2v_{0y}}{g}

Use the expression in the first equation to have

R = \frac{2v^2 \cos\theta\sin\theta}{g}

Using trigonometric identities, you have the answer of the distance.

By doing the ratio for two different angles, you have the second answer. Due to sine function properties, the distances can be the same to complementary angles. Example, for 20° and 70°, the distance is the same.

5 0
3 years ago
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