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.
Answer:
Average acceleration is ![(11.05)g\ m/s^2](https://tex.z-dn.net/?f=%2811.05%29g%5C%20m%2Fs%5E2)
Explanation:
It is given that,
Initial velocity, u = 0
Final velocity, v = 6.5 km/s = 6500 m/s
Time taken, t = 60 s
Acceleration, ![a=\dfrac{v-u}{t}](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7Bv-u%7D%7Bt%7D)
![a=108.33\ m/s^2](https://tex.z-dn.net/?f=a%3D108.33%5C%20m%2Fs%5E2)
Since, ![g=9.8\ m/s^2](https://tex.z-dn.net/?f=g%3D9.8%5C%20m%2Fs%5E2)
So, ![a=(11.05)g\ m/s^2](https://tex.z-dn.net/?f=a%3D%2811.05%29g%5C%20m%2Fs%5E2)
So, the angular acceleration of the missile is
. Hence, this is the required solution.
Answer: The gravitational
Explanation: The student is pushing the box so u have to have gravitational force so it could move
Answer:
The current is reduced to half of its original value.
Explanation:
- Assuming we can apply Ohm's Law to the circuit, as the internal resistance and the load resistor are in series, we can find the current I₁ as follows:
![I_{1} = \frac{V}{R_{int} +r_{L} }](https://tex.z-dn.net/?f=I_%7B1%7D%20%3D%20%5Cfrac%7BV%7D%7BR_%7Bint%7D%20%2Br_%7BL%7D%20%7D)
- where Rint = r and RL = r
- Replacing these values in I₁, we have:
![I_{1} = \frac{V}{R_{int} +r_{L} } = \frac{V}{2*r} (1)](https://tex.z-dn.net/?f=I_%7B1%7D%20%3D%20%5Cfrac%7BV%7D%7BR_%7Bint%7D%20%2Br_%7BL%7D%20%7D%20%3D%20%5Cfrac%7BV%7D%7B2%2Ar%7D%20%281%29)
- When the battery ages, if the internal resistance triples, the new current can be found using Ohm's Law again:
![I_{2} = \frac{V}{R_{int} +r_{L} } = \frac{V}{(3*r) +r} = \frac{V}{4*r} (2)](https://tex.z-dn.net/?f=I_%7B2%7D%20%3D%20%5Cfrac%7BV%7D%7BR_%7Bint%7D%20%2Br_%7BL%7D%20%7D%20%3D%20%5Cfrac%7BV%7D%7B%283%2Ar%29%20%2Br%7D%20%3D%20%5Cfrac%7BV%7D%7B4%2Ar%7D%20%20%282%29)
- We can find the relationship between I₂, and I₁, dividing both sides, as follows:
![\frac{I_{2} }{I_{1} } = \frac{V}{4*r} *\frac{2*r}{V} = \frac{1}{2}](https://tex.z-dn.net/?f=%5Cfrac%7BI_%7B2%7D%20%7D%7BI_%7B1%7D%20%7D%20%3D%20%5Cfrac%7BV%7D%7B4%2Ar%7D%20%2A%5Cfrac%7B2%2Ar%7D%7BV%7D%20%3D%20%5Cfrac%7B1%7D%7B2%7D)
- The current when the internal resistance triples, is half of the original value, when the internal resistance was r, equal to the resistance of the load.