Increases exponentially is your correct answer
Answer:
The answer is the principal Quantum number (n)
Explanation:
The principal quantum number is one of the four quantum numbers associated with an atom.
It is denoted by a number n=1,2,3,4 etc
It tells both size (directly) and energy (indirectly) of an orbital.
When n=1 means it is the closest to the nucleus and is the smallest orbital and with increase in principal quantum number, it depicts that size of the orbital is increasing.
It tells the energy of the orbital as well as smaller number means less distance from nucleus and having less energy. Since electrons requires to absorb energy to jump into higher orbitals making n=2,3,4 etc. Thus electrons in the orbitals with higher n number indicates higher energy orbitals.
Answer:
So the specific heat of the liquid B is greater than that of A.
Explanation:
Liquid A is hotter than the liquid B after both the liquids are heated identically for the same duration of time from the same initial temperature then according to heat equation,
![Q=m.c.\Delta T](https://tex.z-dn.net/?f=Q%3Dm.c.%5CDelta%20T)
where:
m = mass of the body
c = specific heat of the body
change in temperature of the body
The identical heat source supplies the heat for the same amount of time then the quantity of heat supplied is also equal.
So for constant heat, constant mass the temperature change is inversely proportional to the specific of heat of the liquid.
![\Delta T=\frac{Q}{m} \times \frac{1}{c}](https://tex.z-dn.net/?f=%5CDelta%20T%3D%5Cfrac%7BQ%7D%7Bm%7D%20%5Ctimes%20%5Cfrac%7B1%7D%7Bc%7D)
![\Delta T\propto\frac{1}{c}](https://tex.z-dn.net/?f=%5CDelta%20T%5Cpropto%5Cfrac%7B1%7D%7Bc%7D)
So the specific heat of the liquid B is greater than that of A.
Answer:![0.084 m/s^2](https://tex.z-dn.net/?f=0.084%20m%2Fs%5E2)
Explanation:
Given
Total time=27 min 43.6 s=1663.6 s
total distance=10 km
Initial distance ![d_1=8.13 km](https://tex.z-dn.net/?f=d_1%3D8.13%20km)
time taken=25 min =1500 s
initial speed ![v_1=\frac{8.13\times 1000}{25\times 60}=5.6 m/s](https://tex.z-dn.net/?f=v_1%3D%5Cfrac%7B8.13%5Ctimes%201000%7D%7B25%5Ctimes%2060%7D%3D5.6%20m%2Fs)
after 8.13 km mark steve started to accelerate
speed after 60 s
![v_2=v_1+at](https://tex.z-dn.net/?f=v_2%3Dv_1%2Bat)
![v_2=5.6+a\times 60](https://tex.z-dn.net/?f=v_2%3D5.6%2Ba%5Ctimes%2060)
distance traveled in 60 sec
![d_2=v_1\times 60+\frac{a60^2}{2}](https://tex.z-dn.net/?f=d_2%3Dv_1%5Ctimes%2060%2B%5Cfrac%7Ba60%5E2%7D%7B2%7D)
![d_2=336+1800 a](https://tex.z-dn.net/?f=d_2%3D336%2B1800%20a)
time taken in last part of journey
![t_3=1663.6-1560=103.6 s](https://tex.z-dn.net/?f=t_3%3D1663.6-1560%3D103.6%20s)
distance traveled in this time
![d_3=v_2\times t_3](https://tex.z-dn.net/?f=d_3%3Dv_2%5Ctimes%20t_3)
![d_3=\left ( 5.6+a\times 60\right )103.6](https://tex.z-dn.net/?f=d_3%3D%5Cleft%20%28%205.6%2Ba%5Ctimes%2060%5Cright%20%29103.6)
and total distance![=d_1+d_2+d_3](https://tex.z-dn.net/?f=%3Dd_1%2Bd_2%2Bd_3)
![10000=8.13\times 1000+336+1800 a+\left ( 5.6+a\times 60\right )103.6](https://tex.z-dn.net/?f=10000%3D8.13%5Ctimes%201000%2B336%2B1800%20a%2B%5Cleft%20%28%205.6%2Ba%5Ctimes%2060%5Cright%20%29103.6)
![1870=336+1800 a+\left ( 5.6+a\times 60\right )103.6](https://tex.z-dn.net/?f=1870%3D336%2B1800%20a%2B%5Cleft%20%28%205.6%2Ba%5Ctimes%2060%5Cright%20%29103.6)
![a=0.084 m/s^2](https://tex.z-dn.net/?f=a%3D0.084%20m%2Fs%5E2)
Answer:
Hope i could help!
Explanation:
so all but one light could be burned out, and the last one will still function.