As per energy conservation we know that
Energy enter into the bulb = Light energy + Thermal energy
so now we have
energy enter into the bulb = 100 J
Light energy = 5 J
now from above equation we have
![100 = 5 + heat](https://tex.z-dn.net/?f=100%20%3D%205%20%2B%20heat)
![Heat = (100 - 5) J](https://tex.z-dn.net/?f=Heat%20%3D%20%28100%20-%205%29%20J)
![Heat = 95 J](https://tex.z-dn.net/?f=Heat%20%3D%2095%20J)
<span>The question is asking "which of the following suggestions can help you to start a job productively?" and there are options, so let's go through them:
A. Making your own rules will earn you respect. - this is not true, as your own rules might not be the same as rules of the company
B. Make a commitment to do the best job you can. - this is the best answer! It's impossible to do a job better than you can, but if you do it the best you can, typically you will be very, very good at it.
C. Dress the way you want, not the way coworkers do, so you'll stand out. - This is not true! dressing up does not have an influence on the job productivity, and it can make you be seen negatively at work
D. Asking others about productivity requirements is a sign of a poor employee. - asking others is usually good, it means you want to learn, so it should not be bad for you!</span>
Answer:
![0.6\Omega/s](https://tex.z-dn.net/?f=0.6%5COmega%2Fs)
Explanation:
We are given that
![R_1=150\Omega](https://tex.z-dn.net/?f=R_1%3D150%5COmega)
![R_2=75\Omega](https://tex.z-dn.net/?f=R_2%3D75%5COmega)
![\frac{dR_1}{dt}=1\Omega/s](https://tex.z-dn.net/?f=%5Cfrac%7BdR_1%7D%7Bdt%7D%3D1%5COmega%2Fs)
![\frac{dR_2}{dt}=1.5\Omega/s](https://tex.z-dn.net/?f=%5Cfrac%7BdR_2%7D%7Bdt%7D%3D1.5%5COmega%2Fs)
We have to find the rate at which R is changing.
In parallel
![\frac{1}{R}=\frac{1}{R_1}+\frac{1}{R_2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BR%7D%3D%5Cfrac%7B1%7D%7BR_1%7D%2B%5Cfrac%7B1%7D%7BR_2%7D)
Using the formula
![\frac{1}{R}=\frac{1}{50}+\frac{1}{75}=\frac{3+2}{150}=\frac{5}{150}=\frac{1}{30}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BR%7D%3D%5Cfrac%7B1%7D%7B50%7D%2B%5Cfrac%7B1%7D%7B75%7D%3D%5Cfrac%7B3%2B2%7D%7B150%7D%3D%5Cfrac%7B5%7D%7B150%7D%3D%5Cfrac%7B1%7D%7B30%7D)
![R=30\Omega](https://tex.z-dn.net/?f=R%3D30%5COmega)
![-\frac{1}{R^2}\frac{dR}{dt}=-\frac{1}{R^2_1}\frac{dR_1}{dt}-\frac{1}{R^2_2}\frac{dR_2}{dt}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7BR%5E2%7D%5Cfrac%7BdR%7D%7Bdt%7D%3D-%5Cfrac%7B1%7D%7BR%5E2_1%7D%5Cfrac%7BdR_1%7D%7Bdt%7D-%5Cfrac%7B1%7D%7BR%5E2_2%7D%5Cfrac%7BdR_2%7D%7Bdt%7D)
![\frac{1}{R^2}\frac{dR}{dt}=\frac{1}{R^2_1}\frac{dR_1}{dt}+\frac{1}{R^2_2}\frac{dR_2}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7BR%5E2%7D%5Cfrac%7BdR%7D%7Bdt%7D%3D%5Cfrac%7B1%7D%7BR%5E2_1%7D%5Cfrac%7BdR_1%7D%7Bdt%7D%2B%5Cfrac%7B1%7D%7BR%5E2_2%7D%5Cfrac%7BdR_2%7D%7Bdt%7D)
Substitute the values
![\frac{1}{900}\frac{dR}{dt}=1\times\frac{1}{2500}+1.5\times\frac{1}{(75)^2}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B900%7D%5Cfrac%7BdR%7D%7Bdt%7D%3D1%5Ctimes%5Cfrac%7B1%7D%7B2500%7D%2B1.5%5Ctimes%5Cfrac%7B1%7D%7B%2875%29%5E2%7D)
![\frac{dR}{dt}=900(\frac{1}{2500}+\frac{1.5}{5625}](https://tex.z-dn.net/?f=%5Cfrac%7BdR%7D%7Bdt%7D%3D900%28%5Cfrac%7B1%7D%7B2500%7D%2B%5Cfrac%7B1.5%7D%7B5625%7D)
![\frac{dR}{dt}=0.6\Omega/s](https://tex.z-dn.net/?f=%5Cfrac%7BdR%7D%7Bdt%7D%3D0.6%5COmega%2Fs)
Hi there!
a)
We can use the following equation to solve for the momentum. Remember that momentum is ALWAYS CONSERVED.
p = m · v
Plug in the given values:
p = (.250) · 15 = 3.75 kgm/s
b)
The momentum of the system AFTER the collision will be the same because of a CONSERVATION OF MOMENTUM.
![p_i = p_f](https://tex.z-dn.net/?f=p_i%20%3D%20p_f)
c)
We know that by the conservation of momentum:
m₁v₁ + m₂v₂ = m₁v₁ '+ m₂v₂'
The second object (cart) is originally at rest, and the two objects move together after the collision, so:
m₁v₁ = (m₁ + m₂)vf
Solve by plugging in values:
(.250)(15)/(.250 + .400) = vf = 5.769 m/s