P = Momentum
M = Mass
t = time
v = velocity (speed)
J= Impulse
1.)
F = 100N
t = 5s
Change in momentum(P) = Ft
I see that this is not one of your equations but your equations do not work with this problem.
P = Ft
P = (100)(5)
P = 500 Ns
2.)
See if you can get this one by yourself from what I did on the last problem.
3.)
P = 25 kgm/s
M = 2.0 kg
So, you are looking for velocity (speed). Look for the equation that has momentum, mass and velocity.
Use the equation P = mv
25 = 2(v)
25 = 2v
25/ 2 = 2v/2
12.5 m/s = v
4.) Do the same for number 4.
P = 12 kgm/s
M = 1kg
P = mv
12 = 1v
12/1 = 1v/1
12 m/s = v
6.)
P = 100 kgm/s
V= 7.0m/s
P= mv
100 = m7.0
100/7 = 7.0m/7
14.28 kg = m
Try page 2 by yourself and message me or comment if you need more help. Just identify the variable you have and what you need to find and put them in an equation that fits. You can do this!!!!
Answer:
The position is
![d_{o}=1.706m](https://tex.z-dn.net/?f=d_%7Bo%7D%3D1.706m)
Explanation:
The kinetic energy of the motion is
![E_{K}=\frac{1}{2}*m*v^2](https://tex.z-dn.net/?f=E_%7BK%7D%3D%5Cfrac%7B1%7D%7B2%7D%2Am%2Av%5E2)
![E_{K}=\frac{1}{2}*0.8kg*8(\frac{m}{s})^2](https://tex.z-dn.net/?f=E_%7BK%7D%3D%5Cfrac%7B1%7D%7B2%7D%2A0.8kg%2A8%28%5Cfrac%7Bm%7D%7Bs%7D%29%5E2)
![E_{K}=25.6 J](https://tex.z-dn.net/?f=E_%7BK%7D%3D25.6%20J)
So apply the conservation of energy the force of the spring is the same of the kinetic energy
![F_{s}=\frac{1}{2}*k*d^2](https://tex.z-dn.net/?f=F_%7Bs%7D%3D%5Cfrac%7B1%7D%7B2%7D%2Ak%2Ad%5E2)
![F_{s}=E_{K}](https://tex.z-dn.net/?f=F_%7Bs%7D%3DE_%7BK%7D)
![25.6=\frac{1}{2}*30\frac{N}{m}*d_{o}^2](https://tex.z-dn.net/?f=25.6%3D%5Cfrac%7B1%7D%7B2%7D%2A30%5Cfrac%7BN%7D%7Bm%7D%2Ad_%7Bo%7D%5E2)
Solve to do
![d_{o}=\sqrt{\frac{2*25.6J}{30\frac{N}{m}}}](https://tex.z-dn.net/?f=d_%7Bo%7D%3D%5Csqrt%7B%5Cfrac%7B2%2A25.6J%7D%7B30%5Cfrac%7BN%7D%7Bm%7D%7D%7D)
![d_{o}=1.706m](https://tex.z-dn.net/?f=d_%7Bo%7D%3D1.706m)
Yes but they not supposed to
Answer:
It means that person can do work or can posses energy at a rate of 550 watts.
Or : can posses energy of 550 joules in one second.
Explanation:
![{ \bf{power = \frac{work}{time} }}](https://tex.z-dn.net/?f=%7B%20%5Cbf%7Bpower%20%3D%20%20%5Cfrac%7Bwork%7D%7Btime%7D%20%7D%7D)
Answer:
![\alpha = 5.2 \times 10^{-5} per ^oC](https://tex.z-dn.net/?f=%5Calpha%20%3D%205.2%20%5Ctimes%2010%5E%7B-5%7D%20per%20%5EoC)
Explanation:
At room temperature the thickness of the film is just enough that it will cancel the reflected light
So here we will have
![2\mu t = \frac{\lambda}{2}](https://tex.z-dn.net/?f=2%5Cmu%20t%20%3D%20%5Cfrac%7B%5Clambda%7D%7B2%7D)
so we have
![t = \frac{\lambda}{4/mu}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B%5Clambda%7D%7B4%2Fmu%7D)
here we have
![\lambda = 582.3 nm](https://tex.z-dn.net/?f=%5Clambda%20%3D%20582.3%20nm)
![\mu = 1.750](https://tex.z-dn.net/?f=%5Cmu%20%3D%201.750)
now we have
![t = \frac{582.3 nm}{4(1.750)}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B582.3%20nm%7D%7B4%281.750%29%7D)
![t = 83.2 nm](https://tex.z-dn.net/?f=t%20%3D%2083.2%20nm)
now when temperature is increased to 177 degree C
then in this case we have
![\lambda_2 = 587.2 nm](https://tex.z-dn.net/?f=%5Clambda_2%20%3D%20587.2%20nm)
now we have
![t' = \frac{587.2 nm}{4(1.750)}](https://tex.z-dn.net/?f=t%27%20%3D%20%5Cfrac%7B587.2%20nm%7D%7B4%281.750%29%7D)
![t' = 83.89 nm](https://tex.z-dn.net/?f=t%27%20%3D%2083.89%20nm)
now we know by the formula of thermal expansion
![t' = t (1 + \alpha \Delta T)](https://tex.z-dn.net/?f=t%27%20%3D%20t%20%281%20%2B%20%5Calpha%20%5CDelta%20T%29)
![83.89 = 83.2(1 + \alpha(177 - 18.6))](https://tex.z-dn.net/?f=83.89%20%3D%2083.2%281%20%2B%20%5Calpha%28177%20-%2018.6%29%29)
![\alpha = 5.2 \times 10^{-5} per ^oC](https://tex.z-dn.net/?f=%5Calpha%20%3D%205.2%20%5Ctimes%2010%5E%7B-5%7D%20per%20%5EoC)