You have to solve this by using the equations of motion:
u=3
v=0
s=2.5
a=?
v^2=u^2+2as
0=9+5s
Giving a=-1.8m/s^2
Then using the equation:
F=ma
F is the frictional force as there is no other force acting and its negative as its in the opposite direction to the direction of motion.
-F=25(-1.8)
F=45N
Then use the formula:
F=uR
Where u is the coefficient of friction, R is the normal force and F is the frictional force.
45=u(25g)
45=u(25*10)
Therefore, the coefficient of friction is 0.18
Hope that helps
Just remember
Voltage = current times resistance
current = voltage over resistance
Current = 9/3 = 3
A parsec is a measurement of distance.
To solve this problem we need to apply the corresponding sound intensity measured from the logarithmic scale. Since in the range of intensities that the human ear can detect without pain there are large differences in the number of figures used on a linear scale, it is usual to use a logarithmic scale. The unit most used in the logarithmic scale is the decibel yes described as
![\beta_{dB} = 10log_{10} \frac{I}{I_0}](https://tex.z-dn.net/?f=%5Cbeta_%7BdB%7D%20%3D%2010log_%7B10%7D%20%5Cfrac%7BI%7D%7BI_0%7D)
Where,
I = Acoustic intensity in linear scale
= Hearing threshold
The value in decibels is 17dB, then
![17dB = 10log_{10} \frac{I}{I_0}](https://tex.z-dn.net/?f=17dB%20%3D%2010log_%7B10%7D%20%5Cfrac%7BI%7D%7BI_0%7D)
Using properties of logarithms we have,
![\frac{17}{10} = log_{10} \frac{I}{I_0}](https://tex.z-dn.net/?f=%5Cfrac%7B17%7D%7B10%7D%20%3D%20log_%7B10%7D%20%5Cfrac%7BI%7D%7BI_0%7D)
![log_{10} \frac{I}{I_0} = 1.7](https://tex.z-dn.net/?f=log_%7B10%7D%20%5Cfrac%7BI%7D%7BI_0%7D%20%3D%201.7)
![\frac{I}{I_0} = 10^{1.7}](https://tex.z-dn.net/?f=%5Cfrac%7BI%7D%7BI_0%7D%20%3D%2010%5E%7B1.7%7D)
![\frac{I}{I_0} = 50.12 W/m^2](https://tex.z-dn.net/?f=%5Cfrac%7BI%7D%7BI_0%7D%20%3D%2050.12%20W%2Fm%5E2)
Therefore the factor that the intensity of the sound was ![50.12W/m^2](https://tex.z-dn.net/?f=50.12W%2Fm%5E2)
Answer:m1v1 + m2v2 = (m1f + m2f)vf. 3000kg(10.0m/s) + (15000kg)(0.0m/s) = (18000kg)(vf).
Explanation: