The second law of thermodynamics
Answer:
25,000 grams = 25 kilograms
Explanation:
1 gram = .001 kilogram
Answer:
![v = 10.8 m/s^{2}](https://tex.z-dn.net/?f=v%20%3D%2010.8%20m%2Fs%5E%7B2%7D)
Explanation:
The centripetal force experienced by the steel toolbox is provided by the friction between the toolbox and the bed of the truck.
Therefore,
--------------------------------(1)
Here,
v - speed of truck
m - mass of toolbox
r - radius of curve around which the truck is turning
f - frictional force
μmg
In order to obtain Vmax, frictional force must also be max
Thus,
f = μmg--------------------------------------------------(2)
Substitute (2) in (1)
μmg
μgr
g - 9.8 m/![s^{2}](https://tex.z-dn.net/?f=s%5E%7B2%7D)
The standard value of coefficient of static friction between steel and steel is
μ ≈ 0.7
Therefore,
![v = 10.8 m/s^{2}](https://tex.z-dn.net/?f=v%20%3D%2010.8%20m%2Fs%5E%7B2%7D)
Answer:
The minimum acceleration will be
Explanation:
We have given that coefficient of static friction ![\mu _s=0.860](https://tex.z-dn.net/?f=%5Cmu%20_s%3D0.860)
Frictional force is given by ![F=\mu_s N=\mu _sma](https://tex.z-dn.net/?f=F%3D%5Cmu_s%20N%3D%5Cmu%20_sma)
Weight is given by ![W=mg](https://tex.z-dn.net/?f=W%3Dmg)
For train to allow the bat to remain in place ![mg=\mu _sma](https://tex.z-dn.net/?f=mg%3D%5Cmu%20_sma)
![a=\frac{g}{\mu _s}=\frac{9.81}{0.860}=11.40m/sec^2](https://tex.z-dn.net/?f=a%3D%5Cfrac%7Bg%7D%7B%5Cmu%20_s%7D%3D%5Cfrac%7B9.81%7D%7B0.860%7D%3D11.40m%2Fsec%5E2)
So the minimum acceleration will be ![11.40m/sec^2](https://tex.z-dn.net/?f=11.40m%2Fsec%5E2)