The answer is C. Flammability because if you were to test flammability of an object, you would e making a chemical change to the object.
Zeros between a decimal point and a non zero number are insignificant.
Answer:
Final velocity, v = 25.3 m/s
Explanation:
Initial velocity of a locomotive, u = 19 m/s
Acceleration of the locomotive, a = 0.8 m/s²
Length of station, d = 175 m
We need to find its final velocity (v) when the nose leaves the station. It can be calculated using the third law of motion :
![v^2-u^2=2ad](https://tex.z-dn.net/?f=v%5E2-u%5E2%3D2ad)
![v^2=2ad+u^2](https://tex.z-dn.net/?f=v%5E2%3D2ad%2Bu%5E2)
![v^2=2\times 0.8\ m/s^2\times 175\ m+(19\ m/s)^2](https://tex.z-dn.net/?f=v%5E2%3D2%5Ctimes%200.8%5C%20m%2Fs%5E2%5Ctimes%20175%5C%20m%2B%2819%5C%20m%2Fs%29%5E2)
![v^2=(641)\ m^2](https://tex.z-dn.net/?f=v%5E2%3D%28641%29%5C%20m%5E2)
v = 25.31 m/s
v = 25.3 m/s
When the nose leaves the station, it will move with a velocity of 25.3 m/s. Hence, this is the required solution.
Answer:
The velocity is 19.39 m/s
Solution:
As per the question:
Mass, m = 75 kg
Radius, R = 19.2 m
Now,
When the mass is at the top position in the loop, then the necessary centrifugal force is to keep the mass on the path is provided by the gravitational force acting downwards.
![F_{C} = F_{G}](https://tex.z-dn.net/?f=F_%7BC%7D%20%3D%20F_%7BG%7D)
![\frac{mv^{2}}{R} = mg](https://tex.z-dn.net/?f=%5Cfrac%7Bmv%5E%7B2%7D%7D%7BR%7D%20%3D%20mg)
where
v = velocity
g = acceleration due to gravity
![v = \sqrt{2gR} = \sqrt{2\times 9.8\times 19.2} = 19.39\ m/s](https://tex.z-dn.net/?f=v%20%3D%20%5Csqrt%7B2gR%7D%20%3D%20%5Csqrt%7B2%5Ctimes%209.8%5Ctimes%2019.2%7D%20%3D%2019.39%5C%20m%2Fs)
it has a high density. since density is mass divided by volume, if you are dividing a large mass by a small volume it will have a high density