Answer:
μ = 0.385
Explanation:
Given that,
The mass of the student, m = 69 Kg
The horizontal force applied, F = 260 N
The normal force acting on the body, weight = mg = 69g N
= 676.2 N
The coefficient of friction acting on a body is equal to the force acting on the body to the normal force acting on the body due to gravitation.
The formula for coefficient of friction,
μ = F / N
Substituting the values in the above equation,
μ = 260 N / 676.2 N
= 0.385
Hence, the coefficient of friction, μ = 0.385
Answer : The correct option is, (B) 279.2 Kpa
Solution : Given,
Initial pressure of gas = 475 Kpa
Initial volume of gas = 
Final volume of gas = 
Initial temperature of gas = 290 K
Final temperature of gas = 277 K
Using ideal gas equation :
Formula used :
where,
= initial pressure of gas
= final pressure of gas
= initial volume of gas
= final volume of gas
= initial temperature of gas
= final temperature of gas
Now put all the given values in the above formula, we get the final pressure of the gas.
Therefore, the absolute pressure of the gas after expansion is, 279.2 Kpa
For number one, use the equation of v= v initial+ at. Use v initial as zero for this situation and 9.8 m/s^2 as the acceleration due to gravity
