The andwer of tye question is 3O2
The formula for this problem that we will be using is:
F * cos α = m * g * μs where:F = 800m = 87g = 9.8
cos α = m*g*μs/F= 87*9.8*0.55/800= 0.59 So solving the alpha, find the arccos above.
α = arccos 0.59 = 54 ° is the largest value of alpha
Answer:
Explanation:
A pressure that causes the Hg column to rise 1 millimeter is called a torr. The term 1 mmHg used can replaced by the torr.
1 atm = 760 torr = 14.7 psi.
A.
120 mmHg
Psi:
760 mmHg = 14.7 psi
120 mmHg = 14.7/760 * 120
= 2.32 psi
Pa:
1mmHg = 133.322 Pa
120 mmHg = 120 * 133.322
= 15998.4 Pa
B.
80 mmHg
Psi:
760 mmHg = 14.7 psi
80 mmHg = 14.7/760 * 80
= 1.55 psi
Pa:
1mmHg = 133.322 Pa
80 mmHg = 80 * 133.322
= 10665.6 Pa
Answer:
x = 0.974 L
Explanation:
given,
length of inclination of log = 30°
mass of log = 200 Kg
rock is located at = 0.6 L
L is the length of the log
mass of engineer = 53.5 Kg
let x be the distance from left at which log is horizontal.
For log to be horizontal system should be in equilibrium
∑ M = 0
mass of the log will be concentrated at the center
distance of rock from CM of log = 0.1 L
now,
∑ M = 0



x = 0.974 L
hence, distance of the engineer from the left side is equal to x = 0.974 L