Refer to the figure shown below.
g = 9.8 m/s², the acceleration due to gravity.
W = mg, the weight of the mug.
θ = 17°, the angle of the ramp.
Let μ = the coefficient of static friction.
The force acting down the ramp is
F = W sin θ = W sin(17°) = 0.2924W N
The normal reaction is
N = W cosθ = W cos(17°) = 0.9563W N
The resistive force due to friction is
R = μN = 0.9563μW N
For static equilibrium,
μN = F
0.9563μW =0.2924W
μ = 0.3058
The frictional force is F = μN = 0.2924W
The minimum value of μ required to prevent the mug from sliding satisfies
the condition
R > F
0.9563μW > 0.2924W
μ > 002924/.9563 = 0.306
Answer:
The frictional force is 0.2924mg, where m = the mass of the mug.
The minimum coefficient of static friction is 0.306
Wouldn't it be the employee? Because the employee has to adjust to the needs of his/her supervisor. If the supervisor wants 100 boxes the employee has to make those 100 boxes and so on and so forth.
At the lowest point on the Ferris wheel, there are two forces acting on the child: their weight of 430 N, and an upward centripetal/normal force with magnitude n; then the net force on the child is
∑ F = ma
n - 430 N = (430 N)/g • a
where m is the child's mass and a is their centripetal acceleration. The child has a linear speed of 3.5 m/s at any point along the path of the wheel whose radius is 17 m, so the centripetal acceleration is
a = (3.5 m/s)² / (17 m) ≈ 0.72 m/s²
and so
n = 430 N + (430 N)/g (0.72 m/s²) ≈ 460 N
Explanation:
The size of the force varies inversely as the square of the distance between the two charges. Therefore, if the distance between the two charges is doubled, the attraction or repulsion becomes weaker, decreasing to one-fourth of the original value.
<span>4) Formation of a gas
When methane burns, it produces the gases water vapor and carbon dioxide.
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