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
Answer:
Ethyne consists of a triple bond that is most easier to be broken as compared to that of ethyne with a double bond and benzene single bonds. Therefore ethyne is most reactive among the given options.
Explanation:
Power used by the clock=1.03 W
Explanation:
resistance= 14000 ohm
voltage=120 V
The formula for the power is given by

P=(120)²/14000
P=1.03 W
The minimum velocity of the Salmon jumping at the given angle is 12.3 m/s.
The given parameters;
- height of the waterfall, h = 0.432 m
- distance of the Salmon from the waterfall, s = 3.17 m
- angle of projection of the Salmon, = 30.8º
The time of motion to fall from 0.432 m is calculated as;

The minimum velocity of the Salmon jumping at the given angle is calculated as;

Thus, the minimum velocity of the Salmon jumping at the given angle is 12.3 m/s.
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