By Newton's second law, the net vertical force acting on the object is 0, so that
<em>n</em> - <em>w</em> = 0
where <em>n</em> = magnitude of the normal force of the surface pushing up on the object, and <em>w</em> = weight of the object. Hence <em>n</em> = <em>w</em> = <em>mg</em> = 196 N, where <em>m</em> = 20 kg and <em>g</em> = 9.80 m/s².
The force of static friction exerts up to 80 N on the object, since that's the minimum required force needed to get it moving, which means the coefficient of <u>static</u> friction <em>µ</em> is such that
80 N = <em>µ</em> (196 N) → <em>µ</em> = (80 N)/(196 N) ≈ 0.408
Moving at constant speed, there is a kinetic friction force of 40 N opposing the object's motion, so that the coefficient of <u>kinetic</u> friction <em>ν</em> is
40 N = <em>ν</em> (196 N) → <em>ν</em> = (40 N)/(196 N) ≈ 0.204
And so the closest answer is C.
(Note: <em>µ</em> and <em>ν</em> are the Greek letters mu and nu)
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Answer:
85.556metres
Explanation:
Using pythagorean theorem
C²=A²+B²
we have c as the hypotenuse vector A thus:
93.8²=A²+38.4²
93.8²-38.4²=A²
8794.44-1474.56=A²
7319.88=A²
A=85.556
Answer:
20 cm
Explanation:
Given that a ball is released from a vertical height of 20 cm. It rolls down a "perfectly frictionless" ramp and up a similar ramp. What vertical height on the second ramp will the ball reach before it starts to roll back down?
Since it is perfectly frictionless, the Kinetic energy in which the ball is rolling will be equal to the potential energy at the edge of the ramp.
Therefore, the ball will reach 20 cm before it starts to roll back down.
Answer:
The number of crates is 84580.
Explanation:
mass, m = 30 kg
height, h = 0.9 mm
Power, P = 0.5 hp = 0.5 x 746 W = 373 W
time, t = 1 minute = 60 s
Let the number of crates is n.
Power is given by the rate of doing work.
