Answer:
h = 8.588 m
Explanation:
Given:
Mass of hockey puck: m = 200 g = 0.2 K
g
Angle of incline: θ = 30º (with respect to horizontal)
Coefficient of static and kinetic friction between the puck and ramp:
μ
s = 0.4 and μk = 0.3
initial speed:
v = 3.5 m
/s
Let h be the vertical height reached by the puck, above the ground. This corresponds to a distance of
d = h
*sin 30º = 2
*h
along the incline. As the puck is moving, only kinetic friction comes into play as it climbs up the incline. Let N be the normal reaction exerted by the metal ramp on the ball. Then,
N = m*g
*cos θ = 0.2
*9.8
1*cos 30
º = 1.6991 N
Therefore, kinetic friction acting on the puck is:
F
k = μ
k
*N = 0.3
*1.6974 = 0.50974 N
From work energy theorem, the change in kinetic energy should equal the work done by friction and gravity. Therefore,
0.5* 0.2
*16² = 0.50974
*2h + 0.2*9.81*h
⇒ h = 8.588 m
Answer:
a medical journal/study from a licenced doctor on a website with .org
Explanation:
most medical journals from licenced doctors have .org in the end and are trusted
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Hey there
I believe the correct answer is (D)
Division in Java will return the integer value and disregard the remainder. This line of code will return 4.
:)
