Since the angular acceleration α is constant,
0² - (30 rev/s)² = 2α (60 rev)
Solve for α :
α = -(30 rev/s)²/(120 rev) = -7.5 rev/s²
One revolution corresponds to an angular displacement of 2π radians, so
α = (-7.5 rev/s²) × (2π rad/rev) ≈ -47.12 rad/s²
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Use the equation for momentum:

Plug in the given mass and velocity into the equation:


<span>Let t be the angle made by T3 with the x-axis.
T3 * cos(t) = 20 N ---- (1)
T3 * sin(t) = 60 N ----- (2)
Square both equations and add:
T3^2 = 20^2 + 60^2 = 400 + 3600 = 4000
T3 = sqrt(4000) = 63.25 N
divide (2) by (1):
tan(t) = 60/20 = 3
t = arctan(3) = 71.57 degrees.
</span><span>Since the angle clockwise from the x-axis some books may call it a negative angle and say the angle is -71.57 degrees or round it to -71.6 or even -72 degrees.</span>
Answer:
Conditions under which the belt and pulleys are operating – The friction between the belt and pulley may decrease substantially if the belt happens to be muddy or wet, as it may act as a lubricant between the surfaces.
Explanation:
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