Work occurs when an applied force results in movement of an object in the same direction as the applied force.
A)We know the formula of the angular speed is ω = 2π / TWhere T is the time period.When second hand completes one revolution then the time taken is 60s.So T = 60sThen the angular speed of the second hand is ω= 2π / (60s) = 0.1047 rad/sb)When the minute hand completes one revolution the time taken is T = 1 hr = 3600sThen the angular speed of the minute hand is ω =(2π) / (3600s) = 0.001745 rad/sc)When the hour hand completes one revolution then the timeperiod is T = 12hrs = (12)(3600)sThen the angular speed of the hour hand is ω =(2π) / [(12)(3600)s] = 1.45444 x 10^-4 rad/s
Answer:
T = 2010 N
Explanation:
m = mass of the uniform beam = 150 kg
Force of gravity acting on the beam at its center is given as
W = mg
W = 150 x 9.8
W = 1470 N
T = Tension force in the wire
θ = angle made by the wire with the horizontal = 47° deg
L = length of the beam
From the figure,
AC = L
BC = L/2
From the figure, using equilibrium of torque about point C
T (AC) Sin47 = W (BC)
T L Sin47 = W (L/2)
T Sin47 = W/2
T Sin47 = 1470
T = 2010 N
The right answer is
all of the above
good luck
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