Answer:
12164.4 Nm
Explanation:
CHECK THE ATTACHMENT
Given values are;
m1= 470 kg
x= 4m
m2= 75kg
Cm = center of mass
g= acceleration due to gravity= 9.82 m/s^2
The distance of centre of mass is x/2
Center of mass(1) = x/2
But x= 4 m
Then substitute, we have,
Center of mass(1) = 4/2 = 2m
We can find the total torque, through the summation of moments that comes from both the man and the beam.
τ = τ(1) + τ(2)
But
τ(1)= ( Center of m1 × m1 × g)= (2× 470× 9.81)
= 9221.4Nm
τ(2)= X * m2 * g = ( 4× 75 × 9.81)= 2943Nm
τ = τ(1) + τ(2)
= 9221.4Nm + 2943Nm
= 12164.4 Nm
Hence, the magnitude of the torque about the point where the beam is bolted into place is 12164.4 Nm
Answer:
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Explanation:
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Answer:
0 m/s
Explanation:
Average velocity of an object is given by the net displacement divided by time taken. Displacement is equal to the shortest path covered by the object.
In this problem, a player runs the length of the 30-meter court and back. The player does this three times in 60 seconds.
As the player runs the court and returns to the original point. It would mean that the shortest path covered is 0.
Average velocity = displacement/time
v=0/30
v = 0 m/s
Hence, the correct option is (1).
Answer:
speed =wavelenght x frequency
v=4.5 x 10 to the -7 x 667=0.3 x 10 to the -4 m/s
speed= distance/time
time=distance/speed
t=4 x 10 to the 16/0.3 x 10 to the -4=13.33 x 10 to the 20 seconds
Explanation:
