To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.
If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,
So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb
Answer:
6ms^-1
Explanation:
Given that the frequency difference is
( 563- 544) = 19
So alsoThe wavelength of each wave is = v/f = 344 /544
and there are 19 of this waves
So it is assumed that each motorcycle has moved 0.5 of this distance
in one second thus the speed of the motorcycles will be
=> 19/2 x 344/544 = 6.0 m/s
Thank you for posting your question here at brainly. I hope the answer will help you. Feel free to ask more questions here.
The moment of the resultant of these two forces with respect to O 376 lb-ft CCW which is <span>about moment center point O.</span>
Since each student emits 100 W, so 170 students will emit:
total heat = 100 W * 170 = 17,000 W
Convert minutes to seconds:
time = 50 min * (60 s / min) = 3000 s
The energy is therefore:
E = 17,000 W * 3000 s
<span>E = 51 x 10^6 J = 51 MJ</span>