For a system released from rest at θ = 0° when a constant couple moment M = 100 N.m is applied, the speed of the 10-kg block is mathematically given as
V=4.33m/s
<h3>What is the speed of the 10-kg block?</h3>
Generally, the equation for the workdone is mathematically given as
W=T tehta<dtheta
W=100(90*\pi/180)
W=1.5707*100
W=1.57Nm
The change in potential energy across the pulley
dP=mgh
dp=10*9.81*111.8
dp=10.97J
For the thrid position, potential energy is
dP=mg(0.3)
dP=17.658J
dP'=17.658J-13.125
dP'=-4.532J
For 2nd position dP=0
The change in Kinectic energy across the pulley\
dK.E=0.5mv^2
For 1st
dk.E=0.5m(10)^2
2nd
dK.E=0.5Iw^2
dK.E=0.5(7.5*10^-3)(v^2/0.05)^2
3rd
2nd=3rd
In conclusion,
157.07=dKE+dP.E
157.07=5v^2+ (7.5*10^-3)(v^2/0.05)^2+10.97-4.53
V=4.33m/s
Read more about Kinetic energy
brainly.com/question/999862
3 Newtons toward the right because Newton’s second law:
F = ma
15N - 12N = ma
3N = ma
(With the right being the positive direction and the left the negative direction)
Answer:
Explanation:
We can write the expression here, but the point of the problem seems to be to see if you can manipulate the controls on the answer box to reproduce that expression.
<h3>Answer</h3>
option B)
19N
<h3>Explanation</h3>
If the object is at equilibrium, then the net force acting upon the object should be 0 N. Thus, if all the forces are added together, horizontal and vertical forces separately, then the resultant force (the vector sum) should be 0 Newton.
As we only need to find the magnitude of x-component of force F
so find all x component/horizontal forces acting on the object.
50cos(40) - 40cos(25) + 30cos(55) + x = 0
38.30 - 36.25 + 17.21 + x + = 0
19.26 + x = 0
x = - 19.26
x ≈ 19 (magnitude only)
Answer:
spinal cord becuse it does all of the reflexes
