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Answer:
Magnitude the net torque about its axis of rotation is 2.41 Nm
Solution:
As per the question:
The radius of the wrapped rope around the drum, r = 1.33 m
Force applied to the right side of the drum, F = 4.35 N
The radius of the rope wrapped around the core, r' = 0.51 m
Force on the cylinder in the downward direction, F' = 6.62 N
Now, the magnitude of the net torque is given by:

where
= Torque due to Force, F
= Torque due to Force, F'


Now,


The net torque comes out to be negative, this shows that rotation of cylinder is in the clockwise direction from its stationary position.
Now, the magnitude of the net torque:

Answer: 110000
Explanation:
26/9=30.5555555556
30.5555555556 x 60=1833.33333333
110000 x 60=110000
Answer:
1793.7m
Explanation:
From the principle of conservation of energy; the kinetic energy substended by the object equals the potential energy sustain by the object when it gets to its maximum position.
Now the kinetic energy; is
K.E = 1/2 × m × v2
Where m is mass
v is velocity
Hence.
K.E = 1/2 × 2.25 × (187.5)^2
Now this should be same with the potential energy which is given as;
P.E = m× g× h
Where m is mass of object
g is acceleration of free fall due to gravity = 9.8m/S2
h is maximum height substain by the object.
Hence P.E = 2.25 × 9.8 × h
From the foregoing analysis of energy conversation it implies;
1/2 × 2.25 × (187.5)^2 =2.25 × 9.8 × h
=> 1/2 × (187.5)^2 = 9.8 × h
=>1/2 × (187.5)^2 / 9.8 = h
=> 1793.69m = h
h= 1793.69m
h =1793.7m to 1 decimal place
<span>Active Galactic Nuclei.</span>