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(a) 24.6 Nm
The torque produced by the net thrust about the center of the circle is given by:

where
F is the magnitude of the thrust
r is the radius of the wire
Here we have
F = 0.795 N
r = 30.9 m
Therefore, the torque produced is

(b) 
The equivalent of Newton's second law for a rotational motion is

where
is the torque
I is the moment of inertia
is the angular acceleration
If we consider the airplane as a point mass with mass m = 0.741 kg, then its moment of inertia is

And so we can solve the previous equation to find the angular acceleration:

(c) 
The linear acceleration (tangential acceleration) in a rotational motion is given by

where in this problem we have
is the angular acceleration
r = 30.9 m is the radius
Substituting the values, we find

Answer:

Explanation:
Given: that,
Angle of inclination of the surface, 
mass of the crate, 
Force applied along the surface, 
distance the crate moves after the application of force, 
a) work done = F× s
work done = 230 × 1.1
work done = 253 J
b) Work done by the gravitational force:

where:
g = acceleration due to gravity
h = the vertically downward displacement
Now, we find the height:

So, the work done by the gravity:

∵direction of force and displacement are opposite.
= - 343.54J
c)
The normal reaction force on the crate by the inclined surface:

d)
Total work done on crate is with respect to the worker:

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