The weight of the meterstick is:

and this weight is applied at the center of mass of the meterstick, so at x=0.50 m, therefore at a distance

from the pivot.
The torque generated by the weight of the meterstick around the pivot is:

To keep the system in equilibrium, the mass of 0.50 kg must generate an equal torque with opposite direction of rotation, so it must be located at a distance d2 somewhere between x=0 and x=0.40 m. The magnitude of the torque should be the same, 0.20 Nm, and so we have:

from which we find the value of d2:

So, the mass should be put at x=-0.04 m from the pivot, therefore at the x=36 cm mark.
The pickup accelerates towards right. The box is sticky to the pickup, thus its acceleration is the same, towards right. Its inertia (force) is oposing the acceleration, thus it is towards left. For the box not to move, it is necessary that the truck acts on it with a force towards right. (The two forces of the truck on box and the box inertia (force) equilibrate themselves).
Answer:
the thickness required of a masonry wall L = 375mm
Explanation:
The detailed steps and appropriate use of fourier's law of heat conduction is as shown in the attached file.
The velocity of the aircraft relative to the ground is 240 km/h North
Explanation:
We can solve this problem by using vector addition. In fact, the velocity of the aircraft relative to the ground is the (vector) sum between the velocity of the aircraft relative to the air and the velocity of the air relative to the ground.
Mathematically:

where
v' is the velocity of the aircraft relative to the ground
v is the velocity of the aircraft relative to the air
is the velocity of the air relative to the ground.
Taking north as positive direction, we have:
v = +320 km/h
(since the air is moving from North)
Therefore, we find
(north)
Learn more about vector addition:
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Centripetal acceleration = (speed)² / (radius) .
Force = (mass) · (acceleration)
Centripetal force = (mass) · (speed)² / (radius) .
= (11 kg) · (3.5 m/s)² / (0.6 m)
= (11 kg) · (12.25 m²/s²) / (0.6 m)
= (11 · 12.25) / 0.6 kg-m/s²
= 224.58 newtons. (about 50.5 pounds)
That's the tension in Miguel's arm or leg or whatever part of his body
Jesse is swinging him by. It's the centripetal force that's needed in
order to swing 11 kg in a circle with a radius of 0.6 meter, at 3.5
meters/second. If the force is less than that, then the mass has to
either swing slower or else move out to follow a bigger circle.