Answer:
The magnitude of the torque due to gravity if it is supported at the 28-cm mark is 0.5 N-m.
Explanation:
Given that,
Mass of the meter stick, m = 0.3 kg
Center of mass is located at its 45 cm mark.
We need to find the magnitude of the torque due to gravity if it is supported at the 28-cm mark. Torque acting on the object is given by :

or

So, the magnitude of the torque due to gravity if it is supported at the 28-cm mark is 0.5 N-m.
A
Because leading zeros don’t count
Answer:
Jupiter
Explanation:
Since the mass of Jupiter is the greatest from the given choices, it will exert the most force on any object orbiting 100km above its surface.
This is compliance with the Newton's law of universal gravitation which states that "the force of attraction between two bodies is directly proportional to the magnitude of their masses and inversely proportional to the distances between them".
- Therefore, the more the masses of two bodies, the higher the gravitational attraction
- Since the distance is the same, the planet with the greater mass will exert the most force on the satellite.
From the diagram The value of cos C × sin A =
<h3>Determine the numerical value of cos C × sin A</h3>
First step : determine the values of cos C and sin A
cos C = adjacent / hypotenuse
= a / b
=
= √3/2
sin A = sin 60⁰
= √3/2
Therefore the numerical value of cos C * sin A = 
In conclusion From the diagram The value of cos C × sin A =
Learn more about right angle : brainly.com/question/24323420
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