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.
In the system described above we will have four forces that is acting on the puck. These are the weight, the normal force, the frictional force, and the force applied by the player. To determine the force applied by the player, we need to calculate first for the frictional force which is equal to the product of the coefficient of friction and the normal force. We do as follows:
Summation of forces in the y-direction:
W = Fn
Fn = 1.70 N
Summation of force in the x-direction
F = Fr = 0.06Fn
F = 0.06 (1.70) = 0.102 N
Out of the processes of the water cycle, namely evaporation, precipitation, condensation, transpiration, storage, and flow, the condensation part of the water cycle releases energy. People might argue that it is evaporation, but actually during evaporation, what happens to energy is that it is taken in which makes the environment cooler. During the condensation, it is needed for water vapor to release the energy from evaporation which in return makes the environment warmer. As to the other three, transpiration, storage, flow, there is no known releasing of energy involved.
The equation of D = m/V
Where D = density
m = mass
and V = volume
We are solving for V, so with the manipulation of variables we multiply V on both sides giving us
V(D) = m
now we divide D on both sides giving us
V = m/D
We know our mass which is 600g and our density is 3.00 g/cm^3
so
V = 600g/3.00g/cm^3 = 200cm^3 or 200mL
a cubic centimeter (cm^3) is one of the units for volume. It's exactly like mL. 1 cm^3 = 1 mL
If you wish to change it to L, you'd have to convert.