Asimple pendulum, 2.0 m in length, is released with a push when the support string is at an angle of 25° from the vertical. if the initial speed of the suspended mass is 1.2 m/s when at the release point, what is its speed at the bottom of the swing? (g = 9.8 m/s2)

2.3m/s

Explanation:

mass-energy balance: ke(f) + pe(f) = ke(o) + pe(o)

since we are looking for the point at the bottom of the pendulum, thats the reference point, the lowest in the system. pe(f) is 0, since h

ke(f)=0.5m x v(f)^2

pe(f)=0

ke(o)=0.5m x v(o)^2

pe(o)-mxgxh

find h by: drawing a triangle with the pendulum at the vertical, then displaced by 25 degrees , The difference in height is h, because cos(25)=(adj)/(hyp)=(2-h)/2. I found h=0.187m

In the m-e balance, cancel the masses in all the terms.

.5xv(f)^2 =0.5v(o)^2 +gxh

Given v(o) = 1.2 m/s and g = 9.8 then v(f) = 2.2595 m/s

Therefore V(0) = 2.3 m/s

as we know our atmosphere or air have different levels. every level has other gasses, and every gas has another permeability. when the light comes from lower permeability to higher permeability, it changes its direction. so that means that when the light comes from the moon to the earth, it changes it's direction many more times. it basically means that the shape of the light part of the moon is changed on the earth.

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the temperature in kelvin of 2.0 dm3 of an ideal gas is doubled and its pressure is increased by a factor of four. what is the final volume of the gas