Explanation:
c. if the vector is oriented at 0° from the X -axis.
Solution:
initial sphere mvr = final sphere mvr + Iω
where I = mL²/3 = 2.3g * (2m)² / 3 = 3.07 kg·m²
0.25kg * (12.5 + 9.5)m/s * (4/5)2m = 3.07 kg·m² * ω
where: ω = 2.87 rad/s
So for the rod, initial E = KE = ½Iω² = ½ * 3.07kg·m² * (2.87rad/s)²
E = 12.64 J becomes PE = mgh, so
12.64 J = 2.3 kg * 9.8m/s² * h
h = 0.29 m
h = L(1 - cosΘ) → where here L is the distance to the CM
0.03m = 1m(1 - cosΘ) = 1m - 1m*cosΘ
Θ = arccos((1-0.29)/1) = 44.77 º
Answer:
From question (a) and (b) the pendulum motion is perpendicular to the force so the normal force will do no work and the tension in the string of the pendulum will not work

And
so

c
An example will be a where a stone is attached to the end of a string and is made to move in a circular motion while keeping the other end of the string in a fixed position
d
A dog walking along a surface which has friction, here the frictional force would acting in the direction of the motion and this would do positive work
Explanation:
The electron is accelerated through a potential difference of

, so the kinetic energy gained by the electron is equal to its variation of electrical potential energy:

where
m is the electron mass
v is the final speed of the electron
e is the electron charge

is the potential difference
Re-arranging this equation, we can find the speed of the electron before entering the magnetic field:

Now the electron enters the magnetic field. The Lorentz force provides the centripetal force that keeps the electron in circular orbit:

where B is the intensity of the magnetic field and r is the orbital radius. Since the radius is r=25 cm=0.25 m, we can re-arrange this equation to find B:
Ca(NO3)2(aq) + Na2CO3(aq) → 2NaNO3 + CaCO3⬇. NaNO3 is solution so CaCO3 is the precipitate formed.