Answer:
(A) 0.63 J
(B) 0.15 m
Explanation:
length (L) = 0.75 m
mass (m) =0.42 kg
angular speed (ω) = 4 rad/s
To solve the questions (a) and (b) we first need to calculate the rotational inertia of the rod (I)
I = Ic + m
Ic is the rotational inertia of the rod about an axis passing trough its centre of mass and parallel to the rotational axis
h is the horizontal distance between the center of mass and the rotational axis of the rod
I = ![(\frac{1}{12})(mL^{2} ) + m([tex]\frac{L}{2}](https://tex.z-dn.net/?f=%28%5Cfrac%7B1%7D%7B12%7D%29%28mL%5E%7B2%7D%20%29%20%2B%20m%28%5Btex%5D%5Cfrac%7BL%7D%7B2%7D)
)^{2}[/tex]
I = ![(\frac{1}{12})(0.42 x 0.75^{2} ) + ( 0.42 x ([tex]\frac{0.75}{2}](https://tex.z-dn.net/?f=%28%5Cfrac%7B1%7D%7B12%7D%29%280.42%20x%200.75%5E%7B2%7D%20%29%20%2B%20%28%200.42%20x%20%28%5Btex%5D%5Cfrac%7B0.75%7D%7B2%7D)
)^{2}[/tex])
I = 0.07875 kg.m^{2}
(A) rods kinetic energy = 0.5I
= 0.5 x 0.07875 x 
= 0.63 J 0.15 m
(B) from the conservation of energy
initial kinetic energy + initial potential energy = final kinetic energy + final potential energy
Ki + Ui = Kf + Uf
at the maximum height velocity = 0 therefore final kinetic energy = 0
Ki + Ui = Uf
Ki = Uf - Ui
Ki = mg(H-h)
where (H-h) = rise in the center of mass
0.63 = 0.42 x 9.8 x (H-h)
(H-h) = 0.15 m
guaranteeing freedom of speech, press, assembly, and exercise of religion
Answer:
The work done on the suitcase is, W = 1691 J
Explanation:
Given data,
The force on the suitcase is, F = 89 N
The distance Russell dragged the suitcase, S = 19 m
The work done on the suitcase by Russell is equal to the work done on the suitcase to overcome the friction
The work done on the suitcase by Russell is given by the formula
W = F · S
Substituting the given values,
W = 89 N x 19 m
W = 1691 J
Hence, the work done on the suitcase is, W = 1691 J
Answer:
1.36 x 10^-3 cm
Explanation:
Area = 50 ft^2 = 46451.5 cm^2
mass = 6 oz = 170.097 g
density = 2.70 g/cm^3
Let t be the thickness of foil in cm.
mass = volume x density
mass = area x thickness x density
170.097 = 46451.5 x t x 2.70
t = 1.36 x 10^-3 cm
Thus, the thickness of aluminium foil is 1.36 x 10^-3 cm.
Answer:
It has very high density, so a small bulb of a thermometer can contain much mercury. Mercury remains liquid state over a quite wide range of temperature because it freezes at 39°C and boils at 357°C.
Explanation:
