Answer is: T = -4.706°C
Explanation: Please see the image attached with this answer.
Density + mass / volume = 42 / 22 = 1.909 kg / m^3 ( to the nearest thousandth)
Answer:
24.25m/s
Explanation:
m = 0.5kg
u ( initial velocity) = 0m/s
v ( final velocity) =?
a = 9.8m/s^2
d (displacement) = 30m
Since u don't have time, u only have the choice to use this formula
V^2 = u^2 + 2ad
V^2 = 0 + 2 x 9.8 x 30
V^2 = 588
V = 24.25 m/s
Answer:
The kinetic energy of the more massive ball is greater by a factor of 2.
Explanation:
By conservation of energy, we know that the initial energy = final energy. At first, the balls are dropped from a height with no initial velocity so their initial energy is all potential energy. When they reach the bottom, all their energy is kinetic energy. So all of their energy is changed from potential to kinetic energy. This means that the ball with greater potential energy will have a greater kinetic energy.
Potential energy = mgh. Since g = gravity is a constant and h = height is the same, the only difference is mass. Since mass is directly proportional to potential energy, the greater the mass, the greater the potential energy, so the more massive ball has a greater initial potential energy and will have a greater kinetic energy at the bottom.
Additionally, let B1 = lighter ball with mass m and let B2 = heavier ball with mass m2. Since we know that intial potential energy = final kinetic energy. We can rewrite it as potential energy = kinetic energy = mass * gravity constant * height. For B1, it is mgh and for B2 it is 2mgh, so B2's kinetic energy is twice that of B1.
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
