<span>Answer:
For a disc, the moment of inertia about the perpendicular axis through the center is given by 0.5MR^2.
where M is the mass of the disc and R is the radius of the disc.
For the axis through the edge, use parallel axis theorem.
I = I(axis through center of mass) + M(distance between the axes)^2
= 0.5MR^2 + MR^2 (since the axis through center of mass is the axis through the center)
= 1.5 MR^2</span>
Answer: B. CO
Explanation:
Diatomic molecules are those that are formed by two atoms of the same chemical element (homonuclear diatomic molecule) or different chemical element (heteronuclear diatomic molecule).
In this sense, oxygen is a homonuclear diatomic molecule because it is formed by two atoms of the same element (
) and Carbon monoxide (
) is heteronuclear diatomic molecule.
Sodium Chloride 
is not a diatomic molecule because is a product of ionization, but it can be diatomic in its gas phase with a polar covalent bond.
Answer:
Change in potential energy = 7350 Joules
Explanation:
It is given that,
Side of cube, a = 0.5 m
Density of cube, 
The cube is lifted vertically by a crane to a height of 3 m
We know that, density 
So, m = d × V (V = volume of cube = a³)
m = 250 kg
We have to find the change in potential energy of the cube. At ground level, the potential energy is equal to 0.
Potential energy at height h is given by :
PE = 250 kg × 9.8 m/s² ×3 m
PE = 7350 Joules
So, change in potential energy of the cube is 7350 Joules.
Answer:
0.82 mm
Explanation:
The formula for calculation an 
bright fringe from the central maxima is given as:
so for the distance of the second-order fringe when wavelength 
= 745-nm can be calculated as:
where;
n = 2
= 745-nm
D = 1.0 m
d = 0.54 mm
substituting the parameters in the above equation; we have:
= 0.00276 m
= 2.76 × 10 ⁻³ m
The distance of the second order fringe when the wavelength 
= 660-nm is as follows:
= 1.94 × 10 ⁻³ m
So, the distance apart the two fringe can now be calculated as:
= 2.76 × 10 ⁻³ m - 1.94 × 10 ⁻³ m
= 10 ⁻³ (2.76 - 1.94)
= 10 ⁻³ (0.82)
= 0.82 × 10 ⁻³ m
= 0.82 × 10 ⁻³ m 
= 0.82 mm
Thus, the distance apart the second-order fringes for these two wavelengths = 0.82 mm
Answer:
1.08 s
Explanation:
From the question given above, the following data were obtained:
Height (h) reached = 1.45 m
Time of flight (T) =?
Next, we shall determine the time taken for the kangaroo to return from the height of 1.45 m. This can be obtained as follow:
Height (h) = 1.45 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
1.45 = ½ × 9.8 × t²
1.45 = 4.9 × t²
Divide both side by 4.9
t² = 1.45/4.9
Take the square root of both side
t = √(1.45/4.9)
t = 0.54 s
Note: the time taken to fall from the height(1.45m) is the same as the time taken for the kangaroo to get to the height(1.45 m).
Finally, we shall determine the total time spent by the kangaroo before returning to the earth. This can be obtained as follow:
Time (t) taken to reach the height = 0.54 s
Time of flight (T) =?
T = 2t
T = 2 × 0.54
T = 1.08 s
Therefore, it will take the kangaroo 1.08 s to return to the earth.
