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 =
)^{2}[/tex]
I =
)^{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
Answer:
According to your question although I think an object undergoing uniform circular motion is moving with a constant speed. Nevertheless, it is accelerating due to its change in direction. The direction of the acceleration is inwards,therefore a force perpendicular to an objects velocity change the direction of the velocity but not its magnitude.
Answer:
The total amount of energy that would have been released if the asteroid hit earth = The kinetic energy of the asteroid = 1.29 × 10¹⁵ J = 1.29 PetaJoules = 1.29 PJ
1 PJ = 10¹⁵ J
Explanation:
Kinetic energy = mv²/2
velocity of the asteroid is given as 7.8 km/s = 7800 m/s
To obtain the mass, we get it from the specific gravity and diameter information given.
Density = specific gravity × 1000 = 3 × 1000 = 3000 kg/m³
But density = mass/volume
So, mass = density × volume.
Taking the informed assumption that the asteroid is a sphere,
Volume = 4πr³/3
Diameter = 30 m, r = D/2 = 15 m
Volume = 4π(15)³/3 = 14137.2 m³
Mass of the asteroid = density × volume = 3000 × 14137.2 = 42411501 kg = 4.24 × 10⁷ kg
Kinetic energy of the asteroid = mv²/2 = (4.24 × 10⁷)(7800²)/2 = 1.29 × 10¹⁵ J
A beat is an interference pattern between two sounds of slightly different frequencies, perceived as a periodic variation in volume whose rate is the difference of the two frequencies. Frequency beat is equal to,

The reference frequency in our case would be 392Hz, and since there is the possibility of the upper and lower range for the amount of beats per second that the two possible frequencies are heard would be


Therefore the two possible frequencies the piano wire is vibrating at, would be 396Hz and 388Hz