Answer:
Explanation:
Initial moment of inertia of the earth I₁ = 2/5 MR² , M is mss of the earth and R is the radius . If ice melts , it forms an equivalent shell of mass 2.3 x 10¹⁹ Kg
Final moment of inertia I₂ = 2/5 M R² + 2/3 x 2.3 x 10¹⁹ x R²
For change in period of rotation we shall apply conservation of angular momentum law
I₁ ω₁ = I₂ ω₂ , ω₁ and ω₂ are angular velocities initially and finally .
I₁ / I₂ = ω₂ / ω₁
I₁ / I₂ = T₁ / T₂ , T₁ , T₂ are time period initially and finally .
T₂ / T₁ = I₂ / I₁
(2/5 M R² + 2/3 x 2.3 x 10¹⁹ x R²) / 2/5 MR²
1 + 5 / 3 x 2.3 x 10¹⁹ / M
= 1 + 5 / 3 x 2.3 x 10¹⁹ / 5.97 x 10²⁴
= 1 + .0000064
T₂ = 24 (1 + .0000064)
= 24 hours + .55 s
change in length of the day = .55 s .
Answer:
- tension: 19.3 N
- acceleration: 3.36 m/s^2
Explanation:
<u>Given</u>
mass A = 2.0 kg
mass B = 3.0 kg
θ = 40°
<u>Find</u>
The tension in the string
The acceleration of the masses
<u>Solution</u>
Mass A is being pulled down the inclined plane by a force due to gravity of ...
F = mg·sin(θ) = (2 kg)(9.8 m/s^2)(0.642788) = 12.5986 N
Mass B is being pulled downward by gravity with a force of ...
F = mg = (3 kg)(9.8 m/s^2) = 29.4 N
The tension in the string, T, is such that the net force on each mass results in the same acceleration:
F/m = a = F/m
(T -12.59806 N)/(2 kg) = (29.4 N -T) N/(3 kg)
T = (2(29.4) +3(12.5986))/5 = 19.3192 N
__
Then the acceleration of B is ...
a = F/m = (29.4 -19.3192) N/(3 kg) = 3.36027 m/s^2
The string tension is about 19.3 N; the acceleration of the masses is about 3.36 m/s^2.
Answer:
Einstein extended the rules of Newton for high speeds. For applications of mechanics at low speeds, Newtonian ideas are almost equal to reality. That is the reason we use Newtonian mechanics in practice at low speeds.
Explanation:
<em>But on a conceptual level, Einstein did prove Newtonian ideas quite wrong in some cases, e.g. the relativity of simultaneity. But again, in calculations, Newtonian ideas give pretty close to correct answer in low-speed regimes. So, the numerical validity of Newtonian laws in those regimes is something that no one can ever prove completely wrong - because they have been proven correct experimentally to a good approximation.</em>
The way I do it is suddenly, in the same sort of way that magicians try to pull a table cloth off a table when there's things on the table cloth.The sudden approach acts as an impulse of force and starts to accelerate the roll. But, the piece (assuming it has perforations) is off the roll before the roll can move, due to inertia. Then the roll will acclerate, move, slow down and stop. However, in accelerating, the roll will unravel. The bigger the impulse the more it will unravel.+++++++++++++++++++++++++++++++++++++++If on the other hand, the piece of paper is held firmly, and the roll is pulled, then the impulse is presumably given to the paper and the hand whose inertia is a lot more than that of the roll. So, I think I'd actually go for choice c)+++++++++++++++++++++++++++++++++++++This assumes that the roll is free to rotate.I think that a similar idea is behind the design and use of a "ballistic galvanometer". The charge is passed through the galvanometer quickly, as a current pulse. Then the needle starts to deflect, and the deflection is arranged to depend on the total charge that has passed through in the time of the current pulse.
Answer:
The rate of change of the shadow length of a person is 2.692 ft/s
Solution:
As per the question:
Height of a person, H = 20 ft
Height of a person, h = 7 ft
Rate = 5 ft/s
Now,
From Fig.1:
b = person's distance from the lamp post
a = shadow length
Also, from the similarity of the triangles, we can write:
Differentiating the above eqn w.r.t t:
Now, we know that:
Rate = 
Thus
