Answer:
The decrease is due to the bulge at the equator (putting more distance between the rest of the planet and the surface
Explanation:
<span> We're given that x=25 when t=2: </span>
<span>25 = 3 + 12(2) + (1/2)a(2)^2 </span>
<span>Thus a = -1 cm/sec^2</span>
The force required is 319 N
Explanation:
The force of static friction is a force that acts an object on a surface, when this object is pushed by another force to put it in motion. The direction of the force of friction is opposite to the direction of the force of push, and its value increases as the force of push increases, up to a maximum value given by:
where
is the coefficient of friction
W is the weight of the object
Therefore, in order to put the object in motion, the force applied must be greater than this value.
For the pile of leaves in this problem, we have:
(coefficient of friction)
(weight of the leaves)
Substituting, we find:
Learn more about force of friction:
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<span>Most low-level radioactive waste (LLW) is typically sent to land-based disposal immediately following its packaging for long-term management. This means that for the majority (~90% by volume) of all of the waste types produced by nuclear technologies, a satisfactory disposal means has been developed and is being implemented around the world.
</span>
Radioactive wastes are stored so as to avoid any chance of radiation exposure to people, or any pollution.The radioactivity of the wastes decays with time, providing a strong incentive to store high-level waste for about 50 years before disposal.Disposal of low-level waste is straightforward and can be undertaken safely almost anywhere.Storage of used fuel is normally under water for at least five years and then often in dry storage.<span>Deep geological disposal is widely agreed to be the best solution for final disposal of the most radioactive waste produced.
</span>I suggest this site on this subject http://www.world-nuclear.org/information-library/nuclear-fuel-cycle/nuclear-wastes/storage-and-dispo...
Answer:
the required revolution per hour is 28.6849
Explanation:
Given the data in the question;
we know that the expression for the linear acceleration in terms of angular velocity is;
= rω²
ω² = / r
ω = √( / r )
where r is the radius of the cylinder
ω is the angular velocity
given that; the centripetal acceleration equal to the acceleration of gravity a = g = 9.8 m/s²
so, given that, diameter = 4.86 miles = 4.86 × 1609 = 7819.74 m
Radius r = Diameter / 2 = 7819.74 m / 2 = 3909.87 m
so we substitute
ω = √( 9.8 m/s² / 3909.87 m )
ω = √0.002506477 s²
ω = 0.0500647 ≈ 0.05 rad/s
we know that; 1 rad/s = 9.5493 revolution per minute
ω = 0.05 × 9.5493 RPM
ω = 0.478082 RPM
1 rpm = 60 rph
so
ω = 0.478082 × 60
ω = 28.6849 revolutions per hour
Therefore, the required revolution per hour is 28.6849
