All except C, feelings are not physical.
In
order to determine the mass of a standard baseball if it had the same density
(mass per unit volume) as a proton or neutron, we first determine the volume of
the baseball. The formula to be used is V_sphere = (4/3)*pi*r^3. In this case, the
radius r can be obtained from the circumference C, C = 2*pi*r. After plugging
in C = 23 cm to the equation, we get r = 3.6066 cm. The volume of the baseball
is then equal to 205.4625 cm^3.
Next,
take note of these necessary information:
Mass of a neutron/proton
= 10^-27 kg
Diameter of a
neutron/proton = 10^-15 m
Radius of a
neutron/proton = [(10^-15)/2]*100 = 5x10^-14 cm
<span>Thus,
the density, M/V of the neutron/proton is equal to 1.9099x10^12 kg/cm^3. Finally,
the mass of the baseball if it was a neutron/proton can be determined by
multiplying the density of the neutron/proton with the volume of the baseball. The
final answer is then a large value of 3.9241x10^14 kg.</span>
Answer:
A limestone plateau has no surface water. All the water is pulled underground through cracks and crevices in the surface. What most likely will cause the underground of the plateau to change over time?
Physical weathering due to frost wedging
Physical weathering due to abrasion
Chemical weathering due to oxygen
Chemical weathering due to water Correct Answer
Explanation:
<span>Speed (rms) = sqrt 3KT/m, where:
K = Boltzmann's costant = 1.38*10^-23 joule/K;
T = temperature in Kelvin degrees = 273 - 63 = 210 K;
m = 44 * 1.672*10^-27 Kg;
1.672*10^-27 Kg (good approximation) is the weight of a single a.m.u. (atomic mass unit);
44 is molecular weight of CO2.
Molecular Speed (CO2) = sqrt 3*1.38*10^-23*210/44*1.672*10^-27 = 344 m/sec.
Excuse me for some imperfection in my language, I am from South Italy. My cordiality to You, hello.</span>
the total energy is the
sum of the linear and rotational energy: <span>
K = K_rot + K_lin
first, we find the rotational kinetic energy of a rotating
disc with an angular velocity of w. see the references for the moment of
inertia of a disc.
K_rot = (1/2)(I)(w^2)
I = (1/2)(m)(r^2)
K_rot = (1/4)(m)(r^2)(w^2)
next, we find the linear kinetic energy of a rolling disc:
K_lin = (1/2)(m)(v^2)
v = angular velocity * circumference
= w * (pi * 2 * r)
K_lin = (1/2)(m)(w*2*pi*r)^2
= (2*pi^2)(m)(r^2)(w^2)
we find the total kinetic energy:
K = K_rot + K_lin
= (1/4)(m)(r^2)(w^2) + (2*pi^2)(m)(r^2)(w^2)
and find the rotational contribution:
K_rot = K * [K_rot/K]
K_rot = K * [K_rot/(K_rot+K_lin)]
K_rot = K * (1/4) / [(1/4) + (2*pi^2)]
</span>K_rot = K / (8*pi^2 + 1)