air beomes more dense
below dewpoint = a dew forms ... wet underfoot.
cold enough = frost ???
Answer:
a) I = 0.0198 kg m²
, b) I = 21.85 kg m²
Explanation:
For this exercise we will use the definition of moment of inertia
I = ∫ r² dm
For body with high symmetry they are tabulated
sphere I = 2/5 m r²
bar with respect to center of mass I = 1/12 m L²
let's calculate the mass of each body
bar
ρ = m / V
m = ρ V
m = ρ l w h
where we are given the density of the bar rho = 32840 kg / m³ and its dimensions 1 m, 0.8 cm and 4 cm
m = 32820 1 0.008 0.04
m = 10.5 kg
Sphere
M = ρ V
V = 4/3 pi r³
M = rgo 4/3 π r³
give us the density 37800 kg / m³ and the radius of 5 cm
M = 37800 4/3 π 0.05³
M = 19.8 kg
a) asks us for the moment of inertia of the sphere with respect to its center of mass
I = 2/5 M r²
I = 2/5 19.8 0.05²
I = 0.0198 kg m²
b) the moment of inertia with respect to the turning point, for this we will use the theorem of parallel axes
I = I_cm + M d2
where d is the distance from the body to the point of interest
I_cm = 0.0198 kg m²
the distance to the pivot point is
l = length of the bar + radius of the sphere
l = 1 + 0.05 = 1.005 m
I = 0.0198 + 19.8 1.05²
I = 21.85 kg m²
Answer: T= 715 N
Explanation:
The only external force (neglecting gravity) acting on the swinging mass, is the centripetal force, which. in this case, is represented by the tension in the string, so we can say:
T = mv² / r
At the moment that the mass be released, it wil continue moving in a straight line at the same tangential speed that it had just an instant before, which is the same speed included in the centripetal force expression.
So the kinetic energy will be the following:
K = 1/2 m v² = 15. 0 J
Solving for v², and replacing in the expression for T:
T = 1.9 Kg (3.97)² m²/s² / 0.042 m = 715 N
Complete question:
It is measured that 3/4 of a body's volume is submerged in oil of density 800kg/m³. What is the specific gravity of oil?
Answer:
The specific gravity of the oil is 0.8.
Explanation:
Given;
density of the oil, 
= 800 kg/m³
density of water, 
= 1000 kg/m³
The specific gravity of any substance is the ratio of the substance density to the density of water.
Specific gravity of the oil = density of the oil / density of water
Specific gravity of the oil = 800/1000
Specific gravity of the oil = 0.8
Therefore, the specific gravity of the oil is 0.8.
The researcher can change one isotope into a different isotope of the same element by ADDING OR REMOVING NEUTRONS.
Isotopes refers to two or more forms of the same element, which have the same number of protons but different number of neutrons. The difference in the number of neutrons of a particular element is what brings about isotopes, thus, isotopes can be created by removing or adding neutrons to a particular element.<span />
