Answer:
y = 80.2 mille
Explanation:
The minimum size of an object that can be seen is determined by the diffraction phenomenon, if we use the Rayleigh criterion that establishes that two objects can be distinguished without the maximum diffraction of a body coincides with the minimum of the other body, therefore so much for the pupil of the eye that it is a circular opening
θ = 1.22 λ/ d
in a normal eye the diameter of the pupils of d = 2 mm = 0.002 m, suppose the wavelength of maximum sensitivity of the eye λ = 550 nm = 550 10⁻⁹ m
θ = 1.22 550 10⁻⁹ / 0.002
θ = 3.355 10⁻⁴ rad
Let's use trigonometry to find the distance supported by this angle, the distance from the moon to the Earth is L = 238900 mille = 2.38900 10⁵ mi
tan θ = y / L
y = L tan θ
y = 2,389 10⁵ tan 3,355 10⁻⁴
y = 8.02 10¹ mi
y = 80.2 mille
This is the smallest size of an object seen directly by the eye
Answer:
Option B, Some of the cars' kinetic energy was converted to sound and heat energy.
Explanation:
In an elastic collision, no energy is lost during and after collision. Thus, it can be said that in an elastic collision both momentum and kinetic energy remains conserved.
While in non-elastic collision, kinetic energy of the system is lost. However, the momentum of the system is conserved. Generally, during and after collision some of the kinetic energy is lost as thermal energy, sound energy etc.
Hence, option B is correct
210 Pb ---> -ie + 210 B:
84 8.3
Explanation:
Solution:
Let the time be
t1=35min = 0.58min
t2=10min=0.166min
t3=45min= 0.75min
t4=35min= 0.58min
let the velocities be
v1=100km/h
v2=55km/h
v3=35km/h
a. Determine the average speed for the trip. km/h
first we have to solve for the distance
S=s1+s2+s3
S= v1t1+v2t2+v3t3
S= 100*0.58+55*0.166+35*0.75
S=58+9.13+26.25
S=93.38km
V=S/t1+t2+t3+t4
V=93.38/0.58+0.166+0.75+0.58
V=93.38/2.076
V=44.98km/h
b. the distance is 93.38km
Answer:
c) At a distance greater than r
Explanation:
If G= Gravitational constant
M= Mass of earth
r= distance from earth center
then orbital speed is ;
v = 
==> v²=GM/r
If speed of first satellite = V₁
==> V₁² = GM/r
==> r = GM/V₁²
If speed of second satellite say V₂ is less than V₁ then square of V₂ will be less than square of V₁ , and hence GM will be divided by less number in case of second satellite, and hence will give greater value of r as compared to first satellite.
So our answer is c
