Answer:
times
Explanation:
First of all, we need to write both the age of the universe and the lifetime of the top quark in scientific notation.
Age of the universe:
(1 followed by 17 zeroes)
Lifetime of the top quark:
(we moved the decimal point 24 places to the right)
Therefore, to answer the question, we have to calculate the ratio between the age of the universe and the lifetime of the top quark:
Refer to the diagram shown below.
The initial KE (kinetic energy) of the system is
KE₁ = (1/2)mu²
After an inelastic collision, the two masses stick together.
Conservation of momentum requires that
m*u = 2m*v
Therefore
v = u/2
The final KE is
KE₂ = (1/2)(2m)v²
= m(u/2)²
= (1/4)mu²
= (1/2) KE₁
The loss in KE is
KE₁ - KE₂ = (1/2) KE₁.
Conservation of energy requires that the loss in KE be accounted for as thermal energy.
Answer: 1/2
Answer:
73.13°
Explanation:
According to snell's law,
n1sinθi = n2sinθr
n1/n2 = sinθr/sinθi
Critical angle is the angle of incidence at the denser medium when the angle of incidence at the less dense medium is 90°
This means i=C and r = 90°
The Snell's law formula will become
n1/n2 = sinC/sin90°
n2/n1 = 1/sinC
Where n1 is the refractive index of the less dense medium = 1.473
n2 is the refractive index of the denser medium = 1.540
Substituting the values in the formula,
1.540/1.473 = 1/sinC
1.045 = 1/sinC
SinC = 1/1.045
SinC = 0.957
C = sin^-1(0.957)
C = 73.13°
Answer:
the image is virtual and erect and the lens divergent; therefore the correct answer is C
Explanation:
In a thin lens the magnification given by
m = h '/ h = - q / p
where h ’is the height of the image, h is the height of the object, q is the distance to the image and p is the distance to the object.
It indicates that the object is straight and is placed at a distance p> f
analyze the situation tells us that the magnification is positive so the distance to the image must be negative, that is, that the image is on the same side as the object.
Consequently the lens must be divergent
The magnification value is
0.4 = h ’/ h
h ’= 0.4 h
therefore the erect images
therefore the image is virtual and erect and the lens divergent; therefore the correct answer is C
