Answer:
minimum number of photon is 4.05 ×
Explanation:
given data
energy = 50 keV = 50 × eV = 50 ×
× 1.602×
J
thickness = 10^-3
contrast = 1%
to find out
number of incident photons
solution
we know here equation that is
E = n × h × ν .......................1
put here all these value
50 × = n × 6.6×
× c/ 1×
50 × × 1.602×
= n × 6.6×
×( 3 ×
/ 1×
)
solve it and find n
n = 4.05 ×
so here minimum number of photon is 4.05 ×
Answer:
E) are almost circular, with low eccentricities.
Explanation:
Kepler's laws establish that:
All the planets revolve around the Sun in an elliptic orbit, with the Sun in one of the focus (Kepler's first law).
A planet describes equal areas in equal times (Kepler's second law).
The square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit (Kepler's third law).
Where T is the period of revolution and a is the semi-major axis.
Planets orbit around the Sun in an ellipse with the Sun in one of the focus. Because of that, it is not possible to the Sun to be at the center of the orbit, as the statement on option "C" says.
However, those orbits have low eccentricities (remember that an eccentricity = 0 corresponds to a circle)
In some moments of their orbit, planets will be closer to the Sun (known as perihelion). According with Kepler's second law to complete the same area in the same time, they have to speed up at their perihelion and slow down at their aphelion (point farther from the Sun in their orbit).
Therefore, option A and B can not be true.
In the celestial sphere, the path that the Sun moves in a period of a year is called ecliptic, and planets pass very closely to that path.
Answer :
(1) The density of asphalt is,
(2) (a) Length, width and thickness of sheet in meter is, 0.35 m, 1.1 m and 0.015 m respectively.
(b) The volume and mass of slab is, 0.005775 m³ and 15.59 kg respectively.
Explanation :
<u>Part 1 :</u>
As we are given:
Mass of block = 90 kg
Volume of block =
Formula used :
Now put all the given values in this formula, we get:
Thus, the density of asphalt is,
<u>Part 2(a) :</u>
As we are given that:
Length of aluminium sheet = 35 cm
Width of aluminium sheet = 11 dm
Thickness of aluminium sheet = 15 mm
Now we have to convert these dimensions into meters.
Conversions used:
1 cm = 0.01 m
1 dm = 0.1 m
1 mm = 0.001 m
Length of aluminium sheet = 35 cm = 35 × 0.01 = 0.35 m
Width of aluminium sheet = 11 dm = 11 × 0.1 = 1.1 m
Thickness of aluminium sheet = 15 mm = 15 × 0.001 = 0.015 m
<u>Part 2(b) :</u>
First we have to calculate the volume of aluminium sheet.
Volume of aluminum sheet (cuboid) = Length × Width × Thickness
Volume of aluminum sheet (cuboid) = 0.35 m × 1.1 m × 0.015 m
Volume of aluminum sheet (cuboid) = 0.005775 m³
Now we have to calculate the mass of aluminium sheet.
Thus, the volume and mass of slab is, 0.005775 m³ and 15.59 kg respectively.