Answer:
13.51 nm
Explanation:
To solve this problem, we are going to use angle approximation that sin θ ≈ tan θ ≈ θ where our θ is in radians
y/L=tan θ ≈ θ
and ∆θ ≈∆y/L
Where ∆y= wavelength distance= 2.92 mm =0.00292m
L=screen distance= 2.40 m
=0.00292m/2.40m
=0.001217 rad
The grating spacing is d = (90000 lines/m)^−1
=1.11 × 10−5 m.
the small-angle
approx. Using difraction formula with m = 1 gives:
mλ = d sin θ ≈ dθ →
∆λ ≈ d∆θ = =1.11 × 10^-5 m×0.001217 rad
=0.000000001351m
= 13.51 nm
Answer:
At the high temperatures of the inner solar nebula, the small proto-planets were too hot to hold the volatile gases that dominated the solar nebula. These proto-planets were Earth, Mars, Venus, and Mercury.
Explanation:
The materials that accreted into the early Earth were probably added piecemeal, without and particular order. The early earth was very hot from gravitational compression, impacts and radioactive decay; the earth was partially molted. The denser metallic liquids sank to the center of the Earth and less denser silicate liquids rose to the top. In this way the Earth differentiated very quickly into a metallic, mostly iron core and a rocky silicate mantle.
So, there should be two forces acting on the refrigerator: the applied force and the friction force.
The question mentioned that the friction force was set to zero, so the only effective force now would be the applied force.
We have an applied force of 400 N to the right, this means that:
<span>The magnitude of the net force is 400, directed to the right.</span>
In an alpha decay, an atom emits an alpha particle. An alpha particle consists of 2 protons and 2 neutrons: this means that during this kind of decay, the original atom loses 2 protons and 2 neutrons from its nucleus.
This also means that the atomic number Z of the element (the atomic number is the number of protons in the nucleus) decreases by 2 units in the process, while the mass number A (the mass number is the sum of the number of protons and neutrons) decreases by 4 units.
Classius claperyon equation
In (P2/ P2) = ΔHvap/R) × (1/T2-1/T1)
T2 occurs at normal boiling when vapor pressure P2 = 1 atm.
P1 = 55.1 mmHg, P2 = 1 atm = 760mmHg
T1 = 35°c = 308.15k, T2 =
ΔHvap = 32.1kJ/mol = 32100 J/mol
In (760/55.1) = (-32100/ 8.314) × ( 1/T2 - 1/308.15)
The normal boiling point T2 = 390k = 117°c
