Answer:
<em>The depth will be equal to</em> <em>6141.96 m</em>
<em></em>
Explanation:
pressure on the submarine
= 62 MPa = 62 x 10^6 Pa
we also know that
= ρgh
where
ρ is the density of sea water = 1029 kg/m^3
g is acceleration due to gravity = 9.81 m/s^2
h is the depth below the water that this pressure acts
substituting values, we have
= 1029 x 9.81 x h = 10094.49h
The gauge pressure within the submarine
= 101 kPa = 101000 Pa
this gauge pressure is balanced by the atmospheric pressure (proportional to 101325 Pa) that acts on the surface of the sea, so it cancels out.
Equating the pressure
, we have
62 x 10^6 = 10094.49h
depth h = <em>6141.96 m</em>
Answer:
1050 kg
Explanation:
The formula for kinetic energy is:
KE (kinetic energy) = 1/2 × m × v² where <em>m</em> is the <em>mass in kg </em>and <em>v</em> is the velocity or <em>speed</em> of the object <em>in m/s</em>.
We can now substitute the values we know into this equation.
KE = 472 500 J and v = 30 m/s:
472 500 = 1/2 × m × 30²
Next, we can rearrange the equation to make m the subject and solve for m:
m = 472 500 ÷ (1/2 × 30²)
m = 472 500 ÷ 450
m = 1050 kg
Hope this helps!
Answer:
Index of expansion: 4.93
Δu = -340.8 kJ/kg
q = 232.2 kJ/kg
Explanation:
The index of expansion is the relationship of pressures:
pi/pf
The ideal gas equation:
p1*v1/T1 = p2*v2/T2
p2 = p1*v1*T2/(T2*v2)
500 C = 773 K
20 C = 293 K
p2 = 35*0.1*773/(293*1.3) = 7.1 bar
The index of expansion then is 35/7.1 = 4.93
The variation of specific internal energy is:
Δu = Cv * Δt
Δu = 0.71 * (20 - 500) = -340.8 kJ/kg
The first law of thermodynamics
q = l + Δu
The work will be the expansion work
l = p2*v2 - p1*v1
35 bar = 3500000 Pa
7.1 bar = 710000 Pa
q = p2*v2 - p1*v1 + Δu
q = 710000*1.3 - 3500000*0.1 - 340800 = 232200 J/kg = 232.2 kJ/kg
Answer:
i)20369 photons
ii) 40 ps
Explanation:
Momentum of one Sodium atom:

In other to stop it, it must absorb the same momentum in photons:

Now, for the minimun time, we use the speed of light and the wavelength. For the n photons:
