.
<h3>Explanation</h3>
The Stefan-Boltzmann Law gives the energy radiation <em>per unit area</em> of a black body:

where,
the total power emitted,
the surface area of the body,
the Stefan-Boltzmann Constant, and
the temperature of the body in degrees Kelvins.
.
.
.
Keep as many significant figures in
as possible. The error will be large when
is raised to the power of four. Also, the real value will be much smaller than
since the emittance of a human body is much smaller than assumed.
Answer:
KE = 2.535 x 10⁷ Joules
Explanation:
given,
angular speed of the fly wheel = 940 rad/s
mass of the cylinder = 630 Kg
radius = 1.35 m
KE of flywheel = ?
moment of inertia of the cylinder

=
= 574 kg m²
kinetic energy of the fly wheel

KE = 2.535 x 10⁷ Joules
the kinetic energy of the flywheel is equal to KE = 2.535 x 10⁷ Joules
To solve this task we have to make a proportion, but firstly we have to set up all the main points : so, the distance is s=r(B), that has its <span>r=radius,B=angle in rad
velocity v=ds/dt= w(r)
Do not forget about </span> w = angular speed in rad/s and

Now we can go to proportion




SOLVING FOR A :



or something about <span>10 mph --- SOLVING FOR B.
</span>I'm sure it helps!
The helium may be treated as an ideal gas, so that
(p*V)/T =constant
where
p = pressure
V = volume
T = temperature.
Note that
7.5006 x 10⁻³ mm Hg = 1 Pa
1 L = 10⁻³ m³
Given:
At ground level,
p₁ = 752 mm Hg
= (752 mm Hg)/(7.5006 x 10⁻³ mm Hg/Pa)
= 1.0026 x 10⁵ Pa
V₁ = 9.47 x 10⁴ L = (9.47 x 10⁴ L)*(10⁻³ m³/L)
= 94.7 m³
T₁ = 27.8 °C = 27.8 + 273 K
= 300.8 K
At 36 km height,
P₂ = 73 mm Hg = 73/7.5006 x 10⁻³ Pa
= 9.7326 x 10³ Pa
T₂ = 235 K
If the volume at 36 km height is V₂, then
V₂ = (T₂/p₂)*(p₁/T₁)*V₁
= (235/9.7326 x 10³)*(1.0026 x 10⁵/300.8)*94.7
= 762.15 m³
Answer: 762.2 m³
Satellite. think of the moon.
Hope this helps!
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