Answer:
(a) T = 2987.6 k
(b) T = 19986.2 k
Explanation:
The temperature of a star in terms of peak wavelength can be given by Wein's Displacement Law, which is as follows:
where,
T = Radiated surface temperature
= peak wavelength
(a)
here,
= 970 nm = 9.7 x 10⁻⁷ m
Therefore,
<u>T = 2987.6 k</u>
(b)
here,
= 145 nm = 1.45 x 10⁻⁷ m
Therefore,
<u>T = 19986.2 k</u>
Third
every action has an opposite and equal reaction
Answer:
<em>a. 4.21 moles</em>
<em>b. 478.6 m/s</em>
<em>c. 1.5 times the root mean square velocity of the nitrogen gas outside the tank</em>
Explanation:
Volume of container = 100.0 L
Temperature = 293 K
pressure = 1 atm = 1.01325 bar
number of moles n = ?
using the gas equation PV = nRT
n = PV/RT
R = 0.08206 L-atm-
Therefore,
n = (1.01325 x 100)/(0.08206 x 293)
n = 101.325/24.04 = <em>4.21 moles</em>
The equation for root mean square velocity is
Vrms =
R = 8.314 J/mol-K
where M is the molar mass of oxygen gas = 31.9 g/mol = 0.0319 kg/mol
Vrms = = <em>478.6 m/s</em>
<em>For Nitrogen in thermal equilibrium with the oxygen, the root mean square velocity of the nitrogen will be proportional to the root mean square velocity of the oxygen by the relationship</em>
=
where
Voxy = root mean square velocity of oxygen = 478.6 m/s
Vnit = root mean square velocity of nitrogen = ?
Moxy = Molar mass of oxygen = 31.9 g/mol
Mnit = Molar mass of nitrogen = 14.00 g/mol
=
= 0.66
Vnit = 0.66 x 478.6 = <em>315.876 m/s</em>
<em>the root mean square velocity of the oxygen gas is </em>
<em>478.6/315.876 = 1.5 times the root mean square velocity of the nitrogen gas outside the tank</em>
Answer:
Power factor = 0.87 (Approx)
Explanation:
Given:
Load = 1 Kw = 1000 watt
Current (I) = 5 A
Supply (V) = 230 V
Find:
Power factor.
Computation:
Power factor = watts / (V)(I)
Power factor = 1,000 / (230)(5)
Power factor = 1,000 / (1,150)
Power factor = 0.8695
Power factor = 0.87 (Approx)
Answer:
Explanation:
The electric flux is defined as the multiple of electric field and the area that the electric field passes through, such that
When calculating the electric flux, the angle between the directions of electric field and the area becomes important, especially if the angle is changing with time.
The above formula can be rewritten as follows
where θ is the angle between the electric field and the area of the loop. Note that, the direction of the area of the loop is perpendicular to the plane of the loop.
If the loop is rotating with constant angular velocity ω, then the angle can be written as follows
At t = 0, cos(0) = 1 and the electric flux through the loop is at its maximum value.
Therefore the electric flux can be written as a function of time