A quantity of gas has a volume of 0.20 cubic meter and an absolute temperature of 333 degrees kelvin. When the temperature of th
1 answer:
You might be interested in
As show the figure, peak of each blackbody curves shifts towards higher frequency. Then energy emitted by the black body is directly proportional to the frequency incident radiation.
Hence peak is shifted to higher energies.
Therefore the correct option is B: Toward higher energies.
Answer:
162 km
Explanation:
A diagram can be helpful.
Using the law of cosines, we can find the magnitude of the distance (c) to satisfy ...
c^2 = a^2 +b^2 -2ab·cos(C)
where C is the internal angle of the triangle of vectors and resultant. Its value is ...
180° -39.8° -59.9° = 80.3°
Filling in a=76 and b=156, we get ...
c^2 = 76^2 +156^2 -2·76·156·cos(80.3°) ≈ 26116.78
c ≈ √26116.78 ≈ 161.607
The magnitude of the total displacement is about 162 km.
_____
Please note that in the attached diagram North is to the right and East is up. That alteration of directions does not change the angles or the magnitude of the result.
Answer:
a) the required time is 0.6283 μs
b) the inductor current is 0.5 mA
Explanation:
Given the data in the question;
The capacitor voltage has its maximum value of 25 V at t = 0
i.e V = V₀ = 25 V
we determine the angular velocity;
ω = 1 / √( LC )
ω = 1 / √( ( 20 × 10⁻³ H ) × ( 8.0 × 10⁻¹² F) )
ω = 1 / √( 1.6 × 10⁻¹³ )
ω = 1 / 0.0000004
ω = 2.5 × 10⁶ s⁻¹
a) How much time does it take until the capacitor is fully discharged for the first time?
V = V₀sin( ωt )
we substitute
25V = 25V × sin( 2.5 × 10⁶ s⁻¹ × t )
25V = 25V × sin( 2.5 × 10⁶ s⁻¹ × t )
divide both sides by 25 V
sin( 2.5 × 10⁶ × t ) = 1
( 2.5 × 10⁶ × t ) = π/2
t = 1.570796 / (2.5 × 10⁶)
t = 0.6283 × 10⁻⁶ s
t = 0.6283 μs
Therefore, the required time is 0.6283 μs
b) What is the inductor current at that time?
(t) = V₀√(C/L) sin(ωt)
{ sin(ωt) = 1 )
(t) = V₀√(C/L)
we substitute
(t) = 25V × √( ( 8.0 × 10⁻¹² F ) / ( 20 × 10⁻³ H ) )
(t) = 25 × 0.00002
(t) = 0.0005 A
(t) = 0.5 mA
Therefore, the inductor current is 0.5 mA