Answer:
Explanation:
Inductance L = 1.4 x 10⁻³ H
Capacitance C = 1 x 10⁻⁶ F
a )
current I = 14 .0 t
dI / dt = 14
voltage across inductor
= L dI / dt
= 1.4 x 10⁻³ x 14
= 19.6 x 10⁻³ V
= 19.6 mV
It does not depend upon time because it is constant at 19.6 mV.
b )
Voltage across capacitor
V = ∫ dq / C
= 1 / C ∫ I dt
= 1 / C ∫ 14 t dt
1 / C x 14 t² / 2
= 7 t² / C
= 7 t² / 1 x 10⁻⁶
c ) Let after time t energy stored in capacitor becomes equal the energy stored in capacitance
energy stored in inductor
= 1/2 L I²
energy stored in capacitor
= 1/2 CV²
After time t
1/2 L I² = 1/2 CV²
L I² = CV²
L x ( 14 t )² = C x ( 7 t² / C )²
L x 196 t² = 49 t⁴ / C
t² = CL x 196 / 49
t = 74.8 μ s
After 74.8 μ s energy stored in capacitor exceeds that of inductor.
The star with apparent magnitude 2 is more brighter than 7.
To find the answer, we have to know about apparent magnitude.
<h3>What is apparent magnitude?</h3>
- 100 times as luminous as a star with an apparent brightness of 7 is a star with a magnitude of 2.
- The apparent magnitude of bigger stars is always smaller.
- The brightest star in the night sky is Sirius.
- The brightness of a star or other celestial object perceived from Earth is measured in apparent magnitude (m).
- The apparent magnitude of an object is determined by its inherent luminosity, its distance from Earth, and any light extinction brought on by interstellar dust in the path of the observer's line of sight.
Thus, we can conclude that, the star with apparent magnitude 2 is more brighter than 7.
Learn more about the apparent magnitude here:
brainly.com/question/350008
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Answer:
41.4* 10^4 N.m^2/C
Explanation:
given:
E= 4.6 * 10^4 N/C
electric field is 4.6 * 10^4 N/C and square sheet is perpendicular to electric field so, area of vector is parallel to electric field
then electric flux = ∫ E*n dA
= ∫ 4.6 * 10^4 * 3*3
= 41.4* 10^4 N.m^2/C
Answer:
A measure of the ability of a material to transfer heat.
Explanation:
Please mark me as brainliest please
Answer:
Explanation:
Electric field in a given region is given by equation
as we know the relation between electric field and potential difference is given as
so here we have
here we know that
and 
so we will have
so we will have
