Are we actually touching nothing but electrons in reality or can we actually feel things without the electron barier?
1 answer:
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The heat released by the water when it cools down by a temperature difference 
is
where
m=432 g is the mass of the water
is the specific heat capacity of water
is the decrease of temperature of the water
Plugging the numbers into the equation, we find
and this is the amount of heat released by the water.
where
m=432 g is the mass of the water
Plugging the numbers into the equation, we find
and this is the amount of heat released by the water.
The period of the oscillations.T = 1.2042s
Opposition is the process of any quantity or measure fluctuating repeatedly about its equilibrium value throughout time. This process is referred to as oscillation. Oscillation, a periodic fluctuation of a substance, can also be described as alternating between two values or rotating around a central value.
Typically, the mathematical formula for the moment of inertia is
T = 2 π √(I / mgd)
Therefore, a moment of inertia
I = 9.00×10-3 + md^2 ;
I=9.00*10^{-3}+ 0.5 * 0.3^2
I=0.054
T=2
T=1.2042s
The period of the oscillations.T = 1.2042s
Read more about the period of the oscillations. brainly.com/question/14394641
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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