initially coin is at rest and then it drop for total time t = 1.5 s
so here the speed of the coin at which it will hit the floor is to be find
here we know that
a = 9.8 m/s^2
t = 1.5 s
now from above equation
so it will hit the floor with speed 14.7 m/s
We use the formula,
Here, h is the variable represents the height of the flare in feet when it returns to the sea so, h = 0 and u is the initial velocity of the flare, in feet per second and its value of 192 ft/sec.
Substituting these values in above equation, we get
.
Here, t= 0 neglect because it is the time when the flare is launched.
Thus, flare return to the sea in 12 s.
Answer:
Explanation:
Let i be the angle of incidence and r be the angle of refraction .
From the figure
Tan ( 90 - i ) = 2.5 / 8
cot i = 2.5 / 8
Tan i = 8 / 2.5 = 3.2
i = 72.65°
From snell's law
sini / sin r = refractive index
sin 72.65 / sinr = 1.333
sin r = .9545 / 1.333
= .72
r = 46⁰
From the figure
Tan r = d / 4
Tan 46 = d /4
d = 4 x Tan 46
= 4 x 1.0355
=4.14 m .
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