Answer:
i(t) = (E/R)[1 - exp(-Rt/L)]
Explanation:
E−vR−vL=0
E− iR− Ldi/dt = 0
E− iR = Ldi/dt
Separating te variables,
dt/L = di/(E - iR)
Let x = E - iR, so dx = -Rdi and di = -dx/R substituting for x and di we have
dt/L = -dx/Rx
-Rdt/L = dx/x
interating both sides, we have
∫-Rdt/L = ∫dx/x
-Rt/L + C = ㏑x
x = exp(-Rt/L + C)
x = exp(-Rt/L)exp(C) A = exp(C) we have
x = Aexp(-Rt/L) Substituting x = E - iR we have
E - iR = Aexp(-Rt/L) when t = 0, i(0) = 0. So
E - i(0)R = Aexp(-R×0/L)
E - 0 = Aexp(0) = A × 1
E = A
So,
E - i(t)R = Eexp(-Rt/L)
i(t)R = E - Eexp(-Rt/L)
i(t)R = E(1 - exp(-Rt/L))
i(t) = (E/R)(1 - exp(-Rt/L))
Answer:
Explanation:
Image of distant object will be made at far point or at 52.5 so
object distance u = infinity
image distance v = - 52.5 cm
focal length required = f
Lens formula
1 / v - 1 / u = 1 / f
1 / - 52.5 - 0 = 1 / f
f = -52.5 cm
= -.525 m
Power P = 1 / f = - 1 / .525
= - 1.90
now , for eye with glass we shall find new near point .
v = ?
u = - 17.2 cm
f = - 52.5 cm
1 / v - 1 / u = 1 / f
1 / v + 1 / 17.2 = - 1 / 52.5
1 / v = - 1 / 17.2 - 1 / 52.5
= - .05813 - .019
= - .07713
u = - 12.96 cm
so new near point will be 12.96 cm
1 watt = 1 joule/second
1 horsepower = 746 watts = 746 joule/second
(150 horsepower) x (746 watt/HP) x (1 joule/sec / watt) x (10 sec)
= (150 x 746 x 1 x 10) joule = 1,119,000 joules .
if correct plz mark brainly
Fnet=(115+106)-186= 34 N
mass=Force/g= 186N/9.8m/s^2 = 18.98 kg
a=fnet/mass => 34N/18.98kg = 1.79 m/s^2
so A= 1.8m/s^2
Answer:
10.791 m/s
5.93505 m
Explanation:
m = Mass of ball
= Final velocity
= Initial velocity
= Final time
= Initial time
g = Acceleration due to gravity = 9.81 m/s²
From the momentum principle we have
Force
So,
The speed that the ball had just after it left the hand is 10.791 m/s
As the energy of the system is conserved
The maximum height above your hand reached by the ball is 5.93505 m
