Answer:
205 V
V
= 2.05 V
Explanation:
L = Inductance in Henries, (H) = 0.500 H
resistor is of 93 Ω so R = 93 Ω
The voltage across the inductor is

w = 500 rad/s
IwL = 11.0 V
Current:
I = 11.0 V / wL
= 11.0 V / 500 rad/s (0.500 H)
= 11.0 / 250
I = 0.044 A
Now
V
= IR
= (0.044 A) (93 Ω)
V
= 4.092 V
Deriving formula for voltage across the resistor
The derivative of sin is cos
V
= V
cos (wt)
Putting V
= 4.092 V and w = 500 rad/s
V
= V
cos (wt)
= (4.092 V) (cos(500 rad/s )t)
So the voltage across the resistor at 2.09 x 10-3 s is which means
t = 2.09 x 10⁻³
V
= (4.092 V) (cos (500 rads/s)(2.09 x 10⁻³s))
= (4.092 V) (cos (500 rads/s)(0.00209))
= (4.092 V) (cos(1.045))
= (4.092 V)(0.501902)
= 2.053783
V
= 2.05 V
So n=c/v, n= index, c=speed of light and v= speed of light in diamond. 2.42=c/v so v=c/2.42, c≈<span>3x108 m/sec</span><span> so v=</span><span>1.24x108 m/sec</span>.
<span>Hope this helps.</span>
Water equal to iron is greater than cooper
Supposing that the spring is un stretched when θ = 0, and has a toughness of k = 60 N/m.It seems that the spring has a roller support on the left end. This would make the spring force direction always to the left
Sum moments about the pivot to zero.
10.0(9.81)[(2sinθ)/2] + 50 - 60(2sinθ)[2cosθ] = 0 98.1sinθ + 50 - (120)2sinθcosθ = 0 98.1sinθ + 50 - (120)sin(2θ) = 0
by iterative answer we discover that
θ ≈ 0.465 radians
θ ≈ 26.6º
Answer:
The horizontal range will be 
Explanation:
We have given initial speed of the shell u = 
Angle of projection = 51°
Acceleration due to gravity 
We have to find maximum range
Horizontal range in projectile motion is given by

So the horizontal range will be 