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
In order to find the our own velocity with respect to land,we need to apply the theory of relative velocity.
Now consider the velocity of the ship traveling towards the north with respect to land as A.Consider our own velocity headed northwards as B.
The relative velocity is the velocity that the body A would appear to an observer on the body B and vice versa.
In this case the relative velocity would be arrived by summing up our velocity with the velocity of the ship as the object (I) is travelling in the ship.
Relative velocity = Velocity of Body A+ Velocity of Body B.
Velocity of the ship traveling towards the north with respect to land(A)= 13.0m/s. (Given)
Our own velocity headed northwards(B)= 2.8 m/s.
Relative velocity = Velocity of Body A+ Velocity of Body B.
Relative velocity= 13.0 + 2.8 = 15.8m/s.
Thus our own velocity with respect to the land is 15.8 m/s.
Answer:
serie Ceq=0.678 10⁻⁶ F and the charge Q = 9.49 10⁻⁶ C
Explanation:
Let's calculate all capacity values
a) The equivalent capacitance of series capacitors
1 / Ceq = 1 / C1 + 1 / C2 + 1 / C3 + 1 / C4 + 1 / C5
1 / Ceq = 1 / 1.5 + 1 / 3.3 + 1 / 5.5 + 1 / 6.2 + 1 / 6.2
1 / Ceq = 1 / 1.5 + 1 / 3.3 + 1 / 5.5 + 2 / 6.2
1 / Ceq = 0.666 + 0.3030 +0.1818 +0.3225
1 / Ceq = 1,147
Ceq = 0.678 10⁻⁶ F
b) Let's calculate the total system load
Dv = Q / Ceq
Q = DV Ceq
Q = 14 0.678 10⁻⁶
Q = 9.49 10⁻⁶ C
In a series system the load is constant in all capacitors, therefore, the load in capacitor 5.5 is Q = 9.49 10⁻⁶ C
c) The potential difference
ΔV = Q / C5
ΔV = 9.49 10⁻⁶ / 5.5 10⁻⁶
ΔV = 1,725 V
d) The energy stores is
U = ½ C V²
U = ½ 0.678 10-6 14²
U = 66.4 10⁻⁶ J
e) Parallel system
Ceq = C1 + C2 + C3 + C4 + C5
Ceq = (1.5 +3.3 +5.5 +6.2 +6.2) 10⁻⁶
Ceq = 22.7 10⁻⁶ F
f) In the parallel system the voltage is maintained
Q5 = C5 V
Q5 = 5.5 10⁻⁶ 14
Q5 = 77 10⁻⁶ C
g) The voltage is constant V5 = 14 V
h) Energy stores
U = ½ C V²
U = ½ 22.7 10-6 14²
U = 2.2 10⁻³ J
Answer:
As the wavelength of an electromagnetic wave _decrease__ the frequency of the wave _increase_______.
Explanation:
What is the relationship between frequency and wavelength?
Wavelength and frequency of light are closely related. The higher the frequency, the shorter the wavelength. Because all light waves move through a vacuum at the same speed, the number of wave crests passing by a given point in one second depends on the wavelength.
That number, also known as the frequency, will be larger for a short-wavelength wave than for a long-wavelength wave. The equation that relates wavelength and frequency is:
V= fλ
where v= velocity
f= frequency
λ = wavelength
⇒ f = v/λ
also f ∝ 1/λ
For electromagnetic radiation, the speed is equal to the speed of light, c, and the equation becomes:
C= fλ
where c= Speed of light
f= frequency
λ = wavelength
⇒ f = v/λ
also f ∝ 1/λ
Answer:
3.485e+6 inches. hope this helps
