b) ориентируется вдоль новой оси поляризации.

26.38)

(b)
Let T be the time constant

v(t) = 12*e^(-t/T)

3/12 = e^(-4/T)

ln(0.25) = -4/T = -1.3863

T = 4/1.3863 = 2.885 sec

(a)
T = 2.885 = RC = 3.4M C
C = 2.885 / 3.4M = 0.849 uF

26.40) The charge = Q = CV
C = 12.4 uF
T = RC = (0.895)(12.4) = 11.0 sec
To find the charge Q(t), we need to find V(t) using the basic equation:
V(t) = (60 volts)(1 - 1/e ^ t/T)
For ex, at t = 5.0 sec:
- t/T = - 5.0/11.0 = - 0.4545
e to the -0.4545 = 1.72,
V = (60.0)(1 - 1/1.72) = 60(0.42) = 25.2 volts
Q = CV = (12.4 uF)(25.2 V) = 312 uCoulombs

27.55) Магнитное поле должно проявить силу на монете, равной ее весу + нисходящая сила области E

так q*v*B = m*g + E*q

или B = (m*g + E*q) / (q*v)

Я предполагаю, что обвинение - 2400 микро C

так B = (0.0054*9.8 + 28.0*2400x10^-6) / (2400*x10^-6*13.0) = 3.85T

27.64) F (mag) = q (v cross B)
v = v (k)
B = Bx i + By j + Bz k
F = F_0 (3i+4j)
----------------------
F_0 (3i+4j) = q{v k} cross {Bx i + By j + Bz k}
F_0 (3i+4j) = q[v Bx (kXi} +v By (kXj} + v Bz (kXk)}
F_0 (3i+4j) = q[v Bx (j} +v By (-i} + v Bz (0)}
comparing (i) (j) components
Bx = 4 F_0/qv
By = - 3 F_0/qv
=================
given |B| = √[Bx^2 + By^2 + Bz^2] = 6F_0/qv
[Bx^2 + By^2 + Bz^2] = [6F_0/qv]^2
[4F_0/qv]^2 + [-3F_0/qv]^2 + Bz^2 = [6F_0/qv]^2
Bz^2 = [6F_0/qv]^2 - [4F_0/qv]^2 - [3F_0/qv]^2
Bz^2 = [F_0/qv]^2 *{36 - 16 - 9}
Bz^2 = 11 [F_0/qv]^2
Bz = 3.32 [F_0/qv]

27.73) Магнитное поле направлено из страницы. Это проявляет силу на частях провода, которые перпендикулярны магнитному полю, то есть, части, которые находятся в кругу, которые включают три части. Две горизонтальных части провода в кругу имеют ту же самую длину и несут ток в противоположных направлениях, таким образом, силы, проявленные на них магнитным полем, уравновешиваются. Поэтому нас оставляют с вертикальной частью провода выяснить. Для этого мы используем правое правило: укажите пальцами в направлении магнитного поля и большим пальцем в направлении тока (в этом случае конец страницы). Пальмовые лица налево от страницы, таким образом, сила налево.

B) F=1.80 N

F = ILB

F = (6.00 А) (.450 м) (.666 T)

F = 1.80 N

28.60) По правому правилу вектор был бы восходящим, и оба из магнитных полей от двух проводов добавят.

Область из-за длинного, прямо, проводник с током - B=mu_0I/2 (пи) r, где r - расстояние от провода до пункта P.

Расстояние там - r=sqrt (a^2+x^2) от обоих проводников. При помощи суперположения и добавляющих векторов мы можем найти область Б.

Таким образом Btotal=B1+B2

B1=Mu_0*I / (2*pi*sqrt (a^2+x^2))

B2=Mu_0*I / (2*pi*sqrt (a^2+x^2))

B1+B2 = 2*Mu_0*I / (2*pi*sqrt (a^2+x^2))

B1+B2 = Mu_0*I / (pi*sqrt (a^2+x^2))

14M) Формула для тока в серийном кругообороте индуктивности сопротивления

i = (v/r) (1 - e^ ((-r/l) t))

где

i = ток в амперах

v = напряжение в В = 12

r = сопротивление в Омах = 10

e = постоянные 2.71828...

l = индуктивность в henries = 2

t = время в секундах (мы должны решить это),

Заключительный (максимальный) ток = v/r = 12/10 = 1.2a

В этой проблеме мы нуждаемся в 50% заключительной (максимальной) ценности = 0.5 * 1.2 = 0.6a

Возвращение к формуле

i = (v/r) (1 - e^ ((-r/l) t))

0.6 = (12/10) (1 - e^ ((-10/2) t))

0.6 = 1.2 (1 - e^(-5t))

Мы должны изолировать постоянный термин на одной стороне равного знака и условий числа с другой стороны.

Разделите обе стороны равного знака 1.2

0.6/1.2 = 1 - e^(-5t)

Вычтите 1 из обеих сторон

0.5 - 1 =-e^ (-5t)

- 0.5 =-e^ (-5t)

Общие отрицательные знаки полярности уравновешиваются

0.5 = e^(-5t)

Возьмите естественную регистрацию обеих сторон

ln (0.5) = ln (e^(-5t))

- 0.693147181 =-5t

Разделите обе стороны на-5

0.138629436 = t

Переместить

t = 0.138629436 секунды

Упростить

t = 138.629436 миллисекунд

59) http://www.electronics-tutorials.ws/accircuits/series-circuit.html

65) (a) inductive reactance,

XL = 2 f L

XL = 2 (40)(250 x 10 - 3)

XL = 0.785

(b) capactive reactance,

XC = 1/ 2 f C

XC = 1/[2 (40)(2.0 x 10 - 6)]

XC = 1989

(d) impedance,

Z = SQRT[ R2 + (XL - XC)2]

Z = SQRT[ (150)2 + (1989 - 0.785)2]

Z = 1994

(d) maximum current and

Z = V/I

I = V/Z

I = 210 V/ 1994

I = 0.105 A

(c) phase angle.

= tan - 1 [(XL - XC)/R]

= tan - 1 [(1989 - 0.785)/150]

= tan - 1 [13.25]

= 85.7o

P=I*Ucosfi

Pinput=I*r^2

71)

71) Rc=1/wCmax; Rl=wLmin;

U=IRmax

I=V/Rl min Rl=0

I=V/Rcmax Rc=0

67)

49)

I=power/area

1340 =P/4π(1.496×1011)2

P= 3.78×1026W

46) Так, c = 3x10^8 м/с

ско звука, v = 343 м/с.

t = d/v, таким образом,

t (свет) = (100*1000 м) / (3x10^8 м/с) = 3.333x10-^4 s

и t (звук) = (3 м / (343 м/с)) = 8.75x10^-3 с

t (свет) - t (звук) = 3.333x10^-4 - 8.75x10^-3 = - 8.4x10^-3 с

радио-слушатели слышат его 8.4x10^-3

несколькими секундами ранее

22)

b) S=(1/m0)ExB(cros pr); m0=4pi*10^-7

19) S= P/ (4*pi*distance^2)
Power = 250x10^3

Distance = 5 miles * 1609 = 22526m

S = 250x10^3 / (4 * pi * 22526^2)

is the answer in W/m^2

8) n(a)sinө’=n(o)sin 20

n(0)sin20=n(w)sinө”

16)

z

N1sin0’=n2sin0”

1sin30=1.5sin0”

0”=19,5

Cos0”=2m/h

H=2.12cm

Α=0’-0”=30-19.5=10.5

sin α=d/h

d=2.21cm*sin10,5=0.388cm

21)

Sin28*1=sin0”*1.33

Sin0”=30,67

Tan30,67=depth/d

Depth=tan30.67*3=1.13m

23)

a)1*sin75=1,458sin0”; sin0”=41,55

b)0’”+ beta=90; 0”+alfa=90; alfa+beta+60=180

60-0”-0”’=0; 0’”=60-41,5=18,5;

c) 1,458sin18,5=1sin0””; 0””=27,6;

49)

Если угол между плоскостью поляризации и осью поляризации - θ, свет, который проходит:

a) уменьшен в интенсивности фактором (cosθ)^2; и

b) ориентируется вдоль новой оси поляризации.

Таким образом, после n листы polarizer, каждый ориентируемый под углом θ от предыдущего, интенсивность была уменьшена фактором (cosθ)^ (2n).

Для заключительной поляризации, чтобы простереться вдоль Θ = 45 ° = nθ,

θ = 45 °/n

Таким образом, мы должны потребовать что:

(1 - 10%) = 0.9 ≤ (потому что (45 °/n)) ^ (2n)

На это можно легко найти ответ на крупноформатной таблице: Оказывается, что для n = 6, интенсивность уменьшена фактором 0.902, тогда как для n =5, это 0.883.

a) n = 6

b) θ = 45 °/6 = θ = 7.5 ° = π/24

№58

№42

№41 a) 50% of the original intensity is lost at the polariser.

Fractional pass at the analyser = cos30 ^2

%pass with respect to the original is 50. cos30 ^2 = 3.4%

50% is absorbed by the polariser, 3/4% is transmitted, leaving 50-3/4% absorbed by the analyser.

47)

11.1)

11.2)

1. The problem statement, all variables and given/known data
In a lab frame of reference, an observer finds Newtons second law is valid in the form of Fnet=ma. Show that Newtons second law is not valid in a reference frame moving past the laboratory frame of problem 1 with a constant acceleration a1. assume that mass is an invariant quantity and is constant in time.
2. Relevant equations
f=ma
3. The attempt at a solution
i dont get how to show this, i get how to show that the velocities and displacement would be different but not how to relate that to the force

11.3) southwest = 45deg off south

Vs = (100 sin 45) + 600 = 670.71
Vw = 100 sin 45 = 70.71

now find the Magnitude.

V^2 = Vs^2 + Vw^2
V^2 = (670.71)^2 + (70.71)^2

V = 674.42

direction:
Sin A = 70/674.42

A = S 5.958 deg W bearing based from south

(b)
with the accumulated speed of 674.42 on the resulting direction

D = 674.42 x 10/60 = 1124.03 KM

Distance offset from true course at the time is

x = 1124.03 km sin 5.958 deg

X = 116.67 km off course ******ANSWER

11.4)

I've been trying to figure out the following problems for hours, and hours now, and still, no prevail. If anyone can help me with the step-by-steps, I'd greatly appreciated. Thumbs up to anyone whose answer helps. Thanks, in advance. =]

Two cars approach a street corner at right angles to each other. Car 1 travels at a speed relative to Earth v1E = 29 km/h, and car 2 at v2E = 53 km/h.

What is the relative velocity of car 1 as seen by car 2?
___________ km/h __________ ° (west of north)
What is the velocity of car 2 relative to car 1?
_________ km/h ___________° (south of east)

A rescue plane wants to drop supplies to isolated mountain climbers on a rocky ridge 235 m below. Assume the plane is traveling horizontally with a speed of 235 km/h (65.3 m/s).

(a) How far in advance of the recipients (horizontal distance) must the goods be dropped?
_________________m

(b) Suppose, instead, that the plane releases the supplies a horizontal distance of 425 m in advance of the mountain climbers. What vertical velocity (up or down) should the supplies be given so that they arrive precisely at the climbers' position? (Assume up is positive.)
_________________ m/s
(c) With what speed do the supplies land in the latter case?
__________________ m/s

For the first part it's just a matter of how you look at it. At first it may appear that the velocity of car 1 as seen by car 2 is simply 29km/h. However you have to take into account that car 2 is moving.
Advice: draw a picture... seriously... do it lol
you should see that it is a right triangle. The velocity of car 1 is 29km/h BUT, because car 2 is approaching car 1 it appears that car 1 is also traveling toward car 2 at a speed of (-)53km/h.
So use pythagorean theorem:
√(29² + 53²) = 60.4 m/s "west of north" (supposedly)
and the same can be said from car 2's perspective:
√(53² + 29²) = 60.4 m/s "south of east"
which makes sense. Because 2 objects should generally always be approaching each other at the same speed.

For the rescue plane problem. The first step is to find out how much time it takes for anything dropped from 235m to hit the ground.
acceleration due to gravity = -9.8m/s²
initial velocity (Vi) = 0m/s
final velocity (Vf) =??m/s
time =??s
equation:
acceleration = (Vf - Vi) / time
-9.8m/s² = (Vf - 0m/s) / time
Vf = -9.8m/s² * time

then we need the equation:
(AVERAGE velocity) = distance / time
note that it is the average velocity. So if velocity changes (which it does) then you can average the values (by adding them and dividing by 2). So:
(Vi + Vf) / 2 = distance / time
use the value of Vf that we got above:
(0m/s + (-9.8m/s² * time)) / 2 = -235m / time
-9.8m/s² * time / 2 = -235m / time
-4.9m/s² * (time)² = -235m
(time)² = 48.0 (s²)
time = 6.93 seconds

So... when the "goods" are dropped they will be traveling at the same speed as the plane (65.3m/s). How far does it go in 6.93 seconds?
velocity = distance / time
65.3m/s = distance / 6.93s
distance = 453m
must be dropped 453 meters in advance.

Now we must solve it backwards for part B. It didn't say the velocity of the plane changed.
velocity = distance / time
65.3m/s = 425m / time
time = 6.5 seconds

Now we need to know the initial velocity to give the "goods" so that it will fall 235 meters in 6.5 seconds.
acceleration = (Vf - Vi) / time
-9.8m/s² = (Vf - Vi) / 6.5s
Vf = -63.7m/s + Vi

then we again switch to the equation:
(AVERAGE velocity) = distance / time
(Vi + Vf) / 2 = distance / time
plug in the value of Vf that we got above:
(Vi + (-63.7m/s + Vi)) / 2 = -235m / 6.5s
(2Vi - 63.7m/s) / 2 = -235m / 6.5s
Vi - 31.85m/s = -235m / 6.5s
Vi = -4.30 m/s <--answer: because it is negative you can say. 4.30m/s downward.

Then to find the final velocity:
acceleration = (Vf - Vi) / time
-9.8m/s² = (Vf - (-4.30m/s)) / 6.5s
Vf = -68m/s
OR 68m/s down

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