The Hound The Hound of the
Baskervilles
How do
outside
the for
adjes
5
Com
burs
I Ast
2 Fire
3 TL
4 TL
5 F
6
TOUT
A dense, white fog hung over the moor and it was drifting
slowly in our direction Holmes was watching it
"It's moving towards us, Watson," he said impatiently
'Is that serious?" I asked
"Very serious, indeed. It's the one thing that could ruin my
plans. Our success and even Sir Henry's life may depend on his
coming out of the house before the fog is over the path in half
an hour we won't be able to see our hands in front of us."
Shall we move to higher ground?"
"Yes, I think it would be best," replied Holmes.
So we moved back until we were about half a mile from the house.
Suddenly, the sound of quick steps broke the silence of the moor. Through
the fog came Sir Henry. He walked by quite close to us, but he didn't see
us. Then, we heard another sound coming from the white bank of fog.
"Look out!" cried Holmes. "It's coming!"
We stared into the fog, uncertain what horror was about to break
from the heart of it. Then, a dreadful shape sprung out from the
shadows. It was an enormous coal-black hound. Fire burst from its
open mouth and its eyes were burning in the darkness. With long bounds
the huge creature was leaping down the track, following our friend. Far
away on the path we saw Sir Henry looking back, his face white in the
moonlight, his hands raised in horror, staring helplessly at the thing
which was chasing him.кратко о чём.​

Дано:

d = 80 см = 0,8 м

h = 16 см = 0,16 м

|Δd| = 40 см = 0,4 м

D = -2,5 дптр

H'/H - ?

Оптическая сила отрицательна, значит линза - рассеивающая. Следовательно, изображение - мнимое, и расстояние от него до линзы f будет тоже отрицательным. Найдём его по формуле тонкой линзы:

1/f + 1/d = 1/F = D

1/f = D - 1/d = dD/d - 1/d = (dD - 1)/d

f = d/(dD - 1) = 0,8/(0,8*-2,5 - 1) = 0,8/(-3) = -8/30

Линейное увеличение линзы равно:

Г = H/h = |f/d| => H = h*|f/d| = 0,16*|-8/30 : 8/10| = 16/100 * 8/30 * 10/8 = 16/300 = 8/150 = 4/75

При приближении предмета на Δd:

1/f' + 1/d' = D

d' = d + |Δd| => 1/f' + 1/(d + |Δd|) = D

1/f' = D - 1/(d + |Δd|) = (d + |Δd|)D/(d + |Δd|) - 1/(d + |Δd|) = (D(d + |Δd|) - 1)/(d + |Δd|)

f' = (d + |Δd|) / (D(d + |Δd|) - 1) = (0,8 + -0,4) / (-2,5*(0,8 + -0,4) - 1) = 0,4/(-2,5*0,4 - 1) = -0,2

Г' = H'/h = |f'/d'| => H' = h*|f'/d'| = 0,16*|-0,2/0,4| = 0,16*1/2 = 0,8

H'/H = 8/10 : 4/75 = 8/10 * 75/4 = 2/2 * 15/1 = 15

ответ: в 15 раз.

Стандартная ЭКГ ― это запись электрических потенциалов в 12 отведениях:

1) отведения от конечностей ― электроды размещают чуть выше кисти, на внутренней поверхности правого (красный) и левого (желтый) предплечий, а также немного выше наружной лодыжки на левой (зеленый) и правой (черный ― заземление) голени;

а) двухполюсные (стандартные) ― I, II, III; 

б) однополюсные (усиленные) ― aVL, aVR, aVF; 

2) однополюсные грудные отведения ― V1–V6; расположение электродов на грудной клетке →рис. 25.1-1; отведения Vr3 и Vr4 следует записывать рутинно, если диагностируется инфаркт нижней стенки (вероятным критерием сопутствующего инфаркта правого желудочка является элевация сегмента ST в точке J в отведениях Vr3 и Vr4 ≥0,5 мм).