Дж. Визначити натяг троса і потужність, що розвивається при буксируванні. Автомобиль тянут горизонтально с постоянной скоростью 5 м/с. При перемещении автомобиля на расстояние l = 1,5 км выполнена работа А = [tex] 9 * 10x ^ {5} [/ tex] Дж. Определить натяжение троса и мощность при буксировке.
Объяснение:
Newspapers and magazines play a very important part in our life. Practically there is no family that does not read them. We can learn many things from newspapers. Perhaps that is why many years ago an American humorous writer said: "All I know is what I see in the papers"; and another American author more than half a century ago wrote that "the careful reader of a few good newspapers can learn more in a year than most scientists do in their great libraries".
We can agree or disagree with these statements (better to disagree, because scientific books and magazines have more information than newspapers), but we'll have to say that newspapers help us in many ways. There are a lot of different kinds of newspapers in our country. One can buy them practically everywhere. It is impossible to read all the newspapers and magazines. Everyone has favourite ones.
My favourite newspaper is "Komsomolskaya Pravda". We can read almost about everything in this newspaper. There are articles dealing with home and abroad news, sport events, life of favourite actors and singers and even the weather. We can find many interesting things there. We can read some useful pieces of advice, some stories about our life, and so on. There are puzzles, songs and even anecdotes there. "Komsomolskaya Pravda" is one of the most interesting newspapers, to my mind.
1. ρ=1000 кг/м³
g=10 м/с²
h=12 м
насос должен создавать давление p=ρ·g·h=1000·10·12=120000 Па =120 кПа
2. P=gPh
P=10H/кг×1036кг/м³×0,7м=7252 Па
7252 Па = 7,252 кПа
Округляем 7,25 кПа
3. 47,38 МПа= 47380000 Па
p=ρgh
h=p/(ρg)=47380000/(1030•10)=4600 м= 4,6 км
4. Дано:
g - 10 Н/кг
ρ воды - 1000 кг/м^3
h1 - 249 м
h2 - 20 м
p1 - ?
p2 - ?
S (разность давлений) - ?
p=ρgh >>>
р1=ρgh1=1000*10*249=2 490 000 Па
р2=ρgh2=1000*10*20=200 000 Па
S=p1-p2=2 490 000 - 200 000 = 2 290 000 Па
ответ: отличаются на 2 290 000 Па