Рятуйте будь ласка До центра кулі масою 2 кг прикладено шість сил, які лежать в одній площині та утворюють між собою кути 60º. Модулі сил послідовно дорівнюють 1Н, 2Н. 3Н. 4Н, 5Н, 6Н. У якому напрямі і з яким прискоренням рухатиметься куля? (Приклад відповіді:У напрямі дії сили 1Н; 1м/с2)
Nowadays people have started to realize that good health is very important, much more important than wealth. To be in good health people must follow a healthy life style, to do things that will help keep a healthy body and mind, to feel fit.There are several ways to do it. The state of your body depends on how much time you spend doing sports. At least everybody must do morning exercises every day. The healthiest kinds of sports are swimming, running and cycling. Healthy food is also a very important factor. The daily menu should include meat, fruit and vegetables, milk product, which are rich in vitamins, fat, proteins and etc. On the other hand modern diets are very popular especially among women. Diets may be harmful, if they are used in the wrong way. To be healthy, people should get rid of their bad habits. It’s necessary to stop smoking and drinking much. Everyone should remember that cigarettes, alcohol and drugs destroy both body and brain. In addition it is recommended to watch TV less, avoid anxiety and observe daily routine. )
1)Good health is very significant.
2)All diets are useful and healthy.
3)It is recommended to do yoga to be in good health.
4)Alcohol and cigarettes influence only on body.
5)It is important to get rid of bad habits in order to keep a healthy body and mind.
Объяснение:
Дано:
m=10 кг
c2=4200 дж/кг*с
c1=2100 дж/кг*с
t1=-20 c
t2=0 c
t3=15 c
λ=3,4*10⁵ Дж/кг
Найти:
Q - ?
Решение:
1. Нагревание льда от -20с до 0с: Q1=c1m(-t1-t2) Q1=2100*10*(-(-20)-0)=2100*10*20=420 000 Дж
2. Плавление льда: Q2=λm Q2=3,4*10⁵*10=34*10⁵=3 400 000 Дж
3. Нагревание воды от 0с до 15с: Q3=c2m(t3-t2) Q3=4200*10*(15-0)=4200*10*15=630 000 Дж
Q=Q1+Q2+Q3 Q= 420 000+3 400 000+ 630 000 = 4 450 000 Дж = 4 450 кДж
ответ: На приготовление пошло 4 450 кДж энергии.
Вторая задача:
Дано:
L=2,3*10⁶ Дж/кг
m=10 кг
t1=100 c
t2=20c
c=4200 Дж/кг*с
Найти:
Q-?
Решение:
1.Конденсация водяного пара Q1=-Lm Q1=-2,3*10⁶*10=-23 000 000 Дж
2. Охлаждение воды от 100с до 20с Q2=cm(t2-t1) Q2=4200*10*(20-100)=4200*10*(-80)=-3 360 000 Дж
Q=Q1-Q2 Q=-23 000 000-(3 360 000)= -26 360 000 Дж=-26 360 кДж
ответ: Выделяется 23 360 кДж энергии