Pudełko z CD-ROM-em ma masę 50 g i - Zadanie 2: Świat fizyki 1A

Rozwiązanie

	$m (k g)$	$F_{c} = m g (N)$	$p = \frac{F _{c}}{S} (P a)$
1 pudełko	$0, 05 k g$	$0, 5 N$	$30 P a$
2 pudełka	$0, 1 k g$	$1 N$	$60 P a$
3 pudełka	$0, 15 k g$	$1, 5 N$	$90 P a$
4 pudełka	$0, 2 k g$	$2 N$	$120 P a$

$m = 50 g = 0, 05 k g$

$F_{1} = m g = 0, 05 k g \cdot 10 \frac{N}{k g} = 0, 5 N$

$p_{1} = \frac{F}{S} = \frac{0 , 5 N}{167 c m ^{2}} = \frac{0 , 5 N}{0 , 0167 m ^{2}} = \frac{\frac{1}{2} N}{\frac{167}{10000} m ^{2}} = \frac{1}{2} \cdot \frac{10000}{167} P a = \frac{10000}{334} P a \approx 30 P a$

$m_{2} = 2 \cdot 0, 05 k g = 0, 1 k g$

$F_{2} = m g = 0, 1 k g \cdot 10 \frac{N}{k g} = 1 N$

$p_{2} = \frac{F}{S} = \frac{1 N}{167 c m ^{2}} = \frac{1 N}{0 , 0167 m ^{2}} = \frac{1 N}{\frac{167}{10000} m ^{2}} = \frac{10000}{167} P a \approx 60 P a$

$m_{3} = 0, 05 k g \cdot 3 = 0, 15 k g$

$F_{3} = m g = 0, 15 k g \cdot 10 \frac{N}{k g} = 1, 5 N$

$p_{3} = \frac{F}{S} = \frac{1 , 5 N}{167 c m ^{2}} = \frac{1 , 5 N}{0 , 0167 m ^{2}} = \frac{1 , 5 N}{\frac{167}{10000} m ^{2}} = 1, 5 \cdot \frac{10000}{167} P a = \frac{15000}{167} P a \approx 90 P a$

$m_{4} = 0, 05 k g \cdot 4 = 0, 2 k g$

$F_{4} = m g = 0, 2 k g \cdot 10 \frac{N}{k g} = 2 N$

$p_{3} = \frac{F}{S} = \frac{2 N}{167 c m ^{2}} = \frac{2 N}{0 , 0167 m ^{2}} = \frac{2 N}{\frac{167}{10000} m ^{2}} = 2 \cdot \frac{10000}{167} P a = \frac{20000}{167} P a \approx 120 P a$