1.
a)
Z wykresu odczytujemy:
[tex]W = 300 \ J\\t = 15 \ s\\P = ?\\\\\\P = \frac{W}{t}\\\\P = \frac{300 \ J}{15 \ s}\\\\\boxed{P = 20 \
W}[/tex]
b)
[tex]t = 3 \ min = 3\cdot60 \ s = 180 \ s\\P = 20 \ W\\W = ?\\\\\\P = \frac{W}{t} \ \ \rightarrow \ \ W = P\cdot t\\\\W = 20 \ W\cdot 180 \ s = 20\frac{J}{s}\cdot 180 \ s\\\\\boxed{W = 3 \ 600 \ J = 3,6 \ kJ}[/tex]
2.
[tex]Dane:\\E_{p} = 420 \ J\\h = 7 \ m\\g = 10\frac{N}{kg}\\Szukane:\\m = ?\\\\\\E_{p} = mgh \ \ /:gh\\\\m = \frac{E_{p}}{gh}\\\\m = \frac{420 \ J}{10\frac{N}{kg}\cdot7 \ m} = \frac{420 \ N\cdot m}{70\frac{N\cdot m}{kg}}\\\\\boxed{m = 6 \ kg}[/tex]
