Given:
ΔT = 38 - 26 = 12°C, temperature change
Q = 11.3 J, heat input
c = 0.128 J/(g-°C), specific heat of lead
Let m = the mass of the lead.
Then
Q = m*c*ΔT
(m g)*(0.128 J/(g-°C))*(12 °C) = 11.3 J
1.536m = 11.3
m = 7.357 g
Answer: 7.36 g (2 sig. figs)
Answer:100 meters per min
Explanation:
Answer:
40 miles / hour south
Explanation:
120 miles/3 hours = 40 miles / hour
Stan looga
this is correct answer