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:
because these materials absorb the sound and do not allow them from going out of as, some times the sound can be a trouble to the people.
Explanation: Hope this helps!
Answer:
Into sound and heat energy due to friction.
Efficiency = 45J/120J = 0.375
Or
0.375 * 100% = 37.5 %