Answer:
180 kJ
Explanation:
Given that:
Mass (m) = 5 kg
Initial temperature (T1) = 28°C
Final temperature (T2) = 68°C
The change in temperature (ΔT) = T2 - T1 = 68°C - 28°C = 40°C
Specific heat capacity of aluminium (c) = 900 J/kg°C
The quantity of heat energy required (q) is given by:
q = mcΔT
q = 5 kg × 900 J/kg°C × 40°C
q = 180000 Joules
q = 180 kJ
Therefore 180 kJ is required to raise the temperature of aluminium from 28°C to 68°C.
Answer:
b) v
Explanation:
the distance between pole and position of image can denoted by <u>v</u>
answer
64 is C (c) 1-Bromo-3-methylbutane
63 is D
i is secondary ii is primary
Answer:
B).315mph
Explanation:
Let the speed of the plane = p
Let the speed of the wind = w
Set up the system equation as;
Relative V: Time: Distance:
in wind direction: p + w 2 700
against wind: p - w 2.5 700
2(p + w) = 700
2.5(p - w) = 700
2p + 2w = 700
2.5p - 2.5w = 700
2.5 x: 5p + 5w = 1750
2 x: 5p - 5w = 1400
10p = 3150
p = 315 mph
Therefore, he speed of the plane in still air is 315 mph
