To solve this problem, we should recall the law of
conservation of energy. That is, the heat lost by the aluminium must be equal
to the heat gained by the cold water. This is expressed in change in enthalpies
therefore:
- ΔH aluminium = ΔH water
where ΔH = m Cp (T2 – T1)
The negative sign simply means heat is lost. Therefore we
calculate for the mass of water (m):
- 0.5 (900) (20 – 200) = m (4186) (20 – 0)
m = 0.9675 kg
Using same mass of water and initial temperature, the final
temperature T of a 1.0 kg aluminium block is:
- 1 (900) (T – 200) = 0.9675 (4186) (T – 0)
- 900 T + 180,000 = 4050 T
4950 T = 180,000
T = 36.36°C
The final temperature of the water and block is 36.36°C
D. Both exhibit the same particle-to-particle interaction.Because disturbance is propagated with the help of particles. Other than this,[ <span>light waves are electromagnetic waves. ocean waves and sound waves are mechanical waves. they are able to transfer energy. electromagnetic wave and ocean waves are transverse waves while sound waves are the longitudinal wave. they show wave properties: reflection, refraction, diffraction, interference, and plane-polarization. longitudinal waves such as sound waves cannot be plane-polarized]. The one in the box shows different examples of waves with their examples. Hope it helps.</span>
Answer:
4.5m/s
Explanation:
Linear speed (v) = 42.5m/s
Distance(x) = 16.5m
θ= 49.0 rad
radius (r) = 3.67 cm
= 0.0367m
The time taken to travel = t
Recall that speed = distance / time
Time = distance / speed
t = x/v
t = 16.5/42.5
t = 0.4 secs
tangential velocity is proportional to the radius and angular velocity ω
Vt = rω
Angular velocity (ω) = θ/t
ω = 49/0.4
ω = 122.5 rad/s
Vt = rω
Vt = 0.0367 * 122.5
Vt =4.5 m/s
