<span>Heating food under a heat lamp is an example of heat transfer by
<span>Radiation</span></span>
Answer:
Both Thomson and Rutherford used charged particles in their experiments.
Explanation:
<h3><u>Answer;</u></h3>
A.75°C
<h3><u>Explanation</u>;</h3>
Let the change in temp of cold water be x degrees,
while that of hot water be 100 - x degrees.
Heat exchange = mcΔt
Ice
Δt = x
m = 0.50 kg
c = 4.18 kJ/kg*°C
Hot water
Δt = 100 - x
m = 1.5 kg
c = 4.18
But;
Heat lost = heat gained
0.50 * c * x = 1.5 * c * (100 - x)
0.50 *x = 1.5*(100 - x)
0.5x = 150 - 1.5x
0.5x + 1.5x = 150 - 1.5x + 1.5x
2x = 150
x = <u>75° C</u>
Hence; the equilbrium temperature will be 75° C
Answer:
60 rad/s
Explanation:
∑τ = Iα
Fr = Iα
For a solid disc, I = ½ mr².
Fr = ½ mr² α
α = 2F / (mr)
α = 2 (20 N) / (0.25 kg × 0.30 m)
α = 533.33 rad/s²
The arc length is 1 m, so the angle is:
s = rθ
1 m = 0.30 m θ
θ = 3.33 rad
Use constant acceleration equation to find ω.
ω² = ω₀² + 2αΔθ
ω² = (0 rad/s)² + 2 (533.33 rad/s²) (3.33 rad)
ω = 59.6 rad/s
Rounding to one significant figure, the angular velocity is 60 rad/s.
