The friction between the two objects creates heat.
Answer:
Since this is a linear equation
y = m x + b or
U = m F + b is a linear equation
when ΔF = (212 - 32) = 180
and ΔU = (60 - (-15)) = 75
m = 75 / 180 = 2.4 if converting F to U and a = .417
U = .417 F + b
If F = 32 then U = -15 and
-15 = .417 * 32 + b
b = -15 - 13.3 = -28.3 and our equation becomes
U = .417 F - 28.3
Check: let F = 212
U = .417 * 212 - 28.3 = 60 as it should
Answer:
0.195 m
Explanation:
Speed is distance moved per unit time, expressed as s=d/t and making d the subject of the formula then d=st
Where d is distance/depth moved, s is rhe speed of waves and t is time in seconds.
Substituting s with 1300 m/s and t with 0.00015 s then the depth of metal segment will be
D=1300*0.00015=0.195 m
Therefore, the depth is equivalent to 0.195 m
The question is incomplete, the complete question is;
A body initially at a all 100 degree centigarde cools to 60 degree centigarde in 5 minutes and to 40 degree centigarde in 10 minutes . What is the temperature of surrounding? What will be the temperature in 15 minutes?
Answer:
See explanation
Explanation:
From Newton's law of cooling;
θ1 - θ2/t = K(θ1 + θ2/2 - θo]
Where;
θ1 and θ2 are initial and final temperatures
θo is the temperature of the surroundings
K is the constant
t is the time taken
Hence;
100 - 60/5 = K(100 + 60/2 - θo)
100 - 40/10 = K(100 + 40/2 - θo)
8= (80 - θo)K -----(1)
6= (70 - θo)K -----(2)
Diving (1) by (2)
8/6 = (80 - θo)/(70 - θo)
8(70 - θo) = 6(80 - θo)
560 - 8θo = 480 - θo
560 - 480 = -θo + 8θo
80 = 7θo
θo = 11.4°
Again from Newton's law of cooling;
θ = θo + Ce^-kt
Where;
t= 0, θ = 60° and θo = 11.4°
60 = 11.4 + C e^-K(0)
60 - 11.4 = C
C= 48.6°
To obtain K
40 = 11.4 + 48.6e^-10k
40 -11.4 = 48.6e^-10k
28.6/48.6 = e^-10k
0.5585 = e^-10k
-10k = ln0.5585
k= ln0.5585/-10
K= 0.0583
Hence, the temperature in 15 minutes;
θ= 11.4 + 48.6e^(-0.0583 × 15)
θ= 31.7°
