r1 = 5*10^10 m , r2 = 6*10^12 m
v1 = 9*10^4 m/s
From conservation of energy
K1 +U1 = K2 +U2
0.5mv1^2 - GMm/r1 = 0.5mv2^2 - GMm/r2
0.5v1^2 - GM/r1 = 0.5v2^2 - GM/r2
M is mass of sun = 1.98*10^30 kg
G = 6.67*10^-11 N.m^2/kg^2
0.5*(9*10^4)^2 - (6.67*10^-11*1.98*10^30/(5*10^10)) = 0.5v2^2 - (6.67*10^-11*1.98*10^30/(6*10^12))
v2 = 5.35*10^4 m/s
Social Classical is not a type of system.
Answer:
I'm pretty sure the answer you selected is correct.
Explanation:
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°
Answer:
P=I^2*R=400R where R is the resistance of the wire
Explanation:
