Case A :
A .75 kg 65 N/m 1.2 m
m = mass of car = 0.75 kg
k = spring constant of the spring = 65 N/m
h = height of the hill = 1.2 m
x = compression of spring = 0.25 m
Using conservation of energy between Top of hill and Bottom of hill
Total energy at Top of hill = Total energy at Bottom of hill
spring energy + potential energy = kinetic energy
(0.5) k x² + mgh = (0.5) m v²
(0.5) (65) (0.25)² + (0.75 x 9.8 x 1.2) = (0.5) (0.75) v²
v = 5.4 m/s
Case B :
B .60 kg 35 N/m .9 m
m = mass of car = 0.60 kg
k = spring constant of the spring = 35 N/m
h = height of the hill = 0.9 m
x = compression of spring = 0.25 m
Using conservation of energy between Top of hill and Bottom of hill
Total energy at Top of hill = Total energy at Bottom of hill
spring energy + potential energy = kinetic energy
(0.5) k x² + mgh = (0.5) m v²
(0.5) (35) (0.25)² + (0.60 x 9.8 x 0.9) = (0.5) (0.60) v²
v = 4.6 m/s
Case C :
C .55 kg 40 N/m 1.1 m
m = mass of car = 0.55 kg
k = spring constant of the spring = 40 N/m
h = height of the hill = 1.1 m
x = compression of spring = 0.25 m
Using conservation of energy between Top of hill and Bottom of hill
Total energy at Top of hill = Total energy at Bottom of hill
spring energy + potential energy = kinetic energy
(0.5) k x² + mgh = (0.5) m v²
(0.5) (40) (0.25)² + (0.55 x 9.8 x 1.1) = (0.5) (0.55) v²
v = 5.1 m/s
Case D :
D .84 kg 32 N/m .95 m
m = mass of car = 0.84 kg
k = spring constant of the spring = 32 N/m
h = height of the hill = 0.95 m
x = compression of spring = 0.25 m
Using conservation of energy between Top of hill and Bottom of hill
Total energy at Top of hill = Total energy at Bottom of hill
spring energy + potential energy = kinetic energy
(0.5) k x² + mgh = (0.5) m v²
(0.5) (32) (0.25)² + (0.84 x 9.8 x 0.95) = (0.5) (0.84) v²
v = 4.6 m/s
hence closest is in case C at 5.1 m/s
where
is the rate of heat transferred by conduction,
is the thermal conductivity of the material,
and
are the temperatures of each heat deposit,
is the cross area to the flow of heat, and
is the distance that the flow of heat has to go.
, and if
, then the expression would be:
where
is the diameter of the original cylinder, and
is the length of the original cylinder.
,where
is the heat rate in the second case,
and
are the new diameter and length.
and
, substituting in the expression for
.
.
, then:
.