Answer:
0.16 m
Explanation:
A rectangular gasoline tank can hold 50.0 kg of gasoline when full, and the density of gasoline is 6.8 × 10² kg/m³. We can find the volume occupied by the gasoline (volume of the tank).
50.0 kg × (1 m³/6.8 × 10² kg) = 0.074 m³
The volume of the rectangular tank is:
volume = width × length × depth
depth = volume / width × length
depth = 0.074 m³ / 0.500 m × 0.900 m
depth = 0.16 m
Answer: a) vcar= 7 m/s ; b) a train= 0.65 m/s^2
Explanation: By using the kinematic equation for the car and the train we can determine the above values of the car velocity and the acceletarion of the train, respectively.
We have for the car
distance = v car* t, considering the length of train (81.1 m) travel by the car during the first 11.6 s
the v car = distance/time= 81.1 m/11.6s= 7 m/s
In order to calculate the acceleration we have to use the kinematic equation for the train from the rest
distance train = (a* t^2)/2
distance train : distance travel by the car at constant speed
so distance train= (vcar*36.35)m=421 m
the a traiin= (2* 421 m)/(36s)^2=0.65 m/s^2
A. The proeutectoid phase is Fe₃c because 0.95 wt/c is greater than the eutectoid composition which is 0.76 wt/c
b. We determine how much total territe and cementite form, we apply the lever rule expressions yields.
Wx = (fe₃c-co/cfe₃ c-cx = 6.70- 0.95/6.70- 0.022 = 0.86
The total cementite
Wfe₃C = 10-Cx/ Cfe₃c -Cx = 0.95 - 0.022/6.70 - 0.022 = 0.14
The total cementite which is formed is
(0.14) × (3.5kg) = 0.49kg
c. We calculate the pearule and the procutectoid phase which cementite form the equation
Ci = 0.95 wt/c
Wp = 6.70 -ci/6.70 - 0.76 = 6.70 -0.95/6.70 - 0.76 = 0.97
0.97 corresponds to mass.
W fe₃ C¹ = Ci - 0.76/5.94 = 0.03
∴ It is equivalent to
(0.03) × (3.5) = 0.11kg of total of 3.5kg mass.
Answer:
So then the difference of temperature across the material would be 
Explanation:
For this case we can use the Fourier Law of heat conduction given by the following equation:
(1)
Where k = thermal conductivity = 0.2 W/ mK
A= 1m^2 represent the cross sectional area
Q= 3KW represent the rate of heat transfer
is the temperature of difference that we want to find
represent the thickness of the material
If we solve 
in absolute value from the equation (1) we got:
First we convert 3KW to W and we got:
And we have everything to replace and we got:
So then the difference of temperature across the material would be
Answer:
Given a tube of diameter d, = 3cm = 0.03m
Pressure Balance
Mercury pressure at the tube bottom Pₓ = Pa + ρgh
where
Pa = Atmospheric pressure = 101kpa
ρ = Density of mercury = 13,546kg/m3
g = acceleration due to gravity
h = height of the tube?
Given
Bottom pressure in excess of the atmospheric pressure = 48kPa = Pₓ - Pa
Therefore, 48kPa = ρgh
h = 48(kN/m2)/ρg
h = 48,000kgms⁻²m⁻²/(13546kgm⁻³ x 9.81ms⁻²)
h = 0.36m
the tube is 36cm tall
