Answer: hello some parts of your question is missing attached below is the missing information
The radiator of a car is a type of heat exchanger. Hot fluid coming from the car engine, called the coolant, flows through aluminum radiator tubes of thickness d that release heat to the outside air by conduction. The average temperature gradient between the coolant and the outside air is about 130 K/mm . The term ΔT/d is called the temperature gradient which is the temperature difference ΔT between coolant inside and the air outside per unit thickness of tube
answer : Total surface area = 3/2 * area of old radiator
Explanation:
we will use this relation
K = 
change in T = ΔT
therefore New Area ( A ) = 3/2 * area of old radiator
Given that the thermal conductivity is the same in the new and old radiators
Answer:
<u>The automobile rental prices shall show all taxes (including a 6% state tax).</u>
Explanation:
Im pretty sure
Answer:
d) 9.55 psi
Explanation:
pressure at the bottom is =ρgh
weight density is ρg=55 lb/ft³
h=25ft
pressure at the bottom is =
=1375psf
1 ft = 12 inch
pressure at bottom =
= 9.55 psi
so, answer will be option (d) which is 9.55 psi
Answer:
A.) Find the answer in the explanation
B.) Ua = 7.33 m/s , Vb = 7.73 m/s
C.) Impulse = 17.6 Ns
D.) 49%
Explanation:
Let Ua = initial velocity of the rod A
Ub = initial velocity of the rod B
Va = final velocity of the rod A
Vb = final velocity of the rod B
Ma = mass of rod A
Mb = mass of rod B
Given that
Ma = 2kg
Mb = 1kg
Ub = 3 m/s
Va = 0
e = restitution coefficient = 0.65
The general expression for the velocities of the two rods after impact will be achieved by considering the conservation of linear momentum.
Please find the attached files for the solution
Answer:
The flexural strength of a specimen is = 78.3 M pa
Explanation:
Given data
Height = depth = 5 mm
Width = 10 mm
Length L = 45 mm
Load = 290 N
The flexural strength of a specimen is given by
78.3 M pa
Therefore the flexural strength of a specimen is = 78.3 M pa
