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maw [93]
2 years ago
5

Temperatures of gases inside the combustion chamber of a four‑stroke automobile engine can reach up to 1000°C. To remove this en

ormous amount of heat, the engine utilizes a closed liquid‑cooled system that relies on conduction to transfer heat from the engine block into the liquid and then into the atmosphere by flowing coolant around the outside surface of each cylinder.
Suppose that, in a particular 4‑cylinder engine, each cylinder has a diameter of 8.75 cm, a height of 11.3 cm, and a thickness of 3.14 mm. The temperature on the inside of the cylinders is 195.2°C, and the temperature outside, where the coolant passes, is 130.0°C. The temperature of the incoming liquid (a mixture of water and antifreeze) is maintained at 95.0°C.

What volume flow rate of coolant \frac{V}{t} would be required to cool this engine? Assume that the coolant reaches thermal equilibrium with the outer cylinder walls before exiting the engine. The specific heat of the coolant is 3.75 J/g°C and its density is 1.070×103 kg/m3. The cylinder walls have thermal conductivity of 1.10×102 W/m⋅∘C. Assume that no heat passes through the ends of the cylinders.
Physics
1 answer:
Alona [7]2 years ago
8 0

Answer:

thats a lot, which one u want me to do?

Explanation:

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Please help!!!!
Dafna11 [192]

The intensity of the electric field is 30,000 N/C

Explanation:

The strength of the electric field produced by a single-point charge is given by the equation

E=k\frac{q}{r^2}

where:

k=8.99\cdot 10^9 Nm^{-2}C^{-2} is the Coulomb's constant

q is the magnitude of the charge

r is the distance from the charge

In this problem, we have:

q=3\cdot 10^{-9}C is the magnitude of the charge

r = 3 cm = 0.03 m is the distance at which we are calculating the field intensity

Substituting, we find:

E=(8.99\cdot 10^9)\frac{3\cdot 10^{-9}}{(0.03)^2}=30,000 N/C

Learn more about electric field:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

5 0
3 years ago
A 99.1-kg baseball player slides into second base. The coefficient of kinetic friction between the player and the ground is μk =
Stels [109]

Answer:

628.022466 N

8.61 m/s

Explanation:

m = Mass

\mu = Coefficient of friction

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s²

F_f=\mu mg\\\Rightarrow F_f=0.646\times 99.1\times 9.81\\\Rightarrow F_f=628.022466\ N

Magnitude of frictional force is 628.022466 N

F=ma\\\Rightarrow a=\frac{F_f}{m}\\\Rightarrow a=\frac{628.022466}{99.1}\\\Rightarrow a=6.33726\ m/s^2

v=u+at\\\Rightarrow 0=u-6.33726\times 1.36\\\Rightarrow u=8.61\ m/s

Initial speed of the player is 8.61 m/s

4 0
3 years ago
Pls help! True or False?
jasenka [17]

Answer:

True

Explanation:

i hope it helps

please follow me

7 0
2 years ago
Read 2 more answers
Worth 15 points<br><br> Please answer which one matches for each conversion
FrozenT [24]

Answer:

distance - meters

speed - meters/seconds

time - seconds

velocity - meters/seconds

acceleration - meters/seconds²

5 0
2 years ago
What is the difference between torque and the moment of a force ​
STatiana [176]

Answer:

Torque Of a Force: If The Force has tendency or Bends The Body about Longitudinal axis of the Body it is Torque. Moment Of a Force :If Force has Tendency to or Rotates the Body about Transverse asis the Body It is Moment .

Explanation:

3 0
2 years ago
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