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maw [93]
3 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]3 years ago
8 0

Answer:

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

Explanation:

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Starting from rest, a 2.1x10^-4 kg flea springs straight upward. While the flea is pushing off from the ground, the ground exert
Rainbow [258]

Answer:

a) The flea's speed when it leaves the ground is v=1.88m/s

b) The flea move 18cm upward while it is pushing off

Explanation:

Hi

<u>Knwons</u>

Mass m=2.1\times 10^{-4} kg, Work W=3.7\times 10^{-4} J and Force F=0.41N

a) Here we are going to use W=\frac{1}{2} mv^{2}, so v=\sqrt{\frac{2W}{m} }= \sqrt{\frac{2(3.7\times 10^{-4} J)}{2.1\times 10^{-4} kg} }=1.88m/s

a) Here we are going to use W=mgh, so h=\frac{W}{mg}= \frac{3.7\times 10^{-4} J}{(2.1\times 10^{-4} kg)(9.8m/s^{2})} =0,1797m or 18cm approx.

7 0
3 years ago
What type of friction is using chalk in the summer to draw on the ground in Copley square?
Lera25 [3.4K]
The second option rolling friction
7 0
3 years ago
Starting at 1.3 m/s, a runner accelerates at a constant 0.22 m/s2 for 6.0 s. What is the runner’s displacement during this time
Anettt [7]

Answer:

answer is 11.76 meter

Explanation:

use 2nd equation of motion

S=ut+1/2at^2

7 0
3 years ago
A block of mass m slides on a horizontal frictionless surface. The block is attached to a spring with a spring constant K. At th
Fittoniya [83]

Answer:

b) a = -k / m x , c) d²x / dt² = - A w² cos (wt+Ф) , d) and e)  T = 2π √m / k

h)   a = - A w² cos (wt+Ф)

Explanation:

a) see free body diagram in the attachment

b) We write Newton's second law

          Fe = m a

          -k x = ma

           a = -k / m x

c) the acceleration is

         a = d²x / dt²

     

      If x = A cos wt

        v = dx / dt = -A w sin (wt +Ф)

        a = d²x / dt² = - A w² cos (wt+Ф)

d) we substitute in Newton's second law

        d²x / dt² = -k / m x

   

We call

       w² = k / m

e) substitute to find w

     -A w² cos (wt+Ф) = -k / m A cos (wt+Ф)

      w² = k / m

Angular velocity and frequency are related

       w = 2π f

       f = 1 / T

       

 We substitute

      T = 2π / w

      T = 2π √m / k

g)    v= - A w sin (wt+Ф)

h) acceleration is

       a = - A w² cos (wt+Ф)

8 0
4 years ago
What is the total displacement???
muminat
Displacement (between time 0 and time 25) is the area under the velocity time curve, i.e. &int; vdt.
Here, v(0)=10, v(25)=34 (approx.)
Therefore 
displacement = (1/2)(10+34 m/s)*(25-0) s   [ trapezoid area ]
=550 m
4 0
3 years ago
Read 2 more answers
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