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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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A sports car accelerates from rest for 5 seconds reaching a velocity of 23.0 m/s.
denis-greek [22]

Answer:

<h2>4.6 m/s²</h2>

Explanation:

The acceleration of an object given it's velocity and time taken can be found by using the formula

<h3>a =  \frac{v - u}{t}  \\</h3>

where

v is the final velocity

u is the initial velocity

t is the time taken

a is the acceleration

Since the body is from rest u = 0

From the question we have

a =  \frac{23 - 0}{5}  =  \frac{23}{5}  \\

We have the final answer as

<h3>4.6 m/s²</h3>

Hope this helps you

4 0
2 years ago
PLEASE HELP 100 POINTS! Please fill in the scale distance from sun and diversion factor, listing off the numbers works
steposvetlana [31]

Solved your another question same like this with scaling to Cm this time we go with metre(m)

Scale factor

  • 1km=10³m
  • 1m=10^{-3}km

Mercury

  • 58000m

Ven us

  • 108000m

Earth

  • 150000m

Mars

  • 228000m

Jupiter

  • 778000m

Saturn

  • 1430000m

Uranus

  • 2870000m

Neptune

  • 4500000m

7 0
2 years ago
What happens to a light wave when it travels from air into glass?
tino4ka555 [31]
Light is refracted when it crosses the interface from air to glass in which it moves more slowly.
Since the light speed changes at the interface, the wave length of the light must change too. The wave length decreases as the light enter the medium and the light wave changes direction.
7 0
3 years ago
An Atwood machine is constructed using a hoop with spokes of negligible mass. The 2.3 kg mass of the pulley is concentrated on i
Sergeu [11.5K]

Answer:

a = 2.77~{\rm m/s^2}

Explanation:

Since the pulley has a mass concentrated on its rim, the pulley can be considered as a ring.

The moment of inertia of a ring is

I = mr^2 = (2.3)(23.5\times 10^{-2})^2 = 0.127

The mass on the left is heavier, that is the pulley is rotating counterclockwise.

By Newton's Second Law, the net torque is equal to moment of inertia times angular acceleration.

\tau = I \alpha

Here, the net torque is the sum of the weight on the left and the weight on the right.

\tau = m_1gR - m_2gR = (1.65)(9.8)(23.5\times 10^{-2}) - (1)(9.8)(23.5\times 10^{-2}) = 1.497~{\rm Nm}

Applying Newton's Second Law gives the angular acceleration

\tau = I\alpha\\1.497 = 0.127\alpha\\\alpha = 11.78~{\rm rad/s^2}

The relation between angular acceleration and linear acceleration is

a = \alpha R

Then, the linear acceleration of the masses is

a = 11.78 \times 23.5\times 10^{-2} = 2.77~{\rm m/s^2}

5 0
3 years ago
If I toot 88 times what did that mean
Degger [83]

Answer:Nazi skinhead code for Heil Hitler. H being the 8th letter of the alphabet, therefore HH=88.

Explanation:

8 0
3 years ago
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