Air trapped in the cooling system could create undesirable areas of combustion heat buildup in the engine called _____.

Answers

Answer 1

Answer:

Answer:

Air pockets.

Explanation:

Air pockets in the cooling system are bubbles of air trapped within the lines (hoses and pipes) of the cooling system. This air bubbles enter the cooling system usually during the process of filling the radiator coolant fluid (usually water), or replacing the water pump or the radiator hose during repairs or servicing of the cooling system. The trapped air prevent pressure movement that is needed by the coolant to move the heat generated from the engine cylinder, resulting in heat build up. The solution is to "bleed" the engine through the radiator lid or some air release valves.

Related Questions

Chapter 38, Problem 001 Monochromatic light (that is, light of a single wavelength) is to be absorbed by a sheet of a certain material. Photon absorption will occur if the photon energy equals or exceeds 0.42 eV, the smallest amount of energy needed to dissociate a molecule of the material. (a) What is the greatest wavelength of light that can be absorbed by the material? (b) In what region of the electromagnetic spectrum is this wavelength located?
The _______ principle encourages us to resolve a set of stimuli, such as trees across a ridgeline, into smoothly flowing patternsA.) depth perception. B.) perception. C.) similarity. D.) continuity.
Two resistors are to be combined in parallelto form an equivalent resistance of 400Ω. The resistors are takenfrom available stock on hand as acquired over the years. Readily available are two common resistorsrated at 500±50 Ωand two common resistors rated at 2000 Ω±5%. What isthe uncertainty in an equivalent 400 Ωresistance?(Hint: the equivalent resistance connected in parallel can be obtained by 1212TRRRRR=+)
A 1200-kg cannon suddenly fires a 100-kg cannonball at 35 m/s. what is the recoil speed of the cannon? assume that frictional forces are negligible and the cannon is fired horizontally.
A reducing elbow in a horizontal pipe is used to deflect water flow by an angle θ = 45° from the flow direction while accelerating it. The elbow discharges water into the atmosphere. The cross- sectional area of the elbow is 150 cm2 at the inlet and 25 cm2 at the exit. The elevation difference between the centers of the exit and the inlet is 40 cm. The mass of the elbow and the water in it is 50 kg. Determine the anchoring force needed to hold the elbow in place. Take the momentum flux correction factor to be 1.03 at both the inlet and outlet.

As the moon orbits the Earth which of the following changes (1) a. Speed b. Velocity c. Acceleration d. A, B, and C e. None

Answers

Answer:

B- Velocity

Explanation:

This means gravity makes the Moon accelerate all the time, even though its speed remains constant.

An Australian emu is running due north in a straight line at a speed of 13.0 m/s and slows down to a speed of 9.40 m/s in 3.50 s. (a) What is the magnitude and direction of the bird’s acceleration? (b) Assuming that the acceleration remains the same, what is the bird’s velocity after an additional 1.20 s has elapsed?

Answers

(a) The acceleration of the bird is $a = -1.02\ m/s$ . The negative sign indicated the opposite direction of motion. (b) The final speed is $v = 11.76\ m/s$ .

Given:

Initial speed, $u = 13\ m/s$

Final speed, $9.4\ m/s$

Time, $t = 3.5\ s$

The acceleration can be computed from the velocities and time. The standard unit of acceleration is a meter per second square.

(a)

The acceleration is computed as:

$a = v-u/t\na = 9.4-13/3.5\na = -1.02\ m/s$

Hence, the acceleration of the bird is $a = -1.02\ m/s$ . The negative sign indicated the opposite direction of motion.

(b)

The final speed as the given time can be computed from the first equation of motion. The first equation of motion gives the relation between final and initial speed, acceleration, and time.

The final speed at time 1.2 seconds is equal to:

$v = u+at\nv = 13+(-1.02)*1.20\nv = 11.76\ m/s$

Hence, the final speed is $v = 11.76\ m/s$ .

To learn more about Acceleration, here:

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*PLEASE HELP*A baseball is pitched with a horizontal velocity of 25.21 m/s. Mike Trout hits the ball, sending it in the opposite direction (back toward the pitcher) at a speed of -50.67 m/s. The ball is in contact with the bat for 0.0014 seconds. What is the
acceleration of the ball?

Answers

Answer:

-54,200 m/s^2

Explanation:

a=(vf-vi)/t

What is the velocity at discharge if the nozzle of a hose measures 68 psi? 100.25 ft./sec 10.25 ft./sec 125.2 ft./sec 11.93 ft./sec

Answers

Answer:

The velocity at discharge is 100.46 ft/s

Explanation:

Given that,

Pressure = 68 psi

We need to calculate the pressure in pascal

$P=68*6894.74\ Pa$

$P=468842.32\ Pa$

We need to calculate the velocity

Let the velocity is v.

Using Bernoulli equation

$P=(1)/(2)\rho v^2$

$468842.32=0.5*1000* v^2$

$v=\sqrt{(468842.32)/(0.5*1000)}$

$v=30.62\ m/s$

Now, We will convert m/s to ft/s

$v =30.62*3.281$

$v=100.46\ ft/s$

Hence, The velocity at discharge is 100.46 ft/s

Final answer:

The speed of water discharged from a hose depends on the nozzle pressure and the constriction of the flow, but the specific speed cannot be determined from pressure alone without additional parameters.

Explanation:

The question is asking about the velocity or speed achieved by water when it is forced out of a hose with a nozzle pressure of 68 psi. To understand this, we need to know that the pressure within the hose is directly correlated with the speed of the water's exit. This is due to the constriction of the water flow by the nozzle, causing speed to increase.

However, the specific velocity at discharge can't be straightforwardly calculated from pressure alone without knowing more details, such as the dimensions of the hose and nozzle, and the properties of the fluid. Therefore, based on the provided information, a specific answer in ft/sec can't be given.

Learn more about Pressure and Velocity here:

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How does light move?

Answers

Answer:

Light travels as a wave. But unlike sound waves or water waves, it does not need any matter or material to carry its energy along. This means that light can travel through a vacuum—a completely airless space. It speeds through the vacuum of space at 186,400 miles (300,000 km) per second.

Explanation:

Hope this helps :))

A trombone can produce pitches ranging from 85 Hz to 660 Hz approximately. When the trombone is producing a 357 Hz tone, what is the wavelength of that tone in air at standard conditions?

Answers

Answer:

The wavelength of that tone in air at standard condition is 0.96 m.

Explanation:

Given that, a trombone can produce pitches ranging from 85 Hz to 660 Hz approximately. We need to find the wavelength of that tone in air when the trombone is producing a 357 Hz tone.

We know that the speed of sound in air is approximately 343 m/s. Speed of a wave is given by :

$v=f\lambda\n\n\lambda=(v)/(f)\n\n\lambda=(343\ m/s)/(357\ Hz)\n\n\lambda=0.96\ m$

So, the wavelength of that tone in air at standard condition is 0.96 m. Hence, this is the required solution.

Other Questions

Part 1) A cop car traveling at 25 m/s has a siren producing a frequency of 700 Hz. A felon jumps on his motorcycle and speed off in the opposite direction of 15 m/s. What frequency does the felon hear as he sped away (speed of sound is 343 m/s)?Part 2) The cop does a U-turn and speeds towards the felon at 30 m/s, while the felon speeds up to 20 m/s. What frequency does the felon hear as he sped away (speed of sound is 343 m/s)?Part 3) What if the felon then sped up to 30 m/s and all other conditions remained the same?

Testosterone is an example of what kind of biomolecule?

"For the lowest harmonic of pipe with two open ends, how much of a wavelength fits into the pipe’s length?"

Light with a wavelength of 700 nm (7×〖10〗^(-7) m) is incident upon a double slit with a separation of 0.30 mm (3 x 10-4 m). A screen is located 1.5 m from the double slit. At what distance from the screen will the first bright fringe beyond the center fringe appear?