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Vikki [24]
2 years ago
14

If your engine begins to lose power, run rough, stall, or refuse to start when traveling at high altitudes, your vehicle may be

suffering from _______.
Physics
1 answer:
m_a_m_a [10]2 years ago
3 0

Vapor lock

If your engine begins to lose power, run rough, stall, or refuse to start when traveling at high altitudes, your vehicle may be suffering from Vapor lock.

<h3>How is vapor lock fixed?</h3>
  • By first pressing the accelerator pedal slightly (but not all the way to the floor) while starting the engine, you can aid in the removal of any vapor that may have remained in the fuel system after it has cooled.
  • Hold down the accelerator pedal once the engine fires up until the car moves smoothly, which shows the vapor lock has passed.

<h3>What brings about a vapor lock?</h3>
  • When fuel boils in the carburetor or fuel line, vapor lock results.
  • Fuel that has vaporized stops gas from reaching your engine by creating back pressure in your fuel system.
  • It frequently happens after idling or after turning off and back on a vehicle.

<h3>What is the duration of vapor lock?</h3>
  • The heat builds up toward the gasoline lines as you shut down.
  • Fuel in the lines could begin to evaporate in the absence of airflow through the cowling.
  • Vapor lock typically occurs when you try to restart your aircraft after a sharp turn (shut down and restart after about 30 minutes).

To learn more about vapor lock visit:

brainly.com/question/28121377

#SPJ4

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Write a hypothesis about how the height of the cylinder affects the temperature of the water. Use the "if . . . then . . . becau
AnnyKZ [126]
The statement that can be used to answer this  question is:

"If the cylinder is brought higher then, its temperature when brought down becomes higher because a greater amount of potential energy is converted to thermal energy."

The potential energy is converted to thermal energy when the object is released the velocity becomes higher because of the acceleration due to gravity.
8 0
3 years ago
Read 2 more answers
The speed of a moving bullet can be deter-
Bad White [126]

Answer:

<em>v = 381 m/s</em>

Explanation:

<u>Linear Speed</u>

The linear speed of the bullet is calculated by the formula:

\displaystyle v=\frac{x}{t}

Where:

x = Distance traveled

t = Time needed to travel x

We are given the distance the bullet travels x=61 cm = 0.61 m. We need to determine the time the bullet took to make the holes between the two disks.

The formula for the angular speed of a rotating object is:

\displaystyle \omega=\frac{\theta}{t}

Where θ is the angular displacement and t is the time. Solving for t:

\displaystyle t=\frac{\theta}{\omega}

The angular displacement is θ=14°. Converting to radians:

\theta=14*\pi/180=0.2443\ rad

The angular speed is w=1436 rev/min. Converting to rad/s:

\omega = 1436*2\pi/60=150.3776\ rad/s

Thus the time is:

\displaystyle t=\frac{0.2443\ rad}{150.3776\ rad/s}

t = 0.0016 s

Thus the speed of the bullet is:

\displaystyle v=\frac{0.61}{0.0016}

v = 381 m/s

7 0
2 years ago
A pendulum has 895 J of potential energy at the highest point of its swing. How much kinetic energy will it have at the bottom o
LuckyWell [14K]

Newton's law of conservation states that energy of an isolated system  remains a constant. It can neither be created nor destroyed but can be transformed  from one form to the other.

Implying the above law of conservation of energy in the case of pendulum we can conclude that at the bottom of the swing the entire potential energy gets converted to kinetic energy. Also the potential energy is zero at this point.

Mathematically also potential energy is represented as

Potential energy= mgh

Where m is the mass of the pendulum.

g is the acceleration due to gravity

h is the height from the bottom z the ground.

At the bottom of the swing,the height is zero, hence the potential energy is also zero.

The kinetic energy is represented mathematically as

Kinetic energy= 1/2 mv^2

Where m is the mass of the pendulum

v is the velocity of the pendulum

At the bottom the pendulum has the maximum velocity. Hence the kinetic energy is maximum at the bottom.

Also as it has been mentioned energy can neither be created nor destroyed hence the entire potential energy is converted to kinetic energy at the bottom and would be equivalent to 895 J.

7 0
3 years ago
THREE-DIMENSIONAL THINKING
ANEK [815]

The presence of potential energy  between particles supports the shape of a heating curve.

<h2>Potential energy and heating curve</h2>

The existence of potential energy  between particles supports the shape of a heating curve because potential energy causes the heating curve flat as well as in curve form. The heating curves show how the temperature changes as a substance is heated up.

The potential energy of the molecules will increase anytime energy is being supplied to the system but the temperature is not increasing so when the heating curve go flat it means there is potential energy so we can conclude that the existence of potential energy  between particles supports the shape of a heating curve.

Learn more about heating curve here: brainly.com/question/11991469

Learn more: brainly.com/question/26153233

8 0
2 years ago
A block–spring system vibrating on a frictionless, horizontal surface with an amplitude of 7.0 cm has an energy of 14 J. If the
Bingel [31]

Answer:

E_T= 28J

Explanation:

The energy of Mass-Spring System the sum of the potential energy of the block plus the kinetic energy of the block:

E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} mv^2

Where:

\Delta x=Amplitude\hspace{3}or\hspace{3}d eformation\hspace{3} of\hspace{3} the\hspace{3} spring\\m=Mass\hspace{3}of\hspace{3}the\hspace{3}block\\k=Constant\hspace{3}of\hspace{3}the\hspace{3}spring\\v=Velocity\hspace{3}of\hspace{3}the\hspace{3}block

There are two cases, the first case is when the spring is compressed to its maximum value, in this case the value of the kinetic energy is zero, since there is no speed, so:

E_T=\frac{1}{2} k \Delta x^2\\\\14=\frac{1}{2} k7^2\\\\Solving\hspace{3} for\hspace{3} k\\\\k=\frac{28}{49} =\frac{4}{7}

The second case is when the block passes through its equilibrium position, in this case the elastic potential energy is zero since \Delta x=0, so:

E_T=\frac{1}{2} mv^2\\\\14=\frac{1}{2} mv^2\\\\Solving\hspace{3} for\hspace{3} v\\\\v^2=\frac{28}{m}

Now, let's find the energy of the system when the block is replaced by one whose mass is twice the mass of the original block using the previous data:

E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} m_2v^2

Where in this case:

m_2=New\hspace{3}mass=Twice\hspace{3} the\hspace{3} mass \hspace{3}of\hspace{3} the\hspace{3} original=2m

Therefore:

E_T=\frac{1}{2} (\frac{4}{7} ) (7^2)+\frac{1}{2} (2m)(\frac{28}{m_2})=\frac{1}{2} (\frac{4}{7} ) (7^2)+\frac{1}{2} (2m)(\frac{28}{2m})=14+14=28J

8 0
3 years ago
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