<span>The pure form of an element that is typically lustrous, malleable, and conducts heat and electricity is a/an</span> metal
Flow chart of a system is a diagrammatic representation of how that system performs it's functions.
<h3>The flow chart of internal combustion engine</h3>
The flow chart of internal combustion engine is the diagrammatic representation of how the engine works.
The inlet valve of the engine is the valve that allows for the entry of fuel-air mixture into the engine.
The exhaust valve of the engine is the allows for the outflow of used gases from the engine.
From the attached flow chart, Intake valves are opened to allow the flow of an air/fuel mixture into the engine's cylinders prior to compression and ignition, while exhaust valves open to permit the expulsion of exhaust gases from the combustion process after ignition has occurred.
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Answer: 588.9 m/s
Explanation:
Given that :
θ = 30°
Height, h = 3400m
Time, t = 10 seconds
From trigonometry ;
Tanθ = opposite / hypotenus
Tan 30 = 3400 / x
x tan 30 = 3400
0.5773502x = 3400
x = 3400 / 0.5773502
x = 5888.9727
Recall ;
Speed = Distance / time
Speed = 5888.9727 / 10
Speed = 588.897 m/s
Speed = 588.9 m/s
This question can be solved from the Kepler's law of planetary motion.
As per this law the square of time period of a planet is proportional to the cube of semi major axis.
Mathematically it can be written as 
⇒
Here K is the proportionality constant.
If
and
are the orbital periods of the planets and
and
are the distance of the planets from the sun, then Kepler's law can be written as-

⇒ 
Here we are asked to calculate the the distance of Saturn from sun.It can solved by comparing it with earth.
Let the distance from sun and orbital period of Saturn is denoted as
and
respectively.
Let the distance from sun and orbital period of earth is denoted as
and
respectively.
we are given that
we know that
1 AU and
1 year.
1 AU is the mean distance of earth from the sun which is equal to 150 million kilometre.
Hence distance of Saturn from sun is calculated as -
From Kepler's law as mentioned above-

=![[1 ]^{3} *\frac{[29.46]^{2} }{[1]^{2} } AU](https://tex.z-dn.net/?f=%5B1%20%5D%5E%7B3%7D%20%2A%5Cfrac%7B%5B29.46%5D%5E%7B2%7D%20%7D%7B%5B1%5D%5E%7B2%7D%20%7D%20AU)

⇒![R_{1} =\sqrt[3]{867.8916}](https://tex.z-dn.net/?f=R_%7B1%7D%20%3D%5Csqrt%5B3%5D%7B867.8916%7D)
=9.5386 AU [ans]
Answer:
yes, the potential difference across the terminals of the battery can be equal to its emf.
Explanation:
when the current in the battery is zero, meaning the current though, and hence the potential drop across the internal resistance is zero. This only happens when there is no load placed on the battery.