Answer:
it has 15 horsepower to 300 horsepower and it weighs 2,906 to 3,131
Explanation:
its torque is 142 to 180
it has a inline 4 engine
there's a SE-R which has a turbo
Answer:
Part 1: It would be a straight line, current will be directly proportional to the voltage.
Part 2: The current would taper off and will have negligible increase after the voltage reaches a certain value. Graph attached.
Explanation:
For the first part, voltage and current have a linear relationship as dictated by the Ohm's law.
V=I*R
where V is the voltage, I is the current, and R is the resistance. As the Voltage increase, current is bound to increase too, given that the resistance remains constant.
In the second part, resistance is not constant. As an element heats up, it consumes more current because the free sea of electrons inside are moving more rapidly, disrupting the flow of charge. So, as the voltage increase, the current does increase, but so does the resistance. Leaving less room for the current to increase. This rise in temperature is shown in the graph attached, as current tapers.
Answer:
The power developed by engine is 167.55 KW
Explanation:
Given that

Mean effective pressure = 6.4 bar
Speed = 2000 rpm
We know that power is the work done per second.
So

We have to notice one point that we divide by 120 instead of 60, because it is a 4 cylinder engine.
P=167.55 KW
So the power developed by engine is 167.55 KW
Answer:
The flux (volume of water per unit time) through the hoop will also double.
Explanation:
The flux = volume of water per unit time = flow rate of water through the hoop.
The Flow rate of water through the hoop is proportional to the area of the hoop, and the velocity of the water through the hoop.
This means that
Flow rate = AV
where A is the area of the hoop
V is the velocity of the water through the hoop
This flow rate = volume of water per unit time = Δv/Δt =Q
From all the above statements, we can say
Q = AV
From the equation, if we double the area, and the velocity of the stream of water through the hoop does not change, then, the volume of water per unit time will also double or we can say increases by a factor of 2