Answer:
Distillation, heat
Explanation:
Here in this question, we simply want to look at the best options that could fit in the gaps.
We have a mixture of liquids having boiling points which is far from each other.
Whenever we have a mixture of liquids with boiling points far away from each other, the best technique to use in separating them is to use distillation. That is why we have that as the best fit for the first missing gap.
Now, to get the liquids to separate from each other, we shall be needing the heating mantle for the application of heat. This ensures that the mixture is vaporized. After vaporization, the condensing tube will help to condense the vapor of each of the liquids once we reach the boiling point of either of the two.
Kindly note that the liquid with the lower temperature will evaporate first and will be first obtained. In fact after reaching a little above the boiling point of the lower boiling liquid, we can be sure that what we have left in the mixture pot is the second other liquid with the higher boiling point.
Answer:
Materials -> R1 = 100 Ω, NPN BJT 2N2222, R3 = light-sensitive resistor, R4 = 120 KΩ, 5 V battery, red led
Design -> please see the image attached
Explanation:
We use LTspice to simulate the circuit that consists of a voltage divider in the gate of the transistor to change the operation region of the BJT accordingly to the resistance of the light-sensitive resistor; when R3 (light-sensitive resistor) has low resistance the voltage of the gate and emitter is approximately the same and the transistor is in a Cut-off state, when R3 increases its resistance the voltage of the gate start to rise and as a result, there is in a Forward-active state; a diode is placed on the collector with a small resistance R1 and whenever the transistor is active the led is going to turn on.
Note: If you need more current for the led you could reduce the value of R4, this will produce that lower values of R3 activate the transistor, you can adjust the circuit just by changing the value of R4, therefore, I highly recommend to use a potentiometer for R4 in the range of 10-300 K.
Answer:
(a) Work done is 1.59642 kJ
(b) Useful power supplied = 0.1298 kW
Explanation:
(a) Work done = mass of crate × acceleration due to gravity × distance = 90.5 kg × 9.8 m/s^2 × 1.8 m = 1596.42 kgm^2/s^2 × 1 J/1 kgm^2/s^2 × 1 kJ/1000 J = 1.59642 kJ
(b) Useful power supplied = work done ÷ time = 1.59642 kJ ÷ 12.3 s = 0.1298 kJ/s × 1 kW/1 kJ/s = 0.1298 kW
Answer:
Check the explanation
Explanation:
Kindly check the attached images below to see the step by step explanation to the question above.
Answer:
Sorry for the delayed response- Right now I don't have time to give you the answer, but I really want to help so I'll try to phrase it in a easier way to understand things: Basically what you need to do for this problem is find the area of the base of the figure (which means length x width) and then you would simply find the volume of
by finding the length of each side of the figure, find the length of the figure, find the height of the figure and then find the radius.
Have an amazing day and I hope this can somewhat help :)