When a steadily flowing gas flows from a larger diameter pipe to a smaller diameter pipe the pressure does not decrease, it increases, do to the smaller volume.
Answer:
(4) 8.5 m/s
Explanation:
You add both the meters together and both the seconds together and then divide them both.
Two resistor of 2Ω in series parallel to resistor 5Ω in series to a 2Ω resistor. This configuration gives to us an equivalent resistor of 2.55Ω.
To solve this problem we have to use the rules of conection of resistor in series and parallel.
A resistor R1 in serie with other resistor R2 gives us an equivalent resistor Req= R1 + R2.
A resistor R1 in parallel with other resistor R2 gives us an equivalent resistor Req = R1.R2/R1+R2.
The circuit that show an arregement of resistor which we obtain a equivalent resistor of 2.5Ω from three resistor of 2Ω and 5Ω respectively is attached in the image:
the answer is a= 16m/s2
you can use the formula f=m*a to solve this bb :)