Between centre of curvature and principal focus.
A is the correct answer :D
False, because I doesn’t matter if there is noise pollution the child will still able to learn the way words work
Answer:
a. 4 m/s b. 0.2 V
Explanation:
a. Find the flow rate through a 3.00-cm-diameter pipe if the Hall voltage is 60.0 mV.
The hall voltage V = vBd where v = flow-rate, B = magnetic field strength = 0.500 T and d = diameter of pipe = 3.00 cm = 0.03 m
Since V = vBd
v = V/Bd given that V = 60.0 mV = 0.060 V, substituting the values of the other variables, we have
v = 0.060 V/(0.500 T × 0.03 m)
v = 0.060 V/(0.015 Tm)
v = 4 m/s
b. What would the Hall voltage be for the same flow rate through a 10.0-cm-diameter pipe with the same field applied?
Since the hall voltage, V = vBd and v = flow-rate = 4 m/s, B = magnetic field strength = 0.500 T and d' = diameter of pipe = 10.0 cm = 0.10 m
Substituting the variables into the equation, we have
V = vBd
V = 4 m/s × 0.500 T × 0.10 m
V = 0.2 V
Answer:
Spills out 10cm³ of wather.
Explanation:
The ball being submerged halfway in the full glass of water will spill a volume of water equal to the volume of the ball that is being submerged, because the entire volume of the submerged ball was previously occupied by water and is not compressed should move out of the glass.This volume will be 10cm³ since the total of the ball is 20cm³ and it is submerged halfway.