Answer:
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Answer:
72 m/s
Explanation:
D1 = 3 cm, v1 = 2 m/s
D2 = 0.5 cm,
Let the velocity at narrow end be v2.
By use of equation of continuity
A1 v1 = A2 v2
3.14 × 3 × 3 × 2 = 3.14 × 0.5 ×0.5 × v2
v2 = 72 m/s
Answer:
Explanation:
(b) The initial velocity is added to that due to acceleration by gravity. The velocity is increased linearly by gravity at the rate of 9.8 m/s². The average velocity of the pebble will be its velocity halfway through the 2-second time period.* That is, it will be ...
4 m/s + (9.8 m/s²)(2 s)/2 = 13.8 m/s . . . . average velocity
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(a) The distance covered in 2 seconds at an average velocity of 13.8 m/s is ...
d = vt
d = (13.8 m/s)(2 s) = 27.6 m
The water is about 27.6 m below ground.
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* We have chosen to make use of the fact that the velocity curve is linear, so the average velocity is half the sum of initial and final velocities:
vAvg = (vInit + vFinal)/2 = (vInit + (vInit +at))/2 = vInit +at/2
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If you work this in a straightforward way, you would find distance as the integral of velocity, then find average velocity from the distance and time.

1m is equivalent to 100cm
1m = 100cm
It is a physical change, as only the dimensions changed, not the chemical structure.