Answer:
a= (-g) from the moment the ball is thrown, until it stops in the air.
a = (0) when the ball stops in the air.
a = (g) since the ball starts to fall.
Explanation:
The acceleration is <em>(-g)</em> <em>from the moment the ball is thrown, until it stops in the air</em> because the movement goes in the opposite direction to the force of gravity. In the instant <em>when the ball stops in the air the acceleration is </em><em>(0)</em> because it temporarily stops moving. Then, <em>since the ball starts to fall, the acceleration is </em><em>(g)</em><em> </em>because the movement goes in the same direction of the force of gravity
current, the flow of electrons make the current
2 m/s/s means the velocity increases by 2 m/s every second.
Answer: h = 0.52m
Explanation:
Using the equation of out flow;
A1 × V1 = A2 ×V2
Where A1 = area of the first nozzle
A2 = area of the second nozzle
V1= velocity of flow out from the first nozzle
V2 = velocity of flow out from 2nd nozzle
But AV= area of nozzle × velocity of water = volume of water per second(m³/s).
Now we can set A×V = Area of nozzle × height of rise.
Henceb A1× h1 = A2 × h2 ( note the time cancel on both sides)
D1 = 20mm= 0.02m; h1 = 0.13m
D2 = 10mm = 0.01m; h2= ?
h2 = π(D1/2)²× h1 /π(D2/2)²
h2 = (0.02/2)² × 0.13/(0.01/2)²
= (0.01)² ×0.13 /(0.005)²
= 1.3 × 10^-5/(5 × 10^-3)²
= 1.3 × 10^-5/25 × 10^-6
= (1.3/25) 10^-5 × 10^6
= 0.052× 10
= 0.52m
