Answer:
v₂ = 306.12 m/s
Explanation:
We know that the volume flow rate of the water or any in-compressible liquid remains constant throughout motion. Therefore, from continuity equation, we know that:
A₁v₁ = A₂v₂
where,
A₁ = Area of entrance pipe = πd₁²/4 = π(0.016 m)²/4 = 0.0002 m²
v₁ = entrance velocity = 3 m/s
A₂ = Area of nozzle = πd₂²/4 = π(0.005 m)²/4 = 0.0000196 m²
v₂ = exit velocity = ?
Therefore,
(0.0002 m²)(3 m/s) = (0.0000196 m²)v₂
v₂ = (0.006 m³/s)/(0.0000196 m²)
<u>v₂ = 306.12 m/s</u>
Answer:
320 N/m
Explanation:
From Hooke's law, we deduce that
F=kx where F is applied force, k is spring constant and x is extension or compression of spring
Making k the subject of formula then
Conversion
1m equals to 100cm
Xm equals 25 cm
25/100=0.25 m
Substituting 80 N for F and 0.25m for x then
Therefore, the spring constant is equal to 320 N/m
Answer:
The K.E is maximum when the child is at the vertical position and the P.E is maximum at the extreme deviated position from the vertical.
Explanation:
- A child is swinging on swing up and down has both kinetic and potential energy.
- The total mechanical energy of the system is conserved throughout the system. At any instant the total mechanical energy is given by,
E = K.E + P.E
- The K.E is maximum when the child is at the vertical position.
- The P.E is maximum at the extreme deviated position from the vertical.
- And when K.E is maximum P.E becomes minimum and vice versa as per the law of conservation of energy.
Answer:
Speed; v = 17 m/s
Explanation:
We are given;
Radius; r = 110m
Angle; θ = 15°
Now, we know that in circular motion,
v² = rg•tanθ
Thus,
v = √(rg•tanθ)
Where,
v is velocity
r is radius
g is acceleration due to gravity
θ is the angle
Thus,
v = √(rg•tanθ) = √(110 x 9.8•tan15)
v = √(288.85)
v = 17 m/s
