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>
C) total linear momentum of the ball and cannon is conserved.
Basically it happens that in the beginning before there is a momentum acting on the two bodies, these are a unique system. Here the total momentum of the System is 0. However, when the positive momentum of the cannonball is added, the system will be immediately affected by a negative momentum which will pull back the cannon. Could this be extrapolated as a condition of Newton's third law.
Answer:
5 m/s
Explanation:
Here we can see there is no external force acted on a two masses when we consider the motion. If there is no external forces then momentum is conserved.
Initial momentum = Final momentum
0.5 × 10 = 1 × V
V = 5 m/s
Answer:
During charging by conduction, both objects acquire the same type of charge. If a negative object is used to charge a neutral object, then both objects become charged negatively. ... In this case, electrons are transferred from the neutral object to the positively charged rod and the sphere becomes charged positively.
Answer:
It will require a force of 1/5, answer A.
Explanation:
In the attached image we can see an example of an array of pulleys that will lift a 100 kg-f load.
If we analyze pulleys A,B, and C as in the image we can check a force in the cable of 20kg-f.
In the pulley D we have three forces of 20 kg-f each and those forces plus the forces in the pulley B, sum a total of 100 kg-f (60+40). This matches the mechanical advantage (100/5) = 20 kg-f
