Sam moves a box with a with a force of 400N a distance of 5 meters. How long did it take him to move the box if 20 Watts of powe
2 answers:
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Answer:
The speed of water must be expelled at 6.06 m/s
Explanation:
Neglecting any drag effects of the surrounding water we can assume the linear momentum in this case is conserves, that is, the total initial momentum of the octopus and the water kept in it cavity should be equal to the total final linear momentum. That's known as conservation of momentum, mathematically expressed as:
with Pi the total initial momentum and Pf the final total momentum. The total momentum is the sum of the momentums of the individual objects, in our case the octopus and the mass of water that will be expelled:
with Po the momentum of the octopus and Pw the momentum of expelled water. Linear momentum is defined as mass times velocity:
Note that initially the octopus has the water in its cavity and both are at rest before it sees the predator so :
We should find the final velocity of water if the final velocity of the octopus is 2.70 m/s, solving for :
The minus sign indicates the velocity of the water is opposite the velocity of the octopus.
Answer:
1.
2.
3.The results from part 1 and 2 agree when r = R.
Explanation:
The volume charge density is given as
We will investigate this question in two parts. First r < R, then r > R. We will show that at r = R, the solutions to both parts are equal to each other.
1. Since the cylinder is very long, Gauss’ Law can be applied.
The enclosed charge can be found by integrating the volume charge density over the inner cylinder enclosed by the imaginary Gaussian surface with radius ‘r’. The integration of E-field in the left-hand side of the Gauss’ Law is not needed, since E is constant at the chosen imaginary Gaussian surface, and the area integral is
where ‘h’ is the length of the imaginary Gaussian surface.
2. For r> R, the total charge of the enclosed cylinder is equal to the total charge of the cylinder. So,
3. At the boundary where r = R:
As can be seen from above, two E-field values are equal as predicted.