Two piston hydraulic problems
1 answer:
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Answer:
The magnitude of the large object's momentum change is 3 kilogram-meters per second.
Explanation:
Under the assumption that no external forces are exerted on both the small object and the big object, whose situation is described by the Principle of Momentum Conservation:
(1)
Where:
,
- Initial and final momemtums of the small object, measured in kilogram-meters per second.
,
- Initial and final momentums of the big object, measured in kilogram-meters per second.
If we know that ,
and
, then the final momentum of the big object is:
The magnitude of the large object's momentum change is:
The magnitude of the large object's momentum change is 3 kilogram-meters per second.
Answer:
n=6.56×10¹⁵Hz
Explanation:
Given Data
Mass=9.1×10⁻³¹ kg
Radius distance=5.3×10⁻¹¹m
Electric Force=8.2×10⁻⁸N
To find
Revolutions per second
Solution
Let F be the force of attraction
let n be the number of revolutions per sec made by the electron around the nucleus then the centripetal force is given by
F=mω²r......................where ω=2π n
F=m4π²n²r...............eq(i)
as the values given where
Mass=9.1×10⁻³¹ kg
Radius distance=5.3×10⁻¹¹m
Electric Force=8.2×10⁻⁸N
we have to find n from eq(i)
n²=F/(m4π²r)
Answer:
<u>198785.714286 Joules</u>
Explanation:
<u>To find work:</u>
Now that the force and the distance are known, plug and chug:
Note that the question is asking for the magnitude of work, so the negative can be discarded as it is a directional component.
So, your answer is
<u>198785.714286 Joules</u>