The efforts applied are 320Nm.
What is work or effort?
Work is when we apply force on body and displacement occur are called work .
W= Fs
where F is force and s is displacement .
displacement is 2-1.6
. = 0.4m
By applying the value in formula
We have given F =800N
W= 800× 0.4
W= 320Nm.
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Answer: It was on between 7 and 7.30 and also between 8 and 8.30 (when the line on the graph goes up only). So it was on for an hour or 60 minutes
Since speed (v) is in ft/sec, let's convert our diameters from inches to feet:
1) 5/8in = 0.625in
0.625in × 1ft/12in = 0.0521ft
2) 0.25in × 1ft/12in = 0.021ft
Equation:
![v = 4q \div ( {d}^{2} \pi) \: where \: q = flow \\ v = velocity \: (speed) \: and \: \\ d = diameter \: of \: pipe \: or \: hose \\ and \: \pi = 3.142](https://tex.z-dn.net/?f=v%20%3D%204q%20%5Cdiv%20%28%20%7Bd%7D%5E%7B2%7D%20%5Cpi%29%20%5C%3A%20where%20%5C%3A%20q%20%3D%20flow%20%5C%5C%20v%20%3D%20velocity%20%5C%3A%20%28speed%29%20%5C%3A%20and%20%5C%3A%20%20%5C%5C%20d%20%3D%20diameter%20%5C%3A%20of%20%5C%3A%20pipe%20%5C%3A%20or%20%5C%3A%20hose%20%5C%5C%20and%20%5C%3A%20%5Cpi%20%3D%203.142)
![we \: can \: only \: assume \:that \\ flow \: (q) \:stays \: same \: since \: it \\ isnt \: impeded \: by \: anything \\ thus \:it \: (q)\: stays \: the \: same \: \\ so \: 4q \: can \: be \: removed \: from \: \\ the \: equation](https://tex.z-dn.net/?f=we%20%5C%3A%20can%20%5C%3A%20only%20%5C%3A%20assume%20%5C%3Athat%20%5C%5C%20%20flow%20%5C%3A%20%28q%29%20%5C%3Astays%20%5C%3A%20same%20%5C%3A%20since%20%5C%3A%20it%20%5C%5C%20%20isnt%20%5C%3A%20impeded%20%5C%3A%20by%20%5C%3A%20%20anything%20%5C%5C%20thus%20%5C%3Ait%20%20%5C%3A%20%28q%29%5C%3A%20%20stays%20%5C%3A%20the%20%5C%3A%20same%20%5C%3A%20%20%5C%5C%20so%20%5C%3A%204q%20%5C%3A%20can%20%5C%3A%20be%20%5C%3A%20removed%20%5C%3A%20from%20%5C%3A%20%20%5C%5C%20the%20%5C%3A%20equation)
![then \: we \: can \: assume \: that \: only \\ v \: and \: d \: change \: leading \:us \: to > > \\ (v1 \times {d1}^{2} \pi) = (v2 \times {d2}^{2}\pi)](https://tex.z-dn.net/?f=then%20%5C%3A%20we%20%5C%3A%20can%20%5C%3A%20assume%20%5C%3A%20that%20%5C%3A%20only%20%5C%5C%20v%20%5C%3A%20and%20%5C%3A%20d%20%5C%3A%20change%20%5C%3A%20leading%20%5C%3Aus%20%5C%3A%20to%20%3E%20%20%3E%20%20%5C%5C%20%28v1%20%5Ctimes%20%7Bd1%7D%5E%7B2%7D%20%5Cpi%29%20%3D%20%28v2%20%20%5Ctimes%20%20%20%7Bd2%7D%5E%7B2%7D%5Cpi%29%20)
![both \: \pi \: will \: cancel \: each \: other \: out \: \\ as \: constants \:since \: one \: is \: on \\ each \: side \: of \: the \: =](https://tex.z-dn.net/?f=both%20%5C%3A%20%5Cpi%20%5C%3A%20will%20%5C%3A%20cancel%20%5C%3A%20each%20%5C%3A%20other%20%5C%3A%20out%20%5C%3A%20%20%5C%5C%20as%20%5C%3A%20constants%20%5C%3Asince%20%5C%3A%20one%20%5C%3A%20is%20%5C%3A%20on%20%5C%5C%20each%20%5C%3A%20side%20%5C%3A%20of%20%5C%3A%20the%20%5C%3A%20%20%3D%20%20%20)
![(v1 \times {d1}^{2}) = (v2 \ \times {d2}^{2}) \\ (7.0 \times {0.052}^{2}) = (v2 \times {0.021}^{2}) \\ divide \: both \: sides \: by \: {0.021}^{2} \\ to \: solve \: for \: v2 > >](https://tex.z-dn.net/?f=%28v1%20%20%5Ctimes%20%20%20%7Bd1%7D%5E%7B2%7D%29%20%3D%20%28v2%20%5C%20%5Ctimes%20%7Bd2%7D%5E%7B2%7D%29%20%5C%5C%20%287.0%20%5Ctimes%20%20%20%7B0.052%7D%5E%7B2%7D%29%20%3D%20%28v2%20%20%5Ctimes%20%20%20%7B0.021%7D%5E%7B2%7D%29%20%5C%5C%20divide%20%5C%3A%20both%20%5C%3A%20sides%20%5C%3A%20by%20%5C%3A%20%20%7B0.021%7D%5E%7B2%7D%20%5C%5C%20to%20%5C%3A%20solve%20%5C%3A%20for%20%5C%3A%20v2%20%3E%20%20%3E%20%20)
![v2 = (7.0)( {0.052}^{2} ) \div ( {0.021}^{2}) \\ v2 = (7.0)(.0027) \div (.00043) \\ v2 = 44 \: feet \: per \: second](https://tex.z-dn.net/?f=v2%20%3D%20%287.0%29%28%20%7B0.052%7D%5E%7B2%7D%20%29%20%5Cdiv%20%28%20%7B0.021%7D%5E%7B2%7D%29%20%20%5C%5C%20v2%20%3D%20%287.0%29%28.0027%29%20%5Cdiv%20%28.00043%29%20%5C%5C%20v2%20%3D%2044%20%5C%3A%20feet%20%5C%3A%20per%20%5C%3A%20second)
new velocity coming out of the hose then is
44 ft/sec
Answer:
DU = 20 Joules
Explanation:
Given the following data;
Heat absorbed, Q = 60J
Workdone, W = 40J
To find the internal energy;
We would use first law of thermodynamics given by the formula;
Change in internal energy (DU) = Q - W
where;
U is internal energy.
Q is the heat absorbed.
W is the work done.
Substituting into the equation, we have;
DU = 60 - 40
DU = 20 Joules
Answer:
The distance he moved the soil is 32.97 m.
Explanation:
Given;
force exerted on the wheelbarrow, F = 185 N
work done in moving the soil, w = 6.1 kJ = 6100
The work done in moving the soil is given as;
Work done = force x distance he moved the soil
W = F x d
d = W / F
d = (6100) / (185)
d = 32.97 m
Therefore, the distance he moved the soil is 32.97 m.