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1

force = acceleration * mass, hence acceleration will be $a=50N/22kg \approx 2.27m/s^2$ Distance it moves might be found by integration: $\int_0^{1.2}v(t)dt=\int_0^{1.2}atdt$, since speed $v(t)=at$ Answer to (B) then is 1/2*1.2*(2.27*1.2)=1.63, which seems pretty close to what you have got

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Its just a matter of what you use to call as displacement and as distance . I have seen the usage of: dx ds dr as the displacement too. Wikipedia says : The work done by a constant force of magnitude F on a point that moves a displacement (not distance) s in the direction of the force is the product, W = Fs. Note the usage of s ...

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You use the term "Power" but what you are asking about is "energy" or "work done". Power is an instantaneous measure of the rate of doing work which is the same as the rate of using or supplying energy. The standard relationship at constant Velocity V over a distance D in time T is: D = V x T which can be rearranged as .... V = D/T or .... T = D / V Two ...

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Lyman Alpha absorption systems The Lyman-$\alpha$ Forest and Gunn-Peterson troughs are two extremes on the scale of absorption features that are left by neutral hydrogen in intergalactic space. When ultraviolet light from a background source, typically a Quasar or a young, strongly star forming galaxy, travels through intergalactic space, it is ...

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