First, let's calculate the total mechanical energy when the block is at rest and the spring is compressed 5 cm:
![\begin{gathered} ME=PE+KE\\ \\ ME=\frac{kx^2}{2}+\frac{mv^2}{2}\\ \\ ME=\frac{955\cdot0.05^2}{2}+0\\ \\ ME=1.194\text{ J} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20ME%3DPE%2BKE%5C%5C%20%5C%5C%20ME%3D%5Cfrac%7Bkx%5E2%7D%7B2%7D%2B%5Cfrac%7Bmv%5E2%7D%7B2%7D%5C%5C%20%5C%5C%20ME%3D%5Cfrac%7B955%5Ccdot0.05%5E2%7D%7B2%7D%2B0%5C%5C%20%5C%5C%20ME%3D1.194%5Ctext%7B%20J%7D%20%5Cend%7Bgathered%7D)
Now, let's use this total energy to calculate the velocity when the spring is compressed by 2.5 cm:
![\begin{gathered} ME=PE+KE\\ \\ 1.194=\frac{kx^2}{2}+\frac{mv^2}{2}\\ \\ 2.388=955\cdot0.025^2+1.7v^2\\ \\ 1.7v^2=2.388-0.597\\ \\ 1.7v^2=1.791\\ \\ v^2=\frac{1.791}{1.7}\\ \\ v^2=1.0535\\ \\ v=1.026\text{ m/s} \end{gathered}](https://tex.z-dn.net/?f=%5Cbegin%7Bgathered%7D%20ME%3DPE%2BKE%5C%5C%20%5C%5C%201.194%3D%5Cfrac%7Bkx%5E2%7D%7B2%7D%2B%5Cfrac%7Bmv%5E2%7D%7B2%7D%5C%5C%20%5C%5C%202.388%3D955%5Ccdot0.025%5E2%2B1.7v%5E2%5C%5C%20%5C%5C%201.7v%5E2%3D2.388-0.597%5C%5C%20%5C%5C%201.7v%5E2%3D1.791%5C%5C%20%5C%5C%20v%5E2%3D%5Cfrac%7B1.791%7D%7B1.7%7D%5C%5C%20%5C%5C%20v%5E2%3D1.0535%5C%5C%20%5C%5C%20v%3D1.026%5Ctext%7B%20m%2Fs%7D%20%5Cend%7Bgathered%7D)
Therefore the speed is 1.026 m/s.
Answer:
Energy converted = ![240000\,Joules = 240\, kJoules](https://tex.z-dn.net/?f=240000%5C%2CJoules%20%3D%20240%5C%2C%20kJoules)
Explanation:
Recall that Power is the rate at which energy is transferred therefore defined by the mathematical formula: ![Power\,=\,\frac{Energy\,transferred}{time}](https://tex.z-dn.net/?f=Power%5C%2C%3D%5C%2C%5Cfrac%7BEnergy%5C%2Ctransferred%7D%7Btime%7D)
Since the information on the power of the runner is given, as well as the time the energy conversion takes place, we can then use this equation to find how much energy is been converted. Notice that we just need to change the given time *10 minutes) into the appropriate units (seconds)to get the answer in SI units of energy (Joules). The conversion of 10 minutes into seconds is done by multiplying : 10 minutes * 60 seconds/minute = 600 seconds.
We use this then to find the energy converted by the runner:
![Power\,=\,\frac{Energy\,transferred}{time}\\400 \,W = \frac{E}{600\,sec} \\400 \,W * 600\,sec=E\\E=240000\,Joules = 240\, kJoules](https://tex.z-dn.net/?f=Power%5C%2C%3D%5C%2C%5Cfrac%7BEnergy%5C%2Ctransferred%7D%7Btime%7D%5C%5C400%20%5C%2CW%20%3D%20%5Cfrac%7BE%7D%7B600%5C%2Csec%7D%20%5C%5C400%20%5C%2CW%20%2A%20600%5C%2Csec%3DE%5C%5CE%3D240000%5C%2CJoules%20%3D%20240%5C%2C%20kJoules)
Answer:
A.
Explanation:
NEAR THE CENTER OF TECTONIC PLATES.
By
vector addition.
In fact, velocity is a vector, with a magnitude intensity, a direction and a verse, so we can't simply do an algebraic sum of the two (or more velocities).
First we need to decompose each velocity on both x- and y-axis (if we are on a 2D-plane), then we should do the algebraic sum of all the components on the x- axis and of all the components on the y-axis, to find the resultants on x- and y-axis. And finally, the magnitude of the resultant will be given by
![R= \sqrt{(R_x)^2+(R_y)^2}](https://tex.z-dn.net/?f=R%3D%20%5Csqrt%7B%28R_x%29%5E2%2B%28R_y%29%5E2%7D%20)
where Rx and Rx are the resultants on x- and y-axis. The direction of the resultant will be given by
![\tan \alpha = \frac{R_y}{R_x}](https://tex.z-dn.net/?f=%5Ctan%20%5Calpha%20%3D%20%20%5Cfrac%7BR_y%7D%7BR_x%7D%20)
where
![\alpha](https://tex.z-dn.net/?f=%5Calpha)
is its direction with respect to the x-axis.
The best position for the person would be outside, under a clear sky, standing up. He should do it sometime between sunset and sunrise, from a day before until a day after the moment of Full Moon.