At the player's maximum height, their velocity is 0. Recall that
![{v_f}^2-{v_i}^2=2a\Delta y](https://tex.z-dn.net/?f=%7Bv_f%7D%5E2-%7Bv_i%7D%5E2%3D2a%5CDelta%20y)
which tells us the player's initial velocity
is
![0^2-{v_i}^2=-2g(0.65\,\mathrm m)\implies v_i=3.6\dfrac{\rm m}{\rm s}](https://tex.z-dn.net/?f=0%5E2-%7Bv_i%7D%5E2%3D-2g%280.65%5C%2C%5Cmathrm%20m%29%5Cimplies%20v_i%3D3.6%5Cdfrac%7B%5Crm%20m%7D%7B%5Crm%20s%7D)
The player's height at time
is given by
![y=v_it-\dfrac g2t^2](https://tex.z-dn.net/?f=y%3Dv_it-%5Cdfrac%20g2t%5E2)
so we find their airtime to be
![0.65\,\mathrm m=\left(3.6\dfrac{\rm m}{\rm s}\right)t-\dfrac g2t^2\implies t=0.36\,\mathrm s](https://tex.z-dn.net/?f=0.65%5C%2C%5Cmathrm%20m%3D%5Cleft%283.6%5Cdfrac%7B%5Crm%20m%7D%7B%5Crm%20s%7D%5Cright%29t-%5Cdfrac%20g2t%5E2%5Cimplies%20t%3D0.36%5C%2C%5Cmathrm%20s)
Answer:
82.7 kg
Explanation:
the mass of the boxer remains unchanged, this is because mass is a measure of the quantity of matter in an object irrespective of its location and the gravitational force acting at its location. this means mass is independent of the gravitational acceleration hence it remains the same 82.7 kg. its unit is in kilograms (Kg).
Answer:
![a=2.9\ m/sec^2](https://tex.z-dn.net/?f=a%3D2.9%5C%20m%2Fsec%5E2)
Explanation:
<u>Net Forces and Acceleration</u>
The second Newton's Law relates the net force
acting on an object of mass m with the acceleration a it gets. Both the net force and the acceleration are vector and have the same direction because they are proportional to each other.
![\vec F_r=m\vec a](https://tex.z-dn.net/?f=%5Cvec%20F_r%3Dm%5Cvec%20a)
According to the information given in the question, two forces are acting on the swimming student: One of 256 N pointing to the south and other to the west of 104 N. Since those forces are not aligned, we must add them like vectors as shown in the figure below.
The magnitude of the resulting force
is computed as the hypotenuse of a right triangle
![|F_r|=\sqrt{256^2+104^2}](https://tex.z-dn.net/?f=%7CF_r%7C%3D%5Csqrt%7B256%5E2%2B104%5E2%7D)
![|F_r|=276.32\ Nw](https://tex.z-dn.net/?f=%7CF_r%7C%3D276.32%5C%20Nw)
The acceleration can be obtained from the formula
![F_r=ma](https://tex.z-dn.net/?f=F_r%3Dma)
Note we are using only magnitudes here
![\displaystyle a=\frac{F_r}{m}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20a%3D%5Cfrac%7BF_r%7D%7Bm%7D)
![\displaystyle a=\frac{276.32Nw}{95.3Kg}](https://tex.z-dn.net/?f=%5Cdisplaystyle%20a%3D%5Cfrac%7B276.32Nw%7D%7B95.3Kg%7D)
![\boxed{a=2.9\ m/sec^2}](https://tex.z-dn.net/?f=%5Cboxed%7Ba%3D2.9%5C%20m%2Fsec%5E2%7D)
The needle on a compass always points in the direction of magnetic north because of the magnetic poles of earth. the compass is essentially a magnet itself, so the southern pole of the compass is attracted to the northern pole of earth.