Answer:
The velocity is ![v = 4.76 \ m/s](https://tex.z-dn.net/?f=v%20%3D%204.76%20%5C%20m%2Fs)
Explanation:
From the question we are told that
The first distance is ![d_1 = 4.0 \ km = 4000 \ m](https://tex.z-dn.net/?f=d_1%20%20%3D%20%204.0%20%5C%20km%20%20%3D%20%204000%20%5C%20m)
The first speed is ![v_1 = 5.0 \ m/s](https://tex.z-dn.net/?f=v_1%20%3D%20%205.0%20%5C%20m%2Fs)
The second distance is ![d_2 = 1.0 \ km = 1000 \ m](https://tex.z-dn.net/?f=d_2%20%20%3D%20%201.0%20%5C%20km%20%20%3D%20%201000%20%5C%20m)
The second speed is ![v_2 = 4.0 \ m/s](https://tex.z-dn.net/?f=v_2%20%20%3D%20%204.0%20%5C%20m%2Fs)
Generally the time taken for first distance is
![t_1 = \frac{d_1 }{v_1 }](https://tex.z-dn.net/?f=t_1%20%3D%20%20%5Cfrac%7Bd_1%20%7D%7Bv_1%20%7D)
![t_1 = \frac{4000}{5}](https://tex.z-dn.net/?f=t_1%20%3D%20%20%5Cfrac%7B4000%7D%7B5%7D)
![t_1 = 800 \ s](https://tex.z-dn.net/?f=t_1%20%3D%20%20800%20%5C%20s)
The time taken for second distance is
![t_1 = \frac{d_2 }{v_2 }](https://tex.z-dn.net/?f=t_1%20%3D%20%20%5Cfrac%7Bd_2%20%7D%7Bv_2%20%7D)
![t_1 = \frac{1000}{4}](https://tex.z-dn.net/?f=t_1%20%3D%20%20%5Cfrac%7B1000%7D%7B4%7D)
![t_1 = 250 \ s](https://tex.z-dn.net/?f=t_1%20%3D%20%20250%20%5C%20s)
The total time is mathematically represented as
![t = t_1 + t_2](https://tex.z-dn.net/?f=t%20%3D%20%20t_1%20%2B%20t_2)
=> ![t = 800 + 250](https://tex.z-dn.net/?f=t%20%3D%20%20800%20%2B%20250)
=> ![t = 1050 \ s](https://tex.z-dn.net/?f=t%20%3D%20%201050%20%5C%20s)
Generally the constant velocity that would let her finish at the same time is mathematically represented as
![v = \frac{d_1 + d_2}{t }](https://tex.z-dn.net/?f=v%20%3D%20%20%5Cfrac%7Bd_1%20%2B%20d_2%7D%7Bt%20%7D)
=> ![v = \frac{4000 + 1000}{1050 }](https://tex.z-dn.net/?f=v%20%3D%20%20%5Cfrac%7B4000%20%2B%201000%7D%7B1050%20%7D)
=> ![v = 4.76 \ m/s](https://tex.z-dn.net/?f=v%20%3D%204.76%20%5C%20m%2Fs)
Answer:
a) The rotational inertia when it passes through the midpoints of opposite sides and lies in the plane of the square is 16.8 kg m²
b) I = 50.39 kg m²
c) I = 16.8 kg m²
Explanation:
a) Given data:
m = 0.98 kg
a = 4.14 * 4.14
The moment of inertia is:
![I=mr^{2} \\r=\frac{a}{2} \\I=m(a/2)^{2} \\I=\frac{ma^{2} }{4}](https://tex.z-dn.net/?f=I%3Dmr%5E%7B2%7D%20%5C%5Cr%3D%5Cfrac%7Ba%7D%7B2%7D%20%5C%5CI%3Dm%28a%2F2%29%5E%7B2%7D%20%5C%5CI%3D%5Cfrac%7Bma%5E%7B2%7D%20%7D%7B4%7D)
For 4 particles:
![I=4(\frac{ma^{2} }{4} )=ma^{2} =0.98*(4.14)^{2} =16.8kgm^{2}](https://tex.z-dn.net/?f=I%3D4%28%5Cfrac%7Bma%5E%7B2%7D%20%7D%7B4%7D%20%29%3Dma%5E%7B2%7D%20%3D0.98%2A%284.14%29%5E%7B2%7D%20%3D16.8kgm%5E%7B2%7D)
b) Distance from top left mass = x = a/2
Distance from bottom left mass = x = a/2
Distance from top right mass = x = √5 (a/2)
The total moment of inertia is:
![I=m(\frac{a}{2} )^{2} +m(\frac{a}{2} )^{2}+m(\frac{\sqrt{5a} }{2} )^{2}+m(\frac{\sqrt{5a} }{2} )^{2}=\frac{12ma^{2} }{4} =3ma^{2} =3*0.98*(4.14)^{2} =50.39kgm^{2}](https://tex.z-dn.net/?f=I%3Dm%28%5Cfrac%7Ba%7D%7B2%7D%20%29%5E%7B2%7D%20%2Bm%28%5Cfrac%7Ba%7D%7B2%7D%20%29%5E%7B2%7D%2Bm%28%5Cfrac%7B%5Csqrt%7B5a%7D%20%7D%7B2%7D%20%29%5E%7B2%7D%2Bm%28%5Cfrac%7B%5Csqrt%7B5a%7D%20%7D%7B2%7D%20%29%5E%7B2%7D%3D%5Cfrac%7B12ma%5E%7B2%7D%20%7D%7B4%7D%20%3D3ma%5E%7B2%7D%20%3D3%2A0.98%2A%284.14%29%5E%7B2%7D%20%3D50.39kgm%5E%7B2%7D)
c)
![I=2m(\frac{m}{\sqrt{2} } )^{2} =ma^{2} =0.98*(4.14)^{2} =16.8kgm^{2}](https://tex.z-dn.net/?f=I%3D2m%28%5Cfrac%7Bm%7D%7B%5Csqrt%7B2%7D%20%7D%20%29%5E%7B2%7D%20%3Dma%5E%7B2%7D%20%3D0.98%2A%284.14%29%5E%7B2%7D%20%3D16.8kgm%5E%7B2%7D)
The base unit of time in the metric and SI system is the second.
The most exact answer is 78.4J also in this kind of options we can say answer "d"