1km=1000m; 1hr=3600secs
1km/hr=1000/3600= 5/18m/sec
To convert km/hr into m/sec, multiply the number by 5 and then divide it by 18.
18kmh-1= 18•5=90
90/18=5
5ms-1
Because the effective charge of the nucleus increase from left to eight due to the increasing number of protons.
The greater charge pulls the electrons closer to the nucleus, decreasing the radius.
Answer:
the maximum deformation undergone by the spring = 47.46 cm
Explanation:
Using conservation of momentum:
where:
Then;
v = 19.375 m/s
However ; using conservation of energy to determine the maximum deformation undergone by the spring ; we have:
![\frac{1}{2} [m_Av_A^2 +m_Bv_B^2] =\frac{1}{2}[(m_A+m_B)v^2 + kx^2]](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20%5Bm_Av_A%5E2%20%2Bm_Bv_B%5E2%5D%20%3D%5Cfrac%7B1%7D%7B2%7D%5B%28m_A%2Bm_B%29v%5E2%20%2B%20kx%5E2%5D)
![[m_Av_A^2 +m_Bv_B^2] =[(m_A+m_B)v^2 + kx^2]](https://tex.z-dn.net/?f=%5Bm_Av_A%5E2%20%2Bm_Bv_B%5E2%5D%20%3D%5B%28m_A%2Bm_B%29v%5E2%20%2B%20kx%5E2%5D)
![[\frac{250*10^3}{9.81}*40^2 + \frac{550*10^3}{9.81}*10^2] =[ (\frac{800*10^3}{9.81} )*19.375^2 + 70 *10^6 \ * x^2]](https://tex.z-dn.net/?f=%5B%5Cfrac%7B250%2A10%5E3%7D%7B9.81%7D%2A40%5E2%20%2B%20%5Cfrac%7B550%2A10%5E3%7D%7B9.81%7D%2A10%5E2%5D%20%3D%5B%20%28%5Cfrac%7B800%2A10%5E3%7D%7B9.81%7D%20%29%2A19.375%5E2%20%2B%2070%20%2A10%5E6%20%5C%20%2A%20x%5E2%5D)
x = 0.4746 m
x = 47.46 cm
Thus, the maximum deformation undergone by the spring = 47.46 cm
#1 A) Long wavelength and high frequency.
#2) A) Brightness.
#3) D) They are used to examine things.
#4) B) Compression and visible light waves.
The correct answer is<span> measure the distance to a star.
It is</span> the semi-angle of inclination between two sight-lines to the star, as observed when the Earth is on opposite sides of the Sun in its orbit. This is used to measure the distance to closer stars.
