<span>Because of the orbit of the earth and the sun and the moon. </span>
Explanation:
In my view, when the Object A is attracted to a Charged object B. Object B should be Negatively or Positively charged. So Object B should be the Opposite charged according to the Object B
Example =
If Object B is Negatively Charged, the Object A should be Positively Charged
If the Object B is Positively Charged, the Object A should be Negatively Charged
Sometimes it can Mix as a Neutral as well
Hope this Helps
Answer:
Part a)
![v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s](https://tex.z-dn.net/?f=v%20%3D%20-%288.33%5Chat%20j%20%2B%209.33%5Chat%20i%29%5Ctimes%2010%5E6%20m%2Fs)
Part b)
![E = 4.4 \times 10^{-13} J](https://tex.z-dn.net/?f=E%20%3D%204.4%20%5Ctimes%2010%5E%7B-13%7D%20J)
Explanation:
As per momentum conservation we know that there is no external force on this system so initial and final momentum must be same
So we will have
![m_1v_1 + m_2v_2 + m_3v_3 = 0](https://tex.z-dn.net/?f=m_1v_1%20%2B%20m_2v_2%20%2B%20m_3v_3%20%3D%200)
![(5 \times 10^{-27})(6 \times 10^6\hat j) + (8.4 \times 10^{-27})(4 \times 10^6\hat i) + (3.6 \times 10^{-27}) v = 0](https://tex.z-dn.net/?f=%285%20%5Ctimes%2010%5E%7B-27%7D%29%286%20%5Ctimes%2010%5E6%5Chat%20j%29%20%2B%20%288.4%20%5Ctimes%2010%5E%7B-27%7D%29%284%20%5Ctimes%2010%5E6%5Chat%20i%29%20%2B%20%283.6%20%5Ctimes%2010%5E%7B-27%7D%29%20v%20%3D%200)
![(30\hat j + 33.6\hat i)\times 10^6 + 3.6 v = 0](https://tex.z-dn.net/?f=%2830%5Chat%20j%20%2B%2033.6%5Chat%20i%29%5Ctimes%2010%5E6%20%2B%203.6%20v%20%3D%200)
![v = -(8.33\hat j + 9.33\hat i)\times 10^6 m/s](https://tex.z-dn.net/?f=v%20%3D%20-%288.33%5Chat%20j%20%2B%209.33%5Chat%20i%29%5Ctimes%2010%5E6%20m%2Fs)
Part b)
By equation of kinetic energy we have
![E = \frac{1}{2}m_1v_1^2 + \frac{1}{2}m_2v_2^2 + \frac{1}{2}m_3v_3^2](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B1%7D%7B2%7Dm_1v_1%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7Dm_2v_2%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7Dm_3v_3%5E2)
![E = \frac{1}{2}(5 \times 10^{-27})(6\times 10^6)^2 + \frac{1}{2}(8.4 \times 10^{-27})(4 \times 10^6)^2 + \frac{1}{2}(3.6 \times 10^{-27})(8.33^2 + 9.33^2) \times 10^{12}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B1%7D%7B2%7D%285%20%5Ctimes%2010%5E%7B-27%7D%29%286%5Ctimes%2010%5E6%29%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7D%288.4%20%5Ctimes%2010%5E%7B-27%7D%29%284%20%5Ctimes%2010%5E6%29%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7D%283.6%20%5Ctimes%2010%5E%7B-27%7D%29%288.33%5E2%20%2B%209.33%5E2%29%20%5Ctimes%2010%5E%7B12%7D)
![E = 9\times 10^{-14} + 6.72 \times 10^{-14} + 2.82\times 10^{-13}](https://tex.z-dn.net/?f=E%20%3D%209%5Ctimes%2010%5E%7B-14%7D%20%2B%206.72%20%5Ctimes%2010%5E%7B-14%7D%20%2B%202.82%5Ctimes%2010%5E%7B-13%7D%20)
![E = 4.4 \times 10^{-13} J](https://tex.z-dn.net/?f=E%20%3D%204.4%20%5Ctimes%2010%5E%7B-13%7D%20J)
Acceleration means speeding up, slowing down, or changing direction. The graph doesn't show anything about direction, so we just have to examine it for speeding up or slowing down ... any change of speed.
The y-axis of this graph IS speed. So the height of a point on the line is speed. If the line is going up or down, then speed is changing.
Sections a, c, and d are all going up or down. Section b is the only one where speed is not changing. So we can't be sure about b, because we don't know if the track may be curving ... the graph can't tell us that. But a, c, and d are DEFINITELY showing acceleration.
Answer:
a) v_average = 11 m / s, b) t = 0.0627 s
, c) F = 7.37 10⁵ N
, d) F / W = 35.8
Explanation:
a) truck speed can be found with kinematics
v² = v₀² - 2 a x
The fine speed zeroes them
a = v₀² / 2x
a = 22²/2 0.69
a = 350.72 m / s²
The average speed is
v_average = (v + v₀) / 2
v_average = (22 + 0) / 2
v_average = 11 m / s
b) The average time
v = v₀ - a t
t = v₀ / a
t = 22 / 350.72
t = 0.0627 s
c) The force can be found with Newton's second law
F = m a
F = 2100 350.72
F = 7.37 10⁵ N
.d) the ratio of this force to weight
F / W = 7.37 10⁵ / (2100 9.8)
F / W = 35.8
.e) Several approaches will be made:
- the resistance of air and tires is neglected
- It is despised that the force is not constant in time
- Depreciation of materials deformation during the crash