Answer:
F = 0.257 N
Explanation:
The magnitude of the electric force between the charges can be calculated by using Colomb's Law:
![F = \frac{kq_1q_2}{r^2}](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7Bkq_1q_2%7D%7Br%5E2%7D)
where,
F = Electric Force = ?
k = Colomb's Constant = 9 x 10⁹ Nm²/C²
q₁ = charge on first cork = 6 μC = 6 x 10⁻⁶ C
q₂ = charge on second cork = 4.3 μC = 4.3 x 10⁻⁶ C
r = distance between corks = 0.95 m
Therefore,
![F = \frac{(9\ x\ 10^9\ N.m^2/C^2)(6\ x\ 10^{-6}\ C)(4.3\ x\ ^{-6}\ C)}{(0.95\ m)^2}](https://tex.z-dn.net/?f=F%20%3D%20%5Cfrac%7B%289%5C%20x%5C%2010%5E9%5C%20N.m%5E2%2FC%5E2%29%286%5C%20x%5C%2010%5E%7B-6%7D%5C%20C%29%284.3%5C%20x%5C%20%5E%7B-6%7D%5C%20C%29%7D%7B%280.95%5C%20m%29%5E2%7D)
<u>F = 0.257 N</u>
Answer:
Part a)
![a= 0.32 m/s^2](https://tex.z-dn.net/?f=a%3D%200.32%20m%2Fs%5E2)
Part b)
![F_c = 3.6 N](https://tex.z-dn.net/?f=F_c%20%3D%203.6%20N)
Part c)
![F_c = 5.5 N](https://tex.z-dn.net/?f=F_c%20%3D%205.5%20N)
Explanation:
Part a)
As we know that the friction force on two boxes is given as
![F_f = \mu m_a g + \mu m_b g](https://tex.z-dn.net/?f=F_f%20%3D%20%5Cmu%20m_a%20g%20%2B%20%5Cmu%20m_b%20g)
![F_f = 0.02(10.6 + 7)9.81](https://tex.z-dn.net/?f=F_f%20%3D%200.02%2810.6%20%2B%207%299.81)
![F_f = 3.45 N](https://tex.z-dn.net/?f=F_f%20%3D%203.45%20N)
Now we know by Newton's II law
![F_{net} = ma](https://tex.z-dn.net/?f=F_%7Bnet%7D%20%3D%20ma)
so we have
![F_p - F_f = (m_a + m_b) a](https://tex.z-dn.net/?f=F_p%20-%20F_f%20%3D%20%28m_a%20%2B%20m_b%29%20a)
![9.1 - 3.45 = (10.6 + 7) a](https://tex.z-dn.net/?f=9.1%20-%203.45%20%3D%20%2810.6%20%2B%207%29%20a)
![a = \frac{5.65}{17.6}](https://tex.z-dn.net/?f=a%20%3D%20%5Cfrac%7B5.65%7D%7B17.6%7D)
![a= 0.32 m/s^2](https://tex.z-dn.net/?f=a%3D%200.32%20m%2Fs%5E2)
Part b)
For block B we know that net force on it will push it forward with same acceleration so we have
![F_c - F_f = m_b a](https://tex.z-dn.net/?f=F_c%20-%20F_f%20%3D%20m_b%20a)
![F_c = \mu m_b g + m_b a](https://tex.z-dn.net/?f=F_c%20%3D%20%5Cmu%20m_b%20g%20%2B%20m_b%20a)
![F_c = 0.02(7)(9.8) + 7(0.32)](https://tex.z-dn.net/?f=F_c%20%3D%200.02%287%29%289.8%29%20%2B%207%280.32%29)
![F_c = 3.6 N](https://tex.z-dn.net/?f=F_c%20%3D%203.6%20N)
Part c)
If Alex push from other side then also the acceleration will be same
So for box B we can say that Net force is given as
![F_p - F_f - F_c = m_b a](https://tex.z-dn.net/?f=F_p%20-%20F_f%20-%20F_c%20%3D%20m_b%20a)
![9.1 - 0.02(7)(9.8) - F_c = 7(0.32)](https://tex.z-dn.net/?f=9.1%20-%200.02%287%29%289.8%29%20-%20F_c%20%3D%207%280.32%29)
![F_c = 9.1 - 0.02 (7)(9.8) - 7(0.32)](https://tex.z-dn.net/?f=F_c%20%3D%209.1%20-%200.02%20%287%29%289.8%29%20-%207%280.32%29)
![F_c = 5.5 N](https://tex.z-dn.net/?f=F_c%20%3D%205.5%20N)
Answer: The block of 0.4 Kg travel the same distance that the block of
0.2 Kg
Explanation: Considering the second newton law, we have the following
F= m*a
F= P*sin (θ) where θ is the angle for the incline
so mg sin (θ)= m*a
a=g sin(θ)
both block have the same acceleration in the inclined plane so travel the same distance independent of its mass.
If a tennis player does not swing through, meaning they stop swinging the moment they make contact with the ball, they would lose the majority of their power. <em>The momentum that they had built up during the swing is lost the moment they stop swinging</em>, meaning that the ball is hit with a low amount of power.
<em>If the tennis player swings through the whole time they hit the ball, then they keep their momentum as they hit the ball.</em> There is a much higher power level when swinging through than if you were to stop your swing when you hit the ball.
Torque acting dowward = 6 x 0.5 = 3 Nm
Torque acting to the right = 5 x 1 = 5 Nm
5 - 3 = 2 Nm
inertia = 1/2 mr^2
0.5 x 10 x 1^2 = 5 kg-m^2
2/5 = alpha = 0.4 rad /s^2
Hope this helps