Answer:
(d) Negative.
Explanation:
let's test each at a time.
(a) It can't be 0, because cup would slide back other wise.
(b) Positive, well if forward is positive, than the work done against the forward acceleration must be negative , so it can't be positive.
(c) Equal to non-conservative work done by the car's engine.
well no, because work done by car's engine dosen't go all of it into getting car to move, so it can't be that.
(d) negative, this look like it, because work that friction does must be nagative to counteract positive thrust of car which is positive and in forward direction.
(d) this can't be true.
So the answer is (d) negative.
Answer:
2240.92365 m/s
Explanation:
= Mass of electron = ![9.11\times 10^{−31}\ kg](https://tex.z-dn.net/?f=9.11%5Ctimes%2010%5E%7B%E2%88%9231%7D%5C%20kg)
= Speed of electron = ![5.7\times 10^7\ m/s](https://tex.z-dn.net/?f=5.7%5Ctimes%2010%5E7%5C%20m%2Fs)
= Neutrino has a momentum = ![7.3\times 10^{-24}\ kg m/s](https://tex.z-dn.net/?f=7.3%5Ctimes%2010%5E%7B-24%7D%5C%20kg%20m%2Fs)
M = total mass = ![2.34\times 10^{-26}\ kg](https://tex.z-dn.net/?f=2.34%5Ctimes%2010%5E%7B-26%7D%5C%20kg)
In the x axis as the momentum is conserved
![Mv_x=m_1v_1\\\Rightarrow v_x=\dfrac{m_1v_1}{M}\\\Rightarrow v_x=\dfrac{9.11\times 10^{−31}\times 5.7\times 10^7}{2.34\times 10^{-26}}\\\Rightarrow v_x=2219.10256\ m/s](https://tex.z-dn.net/?f=Mv_x%3Dm_1v_1%5C%5C%5CRightarrow%20v_x%3D%5Cdfrac%7Bm_1v_1%7D%7BM%7D%5C%5C%5CRightarrow%20v_x%3D%5Cdfrac%7B9.11%5Ctimes%2010%5E%7B%E2%88%9231%7D%5Ctimes%205.7%5Ctimes%2010%5E7%7D%7B2.34%5Ctimes%2010%5E%7B-26%7D%7D%5C%5C%5CRightarrow%20v_x%3D2219.10256%5C%20m%2Fs)
In the y axis
![Mv_y=p_2\\\Rightarrow v_y=\dfrac{p_2}{M}\\\Rightarrow v_y=\dfrac{7.3\times 10^{-24}}{2.34\times 10^{-26}}\\\Rightarrow v_y=311.96581\ m/s](https://tex.z-dn.net/?f=Mv_y%3Dp_2%5C%5C%5CRightarrow%20v_y%3D%5Cdfrac%7Bp_2%7D%7BM%7D%5C%5C%5CRightarrow%20v_y%3D%5Cdfrac%7B7.3%5Ctimes%2010%5E%7B-24%7D%7D%7B2.34%5Ctimes%2010%5E%7B-26%7D%7D%5C%5C%5CRightarrow%20v_y%3D311.96581%5C%20m%2Fs)
The resultant velocity is
![R=\sqrt{v_x^2+v_y^2}\\\Rightarrow R=\sqrt{2219.10256^2+311.96581^2}\\\Rightarrow R=2240.92365\ m/s](https://tex.z-dn.net/?f=R%3D%5Csqrt%7Bv_x%5E2%2Bv_y%5E2%7D%5C%5C%5CRightarrow%20R%3D%5Csqrt%7B2219.10256%5E2%2B311.96581%5E2%7D%5C%5C%5CRightarrow%20R%3D2240.92365%5C%20m%2Fs)
The recoil speed of the nucleus is 2240.92365 m/s
Answer:
Explanation:
A 40kg child throw stone of 0.5kg
At a direction of 5m/s
Recoil can be calculated using recoil of a gun formula
m_1•v_1 + m_2•v_2
m_1•v_1 = -m_2•v_2
The negative sign show that the momentum of the boy is directed oppositely to that of the stone
m_1 Is mass of boy
v_1 is the recoil velocity of the boy
m_2 is mass of stone
v_2 is the velocity of stone
Then,
m_1•v_1 = -m_2•v_2
40•v_1 = -0.5 × 5
40•v_1 = -2.5
v_1 = -2.5 / 40
v_1 = -0.0625 m/s
The recoil velocity of the boy is 0.0625 m/s