Tutor profile: Youssef H.
Subject: Physics (Newtonian Mechanics)
A 40,000 kg train car travelling at 8 m/s towards another stationary train car. upon collision, they couple together and move along at 5 m/s. Find the mass of the second train car?
The main concept in this question is conservation of momentum; the summation of momenta of the two train cars before collision equals the momentum of the couple after collision. This can be mathematically represented as follows: $$m_1*v_1+m_2*v_2 = (m_1+m_2)*v_t$$ $$m_1, v_1 $$ and $$v_2$$ are given, and hence, the only unknown in the above equation is $$m_2$$ Substituting the given values for $$m_1, v_1 $$ and $$v_2$$: $$40,000*8+m_2*0 = (40,000+m_2)*5$$ $$320,000 = 200,000 + 5m_2$$ $$320,000 -200,000 = + 5m_2=120,000$$ $$m_2=120,000/5=24,000 kg$$
$$ f(x) = x^4 + x^2 - 8x$$ Find $$f''(2)$$.
The question asks for the second derivative of the given function at 2. First, we should get the first derivative. Then, we differentiate it. Lastly, we plug in 2 for x in the second derivative. $$f'(x) = 4x^3 + 2x - 8$$ $$f''(x) = 12x^2 + 2$$ $$f''(2) = 12(2)^2 + 2=48+2=50$$
$$ x + 2 y = 8 $$ $$ 2x - y = 6 $$ Quantity A Quantity B $$x $$ $$x $$ Given the above two equations, determine whether A is greater than B, B is greater than A, A and B are equal, or there is not enough information to determine.
By multiplying the second equation by 2 and summing up the two equations, $$ 5x = 20 $$, which means $$x = 4$$. By substituting in any of the two equations, $$y=2$$. Hence, quantity A is greater.
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