Tutor profile: Nicholas V.
Balance _NaBr + _Cl2 -> _NaCl + _Br2.
Look at each compound/element separately. Looking at Cl2 on the reactants side of the equation we have 2 Cl atoms present. Only 1 is present on the product side bonded with Na. Therefore we need 2 NaCl to balance our chlorine. Now that we have 2 Na atoms on the product side we must have 2 NaBr on the reactants side to balance our Sodium. Lastly we look at our Br. We satisfy having 2 Br on each side with the previous balancing of Na. The final formula should read 2NaBr + Cl2 -> 2NaCl + Br2.
Find the derivative of f(x) = 3x^2-ln(2x)+4.
The first term of the equation used the power rule. Bringing the exponent to the front and reducing the value by 1, the first term becomes 3*(2x)... or 6x. The second term involves taking the derivative of a logarithmic term. Remembering our rule, we take the derivative of the inside and divide this result by the original function. This gives 2/(2x)... or simply 1/x. The last term is simply a constant, the derivative of which is 0. Therefore f'(x) = 6x +1/x.
A block at rest of mass m = 5 kg is dropped from a building of height H = 25 m. When the block is at a height of 6 m, what is the velocity of the block. Assume air resistance is negligible.
Since air resistance can be neglected, we can solve for the velocity of the block at 6 m using conservation of energy. The initial energy of the system has to equal the final energy of the system. Therefore Ui + KEi = Uf + KEf. Since the block was initially at rest, the initial KE is 0, which reduces our equation to Ui = Uf + KEf. Mathematically speaking, mgHi = mgHf + (1/2)mv^2. Using our given mass and heights, we can plug into this equation and solve for the velocity when the block is at a height of 6 m. Doing so gives a result of 19.3 m/s.
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