Advanced Placement (AP), 11.02.2020 22:54 nicog94

The mass of the bullet is 0.01 kg and the mass of the block is 0.99 kg. The bullet is fired into the block andbecomes imbedded in it. The bullet-block combination rises by 0.05 meters. What potential energy is stored inthe pendulum (bullet-block combination)?PE = mgh = a. 50 joules b. 5 joules c. 0.5 joules d. 0.05 joules2. What was the kinetic energy of the pendulum (bullet-block combination) "immediately" after the bullet becameimbedded (i. e. what was the MAXIMUM K. E. of the combination)?KE = PE= (conservation of energy -- no calculation required!) a. 50 joules b. 5 joules c. 0.5 joules d. 0.05 joules3. What was the velocity of the pendulum (bullet-block combination) "immediately" after the bullet becameimbedded (i. e. when its K. E. was maximum)?KE = 1mv2v2 = 2KEv = 2KE = 2 m m a. 0.5 m/sec b. 1.0 m/sec c. 1.5 m/sec d. 2.0 m/s4. What was the momentum of the pendulum (bullet-block combination) "immediately" after the bullet becameimbedded (i. e. when its K. E. was maximum)?p = mv = a. 0 kg m/sec b. 0.25 kg m/sec c. 0.5 kg m/sec d. 1.0 kg m/sec Ballistic Pendulum5/15/082Don RathjenExploratorium Teacher Institute3601 Lyon St., San Francisco, CA [email protected]© 2008 Exploratorium, www. exploratorium5. What was the momentum of the block before the bullet hit it?pblock= mv = (think about this -- mental calculation -- no calculator required!) = a. 0 kg m/sec b. 0.25 kg m/sec c. 0.5 kg m/sec d. 1.0 kg m/sec6. What was the momentum of the bullet before it hit the block?pbullet = (conservation of momentum -- no calculation required!) = a. 0 kg m/sec b. 0.25 kg m/sec c. 0.5 kg m/sec d. 1.0 kg m/sec7. What was the velocity of the bullet before it hit the block (i. e. the "muzzle velocity)?pbullet = mvv = pbullet = m a. 25 m/sec b. 50 m/sec c. 75 m/sec d. 100 m/sec e. 250 m/sec

Please please please show working and explanation for each question! Please I really need help

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chapters are in the book (this fleeting world)

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Answerpeople,vehicles,roadways.;

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To solve this integral, what you can do is split it up as follows: (1) ∫40(x2−4)dx=∫40(x2)dx−∫40(4)dx. to integrate 4 as an indefinite integral, a good rule of thumb is by an equation below for polynomials; . (2) ∫xa=xa+1a+1+c , where a is any number. c denotes some constant that is crucial when doing

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Answer: the interval on which the curve is concave up is (-∞, -1/2)====== $\displaystyle y = \int_{0}^x \frac{1}{1 + t + t^2} \; dt$ we find the interval where this curve is concave up by finding the interval in which the 2nd derivative is positive.differentiate the curve equation with respect to x $\begin{aligned} \frac{dy}{dx} & = \frac{d}{dx} \int_{0}^x \frac{1}{1 + t + t^2} \; dt \end{aligned}$ let the function f denote be the anti-derivative of the inside expression, i.e. a function such that $(d/dx)f(x) = 1/(1 + t + t^2)$ .then use the fundamental theorem of calculus: $\begin{aligned} \frac{dy}{dx} & = \frac{d}{dx} \int_{0}^x \frac{1}{1 + t + t^2} \; dt \\& = \frac{d}{dx} \big[ f(x) - f(0) \big] & & \text{\footnotesize apply ftc} \\ & = \frac{d}{dx}f(x) - 0 \\ & = \frac{1}{x + x + x^2} & & \text{\footnotesize since $\frac{d}{dx}f(x) = \frac{1}{1 + x + x^2}$ } \end{aligned}$ differentiate again to get second derivative. we can use chain rule with 1/x as the outside function and 1 + x + x² as the inside function $\begin{aligned} \frac{d}{dx} \left( \frac{dy}{dx} \right) & = \frac{d}{dx}\left(\frac{1}{1 + x + x^2} \right) \\ \frac{d^2y}{dx^2} & = \frac{-1}{\left(1 + x + x^2\right)^2} \cdot \tfrac{d}{dx}(1 + x + x^2) \\ & = \frac{-1}{\left(1 + x + x^2\right)^2} \cdot (1 + 2x) \\ & = \frac{-(1 + 2x)}{\left(1 + x + x^2\right)^2} \end{aligned}$ find critical numbers of the second derivative. these numbers are where the 2nd deriv. is undefined or zero.where the 2nd derivative is zero: set (d²y/dx²) to equal 0 and solve for x $\begin{aligned} 0 & = \frac{-(1 + 2x)}{\left(1 + x + x^2\right)^2} \\ 0 & = -(1+2x) \\ -1 & = 2x \\ x & = -1/2 \end{aligned}$ where the 2nd derivative is undefined: if the denominator is equal to zero, the derivative will be undefeined. attempting to solve the equation 1 + x + x² = 0results in a discriminant of $b^2 - 4ac = 1^2 - 4(1)(1) = -3$ meaning that the equal 1 + x + x² = 0 has no real solutions and therefore will never equal zero.======our only critical number is x = -1/2.this forms two intervals to test the sign for: (-∞, -1/2) and (-1/2, ∞)test values into the 2nd derivative for (-∞, -1/2)if we try x = -1, we get $\begin{aligned} \left.\frac{d^2y}{dx^2}\right|_{x = -1} & = \frac{-(1+2(-1))}{\left(1+(-1)+(-1)^2\right)^2} \\ & = \frac{-(1-2)}{1} \\ & = 1 \end{aligned}$ so the since the 2nd derivative of the curve is positive on the interval (-∞, -1/2), the curve is concave up on (-∞, -1/2)test values into the 2nd derivative for (-1/2, ∞)if we try x = 0, we get $\begin{aligned} \left.\frac{d^2y}{dx^2}\right|_{x = 0} & = \frac{-(1+2(0))}{\left(1+(0)+(0)^2\right)^2} \\ & = \frac{-(1)}{1} \\ & = -1 \end{aligned}$ so the since the 2nd derivative of the curve is positive on the interval (-∞, -1/2), the curve is concave down on (-∞, -1/2).the interval on which the curve is concave up is (-∞, -1/2)

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