The de Broglie wavelength is given by \(\displaystyle \lambda = \frac{h}{m v}\).

In this question, we're given a brief description of radiocarbon dating. We're given the amount of \(\displaystyle C^{14}\) that has been found in a fossil sample and we're asked to find the approximate age of the fossil.

First, let's briefly go over radiocarbon dating. This method essentially assumes that the amount of radioactive carbon within an organism remains fairly stable at any given time while the organism is alive. Moreover, this amount of radioactive carbon is related to the amount of radioactive carbon in the atmosphere. Once the organism dies, however, it ceases to gain any radioactive carbon; rather, the \(\displaystyle C^{14}\) that was present now begins to decay into \(\displaystyle C^{12}\). Thus, by measuring the amount of \(\displaystyle C^{14}\) in the organism and comparing it to the amount in the atmosphere, the age at which the organism died can be approximated.

Since we're dealing with the decay of \(\displaystyle C^{14}\), this is a radioactive decay problem. Recall that all radioactive decay reactions follow first-order rate kinetics. What this means is that the rate of decay is only dependent on the amount of radioactive material at any given instant. Hence, we can use the first-order rate equation.

\(\displaystyle A_{t}=A_{0}e^{-kt}\)

We can further rearrange this expression to isolate the variable for time.

\(\displaystyle \frac{A_t}{A_0}=e^{-kt}\)

\(\displaystyle t=-\frac{ln(\frac{A_t}{A_0})}{k}\)

In arriving at this expression, we see that we need to know the rate constant, \(\displaystyle k\), in order to solve for \(\displaystyle t\). To do this, we can use the equation that relates the half-life to the rate constant for a first-order process.

\(\displaystyle t_\frac{1}{2}=\frac{ln(2)}{k}\)

Rearranging, we can find the rate-constant.

\(\displaystyle k=\frac{ln(2)}{t_\frac{1}{2}}\)

\(\displaystyle k=\frac{0.693}{5730\: yr}=1.21\cdot 10^{-4}\: yr^{-1}\)

Now that we have the rate constant, we can plug this value into the previous expression to solve for \(\displaystyle t\).

\(\displaystyle t=-\frac{ln(\frac{0.35}{1})}{1.21\cdot 10^{-4}\: yr^{-1}}=8676\:years\)

For this problem, the first step is to find the rate constant for the decay reaction. Since we're given the half-life, we can calculate this value using the following equation.

\(\displaystyle k=\frac{ln(2)}{t_{\frac{1}{2}}}\)

\(\displaystyle k=\frac{0.693}{20\:min}=0.03465\:min^{-1}\)

Now, we can use the first order rate equation to find out how much drug will be left after the first time interval of \(\displaystyle 15\) minutes.

\(\displaystyle A_{t}=A_{o}e^{-kt}\)

\(\displaystyle A_{15\:min}=(30\:mg)e^{-(0.03465\:min^{-1})(15\:min)}\)

\(\displaystyle A_{15\:min}=17.84\:mg\)

So, after the first \(\displaystyle 15\) minutes, there will be \(\displaystyle 17.84\:mg\) of the drug in the patient's body. But from the question stem we're told that an additional \(\displaystyle 25\:mg\) of the drug is injected. Hence, there is now \(\displaystyle 42.84\:mg\) of the drug present.

With this new amount in mind, we'll need to calculate how much of the drug will be present after more time has elapsed. We can do this by using the same equation as before.

\(\displaystyle A_{10\:min}=(42.84\:mg)e^{-(0.03465\:min^{-1})(10\:min)}\)

\(\displaystyle A_{10\:min}=30.29\:mg\)

This is the final amount of drug that is expected to be present in the patient's body at that instant of time.

Finding the amount is a simple half life problem. Start with setting up what you know for any half-life problem.

\(\displaystyle 100 = 200e^{i*8.02}\), \(\displaystyle i\) is the rate of decay, 200 is the initial amount, and 100 is the half left after 8.02 days.

From there simplify. 

\(\displaystyle 0.5 = e^{8.02 * i}\). To solve this take the natural log of both sides so that 

\(\displaystyle ln (0.5) = 8.02i\)

\(\displaystyle i=-0.086\)

Now plug this into the equation and find what happens after 7 days.

\(\displaystyle S = 200e^{-0.086*7}=109.2g\)

Beta decay is when a neutron releases an electron and becomes a proton. Therefore, the atomic number goes up by one but the atomic mass remains the same. This results in xenon-131.

Kinetic energy is given by\(\displaystyle E = \frac{1}{2}mv^2\). We will begin by calculating the car's initial kinetic energy, in terms of the unknown mass of the car \(\displaystyle m\): 

\(\displaystyle E_{initial} = x = \frac{1}{2}m(30\frac{m}{s})^2\).

Next, we will calculate the final energy of the car, also in terms of the unknown mass of the car:

\(\displaystyle E_{final} = \frac{1}{2}m(35\frac{m}{s})^2\).

To find the ratio of the final to initial kinetic energy, we divide \(\displaystyle E_{final}\) by \(\displaystyle E_{initial}\). We see that this reduces to \(\displaystyle \frac{35^2}{30^2}\) with the both the mass \(\displaystyle m\) and \(\displaystyle \frac{1}{2}\) terms cancelling. \(\displaystyle 35^2/30^2 = 1.4\). Thus, the new kinetic energy is \(\displaystyle 1.4x\).

So in order to solve this problem, we first should think about where all the energy in the system is coming from and going. Specifically when you pull the spring back, the work you do is turned into potential energy stored in the spring. So:

\(\displaystyle W_{pulling spring} = PE\)

and the energy transforms into kinetic when you let go, but some of that energy is lost to friction so

\(\displaystyle PE = KE + W_{friction}\)    and \(\displaystyle PE_{spring} = \frac{1}{2} kx^2\)

So use the equation below, enter in your known quantities and then solve for the work done by friction.

\(\displaystyle \frac{1}{2} kx^2 = \frac{1}{2}mv^2 + W_{friction}\)

Plug in known values and solve.

\(\displaystyle 3.75=2.5+W_{friction}\)

\(\displaystyle 1.25J=W_{friction}\)

This is an example of conservation of energy. The car starts at the top of the ramp, at height \(\displaystyle h\). It has no velocity at this time since it is starting from a rest. Therefore its total energy is \(\displaystyle mgh\) where \(\displaystyle m\) is the mass of the car and \(\displaystyle g\) is the value of gravitational acceleration.

At the bottom of the loop, all of the potential energy will have been converted into kinetic energy.

As the car traverses the loop and rises above the ground, kinetic energy will be converted back into potential energy. The shape of the loop does not matter in this case -- only the vertical distance between the ground and the car.

In the tallest possible loop, all kinetic energy at the bottom is converted to potential energy at the top. This is the maximum height the car can reach -- there is no additional energy left to continue climbing a taller loop. Therefore, the potential energy at the top of the tallest loop we can build is equal to the kinetic energy at the bottom of the loop. But we have already noted that the kinetic energy at the bottom of the loop is equal to the potential energy at the top of the ramp.

Therefore, we set \(\displaystyle mgh_{ramp} = mgh_{loop}\). We see that \(\displaystyle m\) and \(\displaystyle g\) cancel, and we are left with  \(\displaystyle h_{ramp} = h_{loop}\). In other words, the tallest loop you can build is equal to the height of whatever ramp you select. In this example, the tallest loop we can build is \(\displaystyle 25cm\). We do not need to know the specific values of \(\displaystyle m\) or \(\displaystyle g\).

Power is the rate of energy transfer. To raise a \(\displaystyle 600 kg\) object \(\displaystyle 100 m\), a total of \(\displaystyle mgh\) or (\(\displaystyle (600 kg)(9.81 \frac{m}{s^2})(100 m) = 589 kJ\) is required. To find the power in Watts (\(\displaystyle \frac{J}{s}\)), we divide the total energy required by the time over which the energy must be transferred:

\(\displaystyle P=\frac{589 kJ}{20s} = 29.4 kW\)

We know that:

\(\displaystyle V_{horizontal}=33\frac{m}{s}\)

\(\displaystyle V_{vertical}=7\frac{m}{s}\)

and we are looking for the maximum height (vertical displacement) this person can obtain, so we aren't concerned with \(\displaystyle V_{horizontal}\).

We can apply the conservation of energy:

\(\displaystyle \Delta E=0\)

\(\displaystyle U_{grav}=KE\)

\(\displaystyle mgh=\frac{1}{2}mv_{vertical}^2\)

Masses cancel, so 

\(\displaystyle gh=\frac{1}{2}v_{vertical}^2\)

Solve for \(\displaystyle h\):

\(\displaystyle h=\frac{1}{2}\frac{v_{vertical}^2}{g}\)

\(\displaystyle g=10 \frac{m}{s^2}\)  (rounded to simplify our calculations)

so let's plug in what we know

\(\displaystyle h=\frac{1}{2}*\frac{7^2}{10}=2.45m\). This is our final answer.

Let's first write down the information we are given:

\(\displaystyle m=15kg\)

\(\displaystyle KE_{initial}=1750J\)

\(\displaystyle KE_{maxheight}=65J\)

In order to solve this problem we must apply the conservation of energy, which states \(\displaystyle \Delta E=0\) since no friction.

This means that as the project reaches its max height energy is converted from Kinetic Energy (energy of motion) to potential gravitational energy (based off of height).

We can subtract \(\displaystyle KE_{maxheight}\) from \(\displaystyle KE_{initial}\) to get the \(\displaystyle PE_{grav}\) at its max height

\(\displaystyle PE_{grav}\)=\(\displaystyle 1750J-65J=1685J\)

\(\displaystyle PE_{grav}\)\(\displaystyle =mgh\)

so we can solve for the height

\(\displaystyle h=\frac{PE_{grav}}{mg}\)

\(\displaystyle =\frac{1685}{(15*10)}\)

where \(\displaystyle g=10\frac{m}{s^2}\)

therefore \(\displaystyle h=11.233m\)