motion (1 Viewer)

stressedadfff · Jul 19, 2021

hey guys, could anyone send a solution to the following please

Screen Shot 2021-07-19 at 12.56.01 am.png

stressedadfff · Jul 19, 2021

also its for a, c and f please

CM_Tutor · Jul 19, 2021

(a) If upwards is positive, then the acceleration due to gravity (10 m/s²) is in the negative direction (ie. downwards). So, you have as your starting equation of motion

$\ddot{x} = -10$

which, after integration, should give

$v = -10t + 40$

and

$x = -5t^2 + 40t + 45$

(c) is asking for the time at which $v = 0$ and the height ( $x$ ) at that time.

(d) is asking you to show that the time ( $t > 0$ ) at which $x = 0$ is $t = 9$ and (e) asks for $v$ at that time.

(f) asks for $x$ at $t = 1$ and $t = 2$ .

(g) asks for average velocity in the second second, which is the change in displacement ( $\Delta x$ ) divided by the change in time ( $\Delta t = 2 - 1 = 1\ \text{s}$ ).

stressedadfff · Jul 19, 2021

thank you! could you please send a solution for the following

Screen Shot 2021-07-19 at 11.59.13 am.png

CM_Tutor · Jul 19, 2021

stressedadfff said:
thank you! could you please send a solution for the following
View attachment 31172

Radioactive decay is governed by the equation

$\cfrac{dM}{dt} = -kM$

where $M$ is the mass of the isotope present at time $t$ and $k > 0$ is a constant. The solution to this equation is

$M = M_0 e^{-kt}$

where $M_0$ is the mass of the isotope present at time $t=0$ .

So, in this case, we have $M_0 = 300\ \text{g}$ and that, at $t = 30\ \text{min}$ , $M = 208\ \text{g}$ , and so:

$\begin{align*} 208 &= 300 e^{-30k} \\ \cfrac{208}{300} &= e^{-30k} \\ \cfrac{52}{75} &= \cfrac{1}{e^{30k}} \qquad \text{as $e^{-x} = \cfrac{1}{e^x}$} \\ e^{30k} &= \cfrac{75}{52} \\ 30k &= \ln\cfrac{75}{52} \\ k &= \cfrac{1}{30}\ln\cfrac{75}{52} \end{align*}$

The half-life is the time it takes for half of a sample to decay, and thus occurs when $M = \cfrac{1}{2}M_0$

$\begin{align*} \text{Put $M = \cfrac{1}{2}M_0 = 150$ g:} \quad 150 &= 300 e^{-kt} \\ \cfrac{150}{300} &= e^{-kt} \\ \cfrac{1}{2} &= \cfrac{1}{e^{kt}} \\ e^{kt} &= 2 \\ kt &= \ln{2} \\ \cfrac{t}{30}\ln\cfrac{75}{52} &= \ln{2} \qquad \text{using the value of $k$ found above} \\ t &= \cfrac{30\ln{2}}{\ln\cfrac{75}{52}} \\ &\approx 56\ \text{min}\ 47\ \text{s (nearest second} \end{align*}$

$\begin{align*} \text{After $2\frac{1}{2}\ \text{h}\ = 150\ \text{min}$, the mass remaining is} \quad M &= 300 e^{-150k} \\ M &= \cfrac{300}{e^{150k}} \\ &= \cfrac{300}{e^{150 \times \cfrac{1}{30}\ln\cfrac{75}{52}}} \qquad \text{which you could just shove into a calculator...} \\ &= \cfrac{300}{e^{5\ln\cfrac{75}{52}}} \\ &= \cfrac{300}{e^{\ln\left(\cfrac{75}{52}\right)^5}} \\ &= \cfrac{300}{\left(\cfrac{75}{52}\right)^5} \\ &= 300\left(\cfrac{52}{75}\right)^5 \\ &= \cfrac{2^{12}\cdot 13^5}{3^4\cdot 5^8} \\ &= 48\cfrac{2\ 066\ 128}{31\ 640\ 625}\ \text{g} \\ &= 48.06529... \ \text{g} \\ &\approx 48\ \text{g (nearest gram} \end{align*}$

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motion (1 Viewer)

stressedadfff

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