Physics, 30.10.2020 18:10 lola06032003

A 0.26 kg rock is thrown vertically upward from the top of a cliff that is 32 m high. When it hits the ground at the base of the cliff, the rock has a speed of 29 m/s. Assuming that air resistance can be ignore and using conservation of mechanical energy, find (a) the initial speed of the rock and (b) the greatest height of the rock as measured from the base of the cliff.

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Answer from: kdenormandie3122

a) The initial speed of the rock is approximately 14.607 meters per second.

b) The greatest height of the rock from the base of the cliff is 42.878 meters.

Explanation:

a) The rock experiments a free-fall motion, that is a vertical uniform accelerated motion due to gravity, in which air friction and effects of Earth's rotation. By Principle of Energy Conservation we have the following model:

$U_{g,1}+K_{1} = U_{g,2}+K_{2}$ (Eq. 1)

Where:

$U_{g,1}$ , $U_{g,2}$ - Initial and final gravitational potential energies, measured in joules.

$K_{1}$ , $K_{2}$ - Initial and final translational kinetic energies, measured in joules.

By definitions of gravitational potential and translational kinetic energies we expand and simplify the equation above:

$m\cdot g\cdot (y_{1}-y_{2})= \frac{1}{2}\cdot m\cdot (v_{2}^{2}-v_{1}^{2})$

$g\cdot (y_{1}-y_{2}) = \frac{1}{2}\cdot (v_{2}^{2}-v_{1}^{2})$ (Eq. 2)

Where:

- Gravitational acceleration, measured in meters per square second.

$y_{1}$ , $y_{2}$ - Initial and final height, measured in meters.

$v_{1}$ , $v_{2}$ - Initial and final speed of the rock, measured in meters per second.

If we know that $g = 9.807\,\frac{m}{s^{2}}$ , $y_{1} = 32\,m$ , $y_{2} = 0\,m$ and $v_{2} = 29\,\frac{m}{s}$ , then the equation is:

$\left(9.807\,\frac{m}{s^{2}} \right)\cdot (32\,m-0\,m) = \frac{1}{2}\cdot \left[\left(29\,\frac{m}{s} \right)^{2}-v_{1}^{2}\right]$

$313.824 = 420.5-0.5\cdot v_{1}^{2}$

$0.5\cdot v_{1}^{2} = 106.676$

$v_{1} \approx 14.607\,\frac{m}{s}$

The initial speed of the rock is approximately 14.607 meters per second.

b) We use (Eq. 1) once again and if we know that $g =9.807\,\frac{m}{s^{2}}$ , $y_{1} = 32\,m$ , $v_{1} \approx 14.607\,\frac{m}{s}$ and $v_{2} = 0\,\frac{m}{s}$ , then the equation is:

$\left(9.807\,\frac{m}{s^{2}} \right)\cdot (32\,m-y_{2}) = \frac{1}{2}\cdot \left[\left(0\,\frac{m}{s} \right)^{2}-\left(14.607\,\frac{m}{s} \right)^{2}\right]$

$313.824-9.807\cdot y_{2} = -106.682$

$9.807\cdot y_{2} = 420.506$

$y_{2} = 42.878\,m$

The greatest height of the rock from the base of the cliff is 42.878 meters.

Answer from: Quest

answer: the correct answer is 1.

explanation: there are various process in which a radioactive nuclei decays:

1) alpha decay: in this process, a heavier nuclei decays into lighter nuclei by releasing alpha particle. the mass number is reduced by 4 units.

$_a^z\textrm{x}\rightarrow _{a-2}^{z-4}+_2^4\alpha$

2) electron capture: in this decay process, a parent nuclei absorbs an electron and gets converted into a neutron. simply, a proton and an electron combines together to form a neutron. mass number does not change in this process.

$_a^z\textrm{x}+_{-1}^0\textrm{e}\rightarrow _{z-1}^a\textrm{y}$

3) gamma ray emission: it is a decay process in which an unstable nuclei gives excess energy by a spontaneous electromagnetic process. this decay releases $\gamma -radiations$ . this process does not change the mass number.

$_a^z\textrm{x}^*\rightarrow _a^z\textrm{x}+_0^0\gamma$

4) positron emission: it is a type of decay process, in which a proton gets converted to neutron and an electron neutrino. this is also known as $\beta ^+$ -decay. in this the mass number remains same.

$_a^z\textrm{x}\rightarrow _{z-1}^a\textrm{y}+_{+1}^0e$

from the above information, we see that only alpha-decay is producing a change in the mass number of the nucleus.

Answer from: Quest

The magnets would be | n s | and | s n|

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