Physics, 25.01.2020 05:31 bvaughn6477

Aheavy-duty stapling gun uses a 0.179 kg metal rod that rams against the staple to eject it. the rod is attached and pushed by a stiff spring called a ram spring (k = 37107 n/m). the mass of this spring may be ignored. squeezing the handle of the gun first compresses. the ram spring is compressed by 3.20 10-2 m from its unstrained length and then releases from rest. assuming that the ram spring is oriented vertically and is still compressed by 1.35 10-2 m when the downward-moving ram hits the staple, find the speed of the ram at the instant of contact.

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Answer from: Serenitybella

v = 13.22 m/s

Explanation:

mass of rod (M) = 0.179 kg

spring constant (k) = 37107 N/m

initial compression (Ei) = 3.2 x 10^{-2} m

final compression (Ef) = 1.35 x 10^{-2} m

acceleration due to gravity (g) = 9.8 m/s^{2}

from the conservation of energy, the total energy in the system before impact = the total energy in the system after impact

$0.5.mv^{2} + mg(hf) + 0.5k(Ef)^{2} = 0.5.mu^{2} + mg(hi) + 0.5k(Ei)^{2}$

where

m = mass = 0.179 kgv = final speed at the instant of contactu = initial speed = 0 since it was initially at resthf = final heightHi = initial heightEf = final compression = 1.35 x 10^{-2} mEi = initial compression = 3.2 x 10^{-2} mk = spring constant = 37107 N/mg = acceleration due to gravity = 9.8 m/s^{2}since u = 0, the equation now becomes

$0.5.mv^{2} + mg(hf) + 0.5k(Ef)^{2} = mg(hi) + 0.5k(Ei)^{2}$

now we rearrange the equation above to make v the subject of the formula

$v= \sqrt{\frac{k(Ei^{2}-Ef^{2})}{m} + 2g(hi - hf) }$

hi - hf = change in height = change in extension = $(3.2x10^{-2}) -(1.35x10^{-2}) = 0.0185 m$ now substituting all required values into the equation above we have

$v= \sqrt{\frac{37107((3.2x10^{-2})^{2} -(1.35x10^{-2})^{2} )}{0.179}+2x 9.8 x(0.0185) }$

v = 13.22 m/s

Answer from: Quest

the answer is b

explanation:

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.

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Aheavy-duty stapling gun uses a 0.179 kg metal rod that rams against the staple to eject it. the rod...

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