Physics, 30.12.2019 23:31 delaneymoses21

Aforce of 4.7 n acts on a 14 kg body initially at rest.
(a) compute the work done by the force in thefirst second.
1 j

(b) compute the work done by the force in the second second.
2 j

(c) compute the work done by the force in the third second.
3 j

(d) compute the instantaneous power due to the force at the end ofthe third second.
4 w

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Answer from: eve2055

(a). The work done by the force in the first second is 0.79 J.

(b). The work done by the force in the second second is 2.40 J.

(c). The work done by the force in the third second is 3.995 J.

(d). The instantaneous power due to the force at the end of the third second is 4.8 W.

Explanation:

Given that,

Force = 4.7 N

Mass of body = 14 kg

(a). We need to calculate the acceleration

Using formula of force

F=ma

$a=\dfrac{F}{m}$

Put the value into the formula

$a=\dfrac{4.7}{14}$

$a=0.34\ m/s^2$

We need to calculate the work done by the force in the first second

Using formula of work done

$W_{1}=F\cdot s$

$W_{1}=F\times(\dfrac{1}{2}at^2)$

$W_{1}=4.7\times\dfrac{1}{2}\times0.34\times1^2$

$W_{1}=0.79\ J$

(b). We need to calculate the work done by the force in the second second

Using formula of work done

$W_{2}=F\times(\dfrac{1}{2}at^2)-W_{1}$

$W_{2}=4.7\times\dfrac{1}{2}\times0.34\times2^2-0.79$

$W_{2}=2.406\ J$

(c). We need to calculate the work done by the force in the third second.

Using formula of work done

$W_{3}=F\times(\dfrac{1}{2}at^2)-(W_{2}+W_{1})$

$W_{3}=4.7\times\dfrac{1}{2}\times0.34\times3^2-(2.406+0.79)$

$W_{3}=4.7\times\dfrac{1}{2}\times0.34\times3^2-3.1$

$W_{3}=3.995\ J$

(d). We need to calculate the instantaneous power due to the force at the end of the third second.

Using formula of power

$P=F\times v$

$P=F\times at$

Put the value into the formula

$P=4.7\times\times0.34\times3$

$P=4.8\ W$

Hence, (a). The work done by the force in the first second is 0.79 J.

(b). The work done by the force in the second second is 2.40 J.

(c). The work done by the force in the third second is 3.995 J.

(d). The instantaneous power due to the force at the end of the third second is 4.8 W.

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 five characteristics of adolescence are biological growth and development, and undefined status, increased decision making, increased pressure and the search of self.

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