What is the equation that defines work

Write 3 – 4 sentences predicting the changes in electrostatic force, strong nuclear force, and the stability of a nucleus if the

mrs_skeptik [129]

Electrostatic force arises due the charges at rest.In a isolated atom there is three fundamental particles called electron,proton and neutron.the electron is negatively charged,proton is positively charged and the neutron is neutral.

Generally the force between electron and proton is attractive,there will be also repulsion between proton and proton.due to the strong nuclear forces atom main its stability In an stable atom the the neutron to proton ratio lies between 1 to 1.42.if the number of neutrons are increased more,the stability will be hampered resulting the decrease in nuclear force also.

As neutron is neutral,there will be no change in the electrostatic forces as there is no increment of charge.but when the neutron number is increased at that time the binding energy per nucleon is decreased which in turn affects the nuclear force which is a short range force.the nucleus becomes heavy and unstable which undergoes decay.

one thing we have to remember also that the nuclear force is spin dependent .so the change in neutron number affects the strong nuclear force which is the reason for its decrement.

8 0

3 years ago

A loaded ore car has a mass of 950 kg and rolls on rails with negligible friction. It starts from rest and is pulled up a mine s

Ierofanga [76]

Answer:

a). P=11.04kW

b). Pmax=11.38 kW

c). Wt=6423.166kJ

Explanation:

The power of the motor when the speed is constant is the work in a determinate time.

$P=\frac{W}{t}$

The work is the force the is applicated in a distance so

W=F*d

replacing:

$P=F*\frac{d}{t}$ and $\frac{d}{t}$ determinate distance in time is velocity so

a).

$P=F*v$

$F=m*a\\F=m*g*sin(33.5)$

$P=950kg*9.8\frac{m}{s^{2}}*sin(33.5)*2.15\frac{m}{s}\\P=11047.846 W\\P=11.0478 kW$

b).

The maximum power must the motor provide, is the maximum force with the maximum speed of the motor in this case

The first step is find the acceleration so

$vi=0\frac{m}{s} \\vf=2.15 \frac{m}{s}\\vf=vi+a*t\\vf-vi=a*t\\ a=\frac{vf-vi}{t}= a=\frac{2.15\frac{m}{s}-0\frac{m}{s}}{13s}\\a=0.1653 \frac{m}{s^{2}}$