coulomb's law
f repulsive = q1q2/4 pi epsilon nought r squared
0.1= q1q2/4 pi epsilon nought 0.911 squared
q1+q2= 7.50 µC
0.1= q1(7.5µC-q1)/4 pi epsilon nought 0.911 squared
solve for q1
When a body is moving in circular path at a distance r from its center its velocity at any instant will be directed tangentially. This is what we call tangential velocity. It is calculated as follows:
Vt = r(ω)
where ω = 5 rev/s (2π rad / rev) = 31.42 rad / s
<span>Vt = 0.24(31.42) = 7.5 m/s</span>
Answer:
<h3>
Young modulus of elasticity for a gas is</h3><h2>
<em>Zero</em></h2>
Explanation:
<em>As</em><em> </em><em>the</em><em> </em><em>gas</em><em> </em><em>doesn't</em><em> </em><em>undergo</em><em> </em><em>any</em><em> </em><em>chan</em><em>g</em><em>es</em><em> </em>
<em>so</em><em> </em><em>the</em><em> </em><em>young</em><em> </em><em>modules</em><em> </em><em>of</em><em> </em><em>gas</em><em> </em><em>is</em><em> </em><em>not</em><em> </em><em>defined</em><em>.</em><em>.</em><em>.</em>
I'm going to assume this is over a horizontal distance. You know from Newton's Laws that F=ma --> a = F/m. You also know from your equations of linear motion that v^2=v0^2+2ad. Combining these two equations gives you v^2=v0^2+2(F/m)d. We can plug in the given values to get v^2=0^2+2(20/3)0.25. Solving for v we get v=1.82 m/s!
<span>As the temperature goes down, the chaotic motion (velocity) of atoms start decreasing. If the temperature hits the absolute zero (which, in reality, is impossible to achieve), the atoms of the body would freeze, making the body still and stiff. One thing to note here is that the atoms do not get destroyed when the temperature reaches the absolute zero. That is the reason why the object can still be seen when it is at absolute zero.</span>
