Answer: True
Explanation:
As the spring is compressed, it acumulates energy, and the spring "wants to release that energy". This acumulated energy, (potential energy) is called "elastic potential energy" because of the elastical nature of the spring, that when compressed it wants to return to the original shape. So the sentence is true
Answer:
![q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C](https://tex.z-dn.net/?f=q%3D10%5E%7B-6%7D%28e%5E%7B-4000t%7D-4e%5E%7B-1000t%7D%2B3%29C)
Explanation:
Given that
, we use Kirchhoff's 2nd Law to determine the sum of voltage drop as:
![E(t)=\sum{Voltage \ Drop}\\\\L\frac{d^2q}{dt^2}+R\frac{dq}{dt}+\frac{1}{C}q=E(t)\\\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+\frac{1}{0.25\times10^{-6}}q=12\\\\\frac{d^2q}{dt^2}+5000\frac{dq}{dt}+4000000q=12\\\\m^2+5000m+4000000=0\\\\(m+4000)(m+1000)=0\\\\m=-4000 \ or \ m=-1000\\\\q_c=c_1e^{-4000t}+c_2e^{-1000t}](https://tex.z-dn.net/?f=E%28t%29%3D%5Csum%7BVoltage%20%5C%20Drop%7D%5C%5C%5C%5CL%5Cfrac%7Bd%5E2q%7D%7Bdt%5E2%7D%2BR%5Cfrac%7Bdq%7D%7Bdt%7D%2B%5Cfrac%7B1%7D%7BC%7Dq%3DE%28t%29%5C%5C%5C%5C%5C%5C%5Cfrac%7Bd%5E2q%7D%7Bdt%5E2%7D%2B5000%5Cfrac%7Bdq%7D%7Bdt%7D%2B%5Cfrac%7B1%7D%7B0.25%5Ctimes10%5E%7B-6%7D%7Dq%3D12%5C%5C%5C%5C%5Cfrac%7Bd%5E2q%7D%7Bdt%5E2%7D%2B5000%5Cfrac%7Bdq%7D%7Bdt%7D%2B4000000q%3D12%5C%5C%5C%5Cm%5E2%2B5000m%2B4000000%3D0%5C%5C%5C%5C%28m%2B4000%29%28m%2B1000%29%3D0%5C%5C%5C%5Cm%3D-4000%20%20%5C%20or%20%5C%20m%3D-1000%5C%5C%5C%5Cq_c%3Dc_1e%5E%7B-4000t%7D%2Bc_2e%5E%7B-1000t%7D)
#To find the particular solution:
![Q(t)=A,\ Q\prime(t)=0,Q\prime \prime(t)=0\\\\0+0+4000000A=12\\\\A=3\times10^{-6}\\\\Q(t)=3\times10^{-6},\\\\q=q_c+Q(t)\\\\q=c_1e^{-4000t}+c_2e^{-1000t}+3\times10^{-6}\\\\q\prime=-4000c_1e^{-4000t}-1000c_2e^{-1000t}\\q\prime(0)=0\\\\-4000c_1-1000c_2=0\\c_1+c_2+3\times10^{-6}=0\\\\#solving \ simultaneously\\\\c_1=10^{-6},c_2=-4\times10^{-6}\\\\q=10^{-6}e^{-4000t}-4\times10^{-6}e^{-1000t}+3\times10^{-6}\\\\q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C](https://tex.z-dn.net/?f=Q%28t%29%3DA%2C%5C%20Q%5Cprime%28t%29%3D0%2CQ%5Cprime%20%5Cprime%28t%29%3D0%5C%5C%5C%5C0%2B0%2B4000000A%3D12%5C%5C%5C%5CA%3D3%5Ctimes10%5E%7B-6%7D%5C%5C%5C%5CQ%28t%29%3D3%5Ctimes10%5E%7B-6%7D%2C%5C%5C%5C%5Cq%3Dq_c%2BQ%28t%29%5C%5C%5C%5Cq%3Dc_1e%5E%7B-4000t%7D%2Bc_2e%5E%7B-1000t%7D%2B3%5Ctimes10%5E%7B-6%7D%5C%5C%5C%5Cq%5Cprime%3D-4000c_1e%5E%7B-4000t%7D-1000c_2e%5E%7B-1000t%7D%5C%5Cq%5Cprime%280%29%3D0%5C%5C%5C%5C-4000c_1-1000c_2%3D0%5C%5Cc_1%2Bc_2%2B3%5Ctimes10%5E%7B-6%7D%3D0%5C%5C%5C%5C%23solving%20%5C%20simultaneously%5C%5C%5C%5Cc_1%3D10%5E%7B-6%7D%2Cc_2%3D-4%5Ctimes10%5E%7B-6%7D%5C%5C%5C%5Cq%3D10%5E%7B-6%7De%5E%7B-4000t%7D-4%5Ctimes10%5E%7B-6%7De%5E%7B-1000t%7D%2B3%5Ctimes10%5E%7B-6%7D%5C%5C%5C%5Cq%3D10%5E%7B-6%7D%28e%5E%7B-4000t%7D-4e%5E%7B-1000t%7D%2B3%29C)
Hence the charge at any time, t is ![q=10^{-6}(e^{-4000t}-4e^{-1000t}+3)C](https://tex.z-dn.net/?f=q%3D10%5E%7B-6%7D%28e%5E%7B-4000t%7D-4e%5E%7B-1000t%7D%2B3%29C)
Answer:
Explanation:
As the current in the ire is towards right and the charge particle is above the wire, the direction of magnetic field due to the current carrying wire is perpendicularly outwards to the plane of paper. It is calculated by the Maxwell's right hand thumb rule. Now by using the Fleming's left hand rule, the direction of force is upwards.
The metal’s feely moving sea of electrons