Answer:
Explanation:
Area of electrodes, A = 2 cm x 2 cm = 4 cm²
Separation between electrodes, d = 1 mm
Voltage, V = 9 V
(a)
Let C is the capacitance between the electrodes
![C = \frac{\epsilon _{0}A}{d}](https://tex.z-dn.net/?f=C%20%3D%20%5Cfrac%7B%5Cepsilon%20_%7B0%7DA%7D%7Bd%7D)
![C = \frac{8.854\times 10^{-12}\times 4\times 10^{-4}}{1\times 10^{-3}}](https://tex.z-dn.net/?f=C%20%3D%20%5Cfrac%7B8.854%5Ctimes%2010%5E%7B-12%7D%5Ctimes%204%5Ctimes%2010%5E%7B-4%7D%7D%7B1%5Ctimes%2010%5E%7B-3%7D%7D)
C = 3.54 x 10^-12 F
Let q be the charge on each of the electrode
q = C x V
q = 3.54 x 10^-12 x 9 = 3.2 x 10^-11 C
(b)
As, the battery is disconnected the charge on the electrodes remains same.
(c)
As the battery is connected the voltage is same.
capacitance is change.
As the distance is doubled, the capacitance becomes half and the charge is also halved. q' = q/2 = 1.6 x 10^-11 C
Answer:
The magnitude of the resultant decreases from A+B to A-B
Explanation:
The magnitude of the resultant of two vectors is given by
![R=\sqrt{A^2 +B^2 +2AB cos \theta}](https://tex.z-dn.net/?f=R%3D%5Csqrt%7BA%5E2%20%2BB%5E2%20%2B2AB%20cos%20%5Ctheta%7D)
where
A is the magnitude of the first vector
B is the magnitude of the second vector
is the angle between the directions of the two vectors
In the formula, A and B are constant, so the behaviour depends only on the function
. The value of
are:
- 1 (maximum) when the angle is 0, so the magnitude of the resultant in this case is
![R=\sqrt{A^2 +B^2+2AB}=\sqrt{(A+B)^2}=A+B](https://tex.z-dn.net/?f=R%3D%5Csqrt%7BA%5E2%20%2BB%5E2%2B2AB%7D%3D%5Csqrt%7B%28A%2BB%29%5E2%7D%3DA%2BB)
- then it decreases, until it becomes 0 when the angle is 90 degrees, where the magnitude of the resultant is
![R=\sqrt{A^2 +B^2+0}=\sqrt{A^2+B^2}](https://tex.z-dn.net/?f=R%3D%5Csqrt%7BA%5E2%20%2BB%5E2%2B0%7D%3D%5Csqrt%7BA%5E2%2BB%5E2%7D)
- then it becomes negative, and continues to decrease, until it reaches a value of -1 when the angle is 180 degrees, and the magnitude of the resultant is
![R=\sqrt{A^2 +B^2-2AB}=\sqrt{(A-B)^2}=A-B](https://tex.z-dn.net/?f=R%3D%5Csqrt%7BA%5E2%20%2BB%5E2-2AB%7D%3D%5Csqrt%7B%28A-B%29%5E2%7D%3DA-B)
Answer:
A. It does not exhibit projectile motion and follows a straight path down the ramp.
Answer:
Why are continental rises and abyssal plains relatively rare in the Pacific? This is because the extensive system of trenches along the active margins of the Pacific, trap much of the sediments flowing off the continents, preventing them from building the broad, flat abyssal plains typical of the Atlantic ocean basins.
(a) The final velocity of the two vehicles if the collision was inelastic is 1.1 m/s.
(b) For the elastic collision, the final velocity of the car is 9.81 m/s backwards and the final velocity of the truck is 2.19 m/s forward.
(c) The time taken to exert the given force is 0.00625 m (s).
The given parameters;
- <em>Initial velocity of the car, u₁ = 12 m/s</em>
- <em>Initial velocity of the truck, u₂ = 0</em>
- <em>Mass of the car, = m</em>
- <em>Mass of the truck, = 10m</em>
(a) The final velocity of the two vehicles if the collision was inelastic is calculated as follows;
![m_1 u_1 + m_2u_2 = v(m_1+ m_2)\\\\12m + 10m(0) = v(m + 10m)\\\\12m = v(11m)\\\\v = \frac{12m}{11m} \\\\v = 1.1 \ m/s](https://tex.z-dn.net/?f=m_1%20u_1%20%2B%20m_2u_2%20%3D%20v%28m_1%2B%20m_2%29%5C%5C%5C%5C12m%20%2B%2010m%280%29%20%3D%20v%28m%20%2B%2010m%29%5C%5C%5C%5C12m%20%3D%20v%2811m%29%5C%5C%5C%5Cv%20%3D%20%5Cfrac%7B12m%7D%7B11m%7D%20%5C%5C%5C%5Cv%20%3D%201.1%20%5C%20m%2Fs)
(b) The final velocity of the two vehicles if the collision was elastic is calculated as follows;
![m_1 u_1 + m_2u_2 = m_1v_1 + m_2v_2\\\\\12m \ + \ 10m(0) = mv_1 + 10mv_2\\\\12m = m(v_1 + 10v_2)\\\\12 = v_1 + 10 v_2\ \ - --(1)](https://tex.z-dn.net/?f=m_1%20u_1%20%2B%20m_2u_2%20%3D%20m_1v_1%20%2B%20m_2v_2%5C%5C%5C%5C%5C12m%20%5C%20%2B%20%5C%2010m%280%29%20%3D%20mv_1%20%2B%2010mv_2%5C%5C%5C%5C12m%20%3D%20m%28v_1%20%2B%2010v_2%29%5C%5C%5C%5C12%20%3D%20v_1%20%2B%2010%20v_2%5C%20%5C%20-%20--%281%29)
Apply one-directional velocity equation:
![u_1 +v_1 = u_2 + v_2\\\\12 + v_1 = 0 + v_2\\\\12+ v_1 = v_2 \ \ --- (2)](https://tex.z-dn.net/?f=u_1%20%2Bv_1%20%3D%20u_2%20%2B%20v_2%5C%5C%5C%5C12%20%2B%20v_1%20%3D%200%20%2B%20v_2%5C%5C%5C%5C12%2B%20v_1%20%3D%20v_2%20%5C%20%5C%20---%20%282%29)
<em>Substitute </em><em>the value of </em>
<em> into equation (1);</em>
![12 = v_1 + 10(12 + v_1)\\\\12= v_1 + 120 + 10v_1\\\\12- 120 = 11v_1\\\\-108 = 11v_1\\\\v_1 = \frac{-108}{11} \\\\v_1 = -9.81 \ m/s\\\\](https://tex.z-dn.net/?f=12%20%3D%20v_1%20%2B%2010%2812%20%2B%20v_1%29%5C%5C%5C%5C12%3D%20v_1%20%2B%20120%20%2B%2010v_1%5C%5C%5C%5C12-%20120%20%3D%2011v_1%5C%5C%5C%5C-108%20%3D%2011v_1%5C%5C%5C%5Cv_1%20%3D%20%5Cfrac%7B-108%7D%7B11%7D%20%5C%5C%5C%5Cv_1%20%3D%20-9.81%20%5C%20m%2Fs%5C%5C%5C%5C)
Solve for
<em>;</em>
<em />
<em />
<em />
Thus, for the elastic collision, the final velocity of the car is 9.81 m/s backwards and the final velocity of the truck is 2.19 m/s forward.
(c)
The change in the momentum of the truck is calculated as;
![\Delta P = m_2(v_2 - u_2)\\\\\Delta P = 10m(2.19)\\\\\Delta P = 21.9m](https://tex.z-dn.net/?f=%5CDelta%20P%20%3D%20m_2%28v_2%20-%20u_2%29%5C%5C%5C%5C%5CDelta%20P%20%3D%2010m%282.19%29%5C%5C%5C%5C%5CDelta%20P%20%3D%2021.9m)
The time taken to exert the given force is calculated as follows;
![Ft = \Delta P\\\\t = \frac{\Delta P}{F} \\\\t = \frac{21.9 \ m}{3500} \\\\t = 0.00625 \ m (seconds)](https://tex.z-dn.net/?f=Ft%20%3D%20%5CDelta%20P%5C%5C%5C%5Ct%20%3D%20%5Cfrac%7B%5CDelta%20P%7D%7BF%7D%20%5C%5C%5C%5Ct%20%3D%20%5Cfrac%7B21.9%20%5C%20m%7D%7B3500%7D%20%5C%5C%5C%5Ct%20%3D%200.00625%20%5C%20m%20%28seconds%29)
Learn more about elastic and inelastic collision here: brainly.com/question/12497950