Drift velocity is equal to displacement of the moving object per unit time. The SI unit for displacement is meters while that of time is second. Hence the derived SI unit of velocity is meter per second. This also applies to electron mobility which relates to the displacement per unit time of a moving electron
c.charge due to the reaction process between the two
Answer:
A-Caclcuate the potential energy of the ball at that height
Explanation:
(a). Mass of the Body = 10 kg.
Height = 10 m.
Acceleration due to gravity = 9.8 m/s².
Using the Formula,Potential Energy = mgh
= 10 × 9.8 × 10 = 980 J.
(b). Now, By the law of the conservation of the Energy, Total amount of the energy of the system remains constant.
∴ Kinetic Energy before the body reaches the ground is equal to the Potential Energy at the height of 10 m.
∴ Kinetic Energy = 980 J.
(c). Kinetic Energy = 980 J.
Mass of the ball = 10 kg.
∵ K.E. = 1/2 × mv²
∴ 980 = 1/2 × 10 × v²
∴ v² = 980/5
⇒ v² = 196
∴ v = 14 m/s.
The speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.
<h3>Angular Speed of the pulley </h3>
The angular speed of the pulley after the block m1 fall through a distance, d, is obatined from conservation of energy and it is given as;
K.E = P.E
![\frac{1}{2} mv^2 + \frac{1}{2} I\omega^2 = mgh\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2(m_1R^2_2 + m_2R_2^2) + \frac{1}{2} \omega^2( \frac{1}{2} MR_1^2 + \frac{1}{2} MR_2^2) = m_1gd- \mu_km_2gd\\\\\frac{1}{2} m_2v_0^2 + \frac{1}{2} \omega^2[R_2^2(m_1 + m_2)+ \frac{1}{2} M(R_1^2 + R_2^2)] = gd(m_1 - \mu_k m_2)\\\\](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20mv%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20I%5Comega%5E2%20%3D%20mgh%5C%5C%5C%5C%5Cfrac%7B1%7D%7B2%7D%20m_2v_0%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%5Comega%5E2%28m_1R%5E2_2%20%2B%20m_2R_2%5E2%29%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%5Comega%5E2%28%20%5Cfrac%7B1%7D%7B2%7D%20MR_1%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20MR_2%5E2%29%20%3D%20m_1gd-%20%5Cmu_km_2gd%5C%5C%5C%5C%5Cfrac%7B1%7D%7B2%7D%20m_2v_0%5E2%20%2B%20%5Cfrac%7B1%7D%7B2%7D%20%5Comega%5E2%5BR_2%5E2%28m_1%20%2B%20m_2%29%2B%20%5Cfrac%7B1%7D%7B2%7D%20M%28R_1%5E2%20%2B%20R_2%5E2%29%5D%20%3D%20gd%28m_1%20-%20%5Cmu_k%20m_2%29%5C%5C%5C%5C)
![\frac{1}{2} m_2v_0 + \frac{1}{4} \omega^2[2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = gd(m_1 - \mu_k m_2)\\\\2m_2v_0 + \omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2)\\\\\omega^2 [2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)] = 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2\\\\\omega^2 = \frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)} \\\\\omega = \sqrt{\frac{ 4gd(m_1 - \mu_k m_2) - 2m_2v_0^2}{2R_2^2(m_1 + m_2) + M(R^2_1 + R^2_2)}} \\\\](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%20m_2v_0%20%2B%20%5Cfrac%7B1%7D%7B4%7D%20%5Comega%5E2%5B2R_2%5E2%28m_1%20%2B%20m_2%29%20%2B%20M%28R%5E2_1%20%2B%20R%5E2_2%29%5D%20%3D%20gd%28m_1%20-%20%5Cmu_k%20m_2%29%5C%5C%5C%5C2m_2v_0%20%2B%20%5Comega%5E2%20%5B2R_2%5E2%28m_1%20%2B%20m_2%29%20%2B%20M%28R%5E2_1%20%2B%20R%5E2_2%29%5D%20%3D%204gd%28m_1%20-%20%5Cmu_k%20m_2%29%5C%5C%5C%5C%5Comega%5E2%20%5B2R_2%5E2%28m_1%20%2B%20m_2%29%20%2B%20M%28R%5E2_1%20%2B%20R%5E2_2%29%5D%20%3D%20%204gd%28m_1%20-%20%5Cmu_k%20m_2%29%20-%202m_2v_0%5E2%5C%5C%5C%5C%5Comega%5E2%20%3D%20%5Cfrac%7B%204gd%28m_1%20-%20%5Cmu_k%20m_2%29%20-%202m_2v_0%5E2%7D%7B2R_2%5E2%28m_1%20%2B%20m_2%29%20%2B%20M%28R%5E2_1%20%2B%20R%5E2_2%29%7D%20%5C%5C%5C%5C%5Comega%20%3D%20%5Csqrt%7B%5Cfrac%7B%204gd%28m_1%20-%20%5Cmu_k%20m_2%29%20-%202m_2v_0%5E2%7D%7B2R_2%5E2%28m_1%20%2B%20m_2%29%20%2B%20M%28R%5E2_1%20%2B%20R%5E2_2%29%7D%7D%20%5C%5C%5C%5C)
Substitute the given parameters and solve for the angular speed;
<h3>Linear speed of the block</h3>
The linear speed of the block after travelling 0.7 m;
v = ωR₂
v = 35.39 x 0.03
v = 1.1 m/s
Thus, the speed of the block after it has moved the given distance away from the initial position is 1.1 m/s.
Learn more about conservation of energy here: brainly.com/question/24772394
