Answer:

Explanation:
The gravitational force between the proton and the electron is given by

where
G is the gravitational constant
is the proton mass
is the electron mass
r = 3 m is the distance between the proton and the electron
Substituting numbers into the equation,

The electrical force between the proton and the electron is given by

where
k is the Coulomb constant
is the elementary charge (charge of the proton and of the electron)
r = 3 m is the distance between the proton and the electron
Substituting numbers into the equation,

So, the ratio of the electrical force to the gravitational force is

So, we see that the electrical force is much larger than the gravitational force.
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
#1.
<em>Car </em>1<em> weighs </em>300 kilograms<em> and is moving right at </em>3 meters per second (m/s)
#2.
Law of conservation of momentum
momentum before collorion = momentim after collosion
MV + mv = MV' + mv'
1500x25+ 1000x5
37500 + 15000
Answer:
Workdone = 600 Kilojoules
Explanation:
Given the following data:
Time = 8 seconds
Power = 75,000 Watts
Distance = 58 m
To find the work done;
Power can be defined as the energy required to do work per unit time.
Mathematically, it is given by the formula;
Thus, work done is given by the formula;
Workdone = power * time
Workdone = 75000 * 8
Workdone = 600,000 = 600 KJ