Answer:
A first-class lever: fulcrum is between input and output force; second-class lever: output force is between input force and fulcrum; third-class lever: input force is between fulcrum and output force
The state of matter that the particles move independently of one another with very little attraction is, I believe, gas
Answer:
Explanation:
a ) Let let the frictional force needed be F
Work done by frictional force = kinetic energy of car
F x 107 = 1/2 x 1400 x 35²
F = 8014 N
b )
maximum possible static friction
= μ mg
where μ is coefficient of static friction
= .5 x 1400 x 9.8
= 6860 N
c )
work done by friction for μ = .4
= .4 x 1400 x 9.8 x 107
= 587216 J
Initial Kinetic energy
= .5 x 1400 x 35 x 35
= 857500 J
Kinetic energy at the at of collision
= 857500 - 587216
= 270284 J
So , if v be the velocity at the time of collision
1/2 mv² = 270284
v = 19.65 m /s
d ) centripetal force required
= mv₀² / d which will be provided by frictional force
= (1400 x 35 x 35) / 107
= 16028 N
Maximum frictional force possible
= μmg
= .5 x 1400 x 9.8
= 6860 N
So this is not possible.
Answer:
add text fully or atleast add book name edition . solution is in picture. chk pictures ignore b part
Explanation:
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
