Thank you for the message my friend, yoU have a good day as well :)
Answer:
b. 0.034
Explanation:
The heat transfer coefficient of a material (U-value) is equal to the reciprocal of its R-value, therefore:

where
R is the R-value of the material
For the insulator in this problem,
R = 29
Substituting into the equation, we find the heat transfer coefficient:

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
Answer:
Im pretty sure its number 2
Explanation:
Explanation:
Yes, in order to determine whether two geometric figures are identical or not we tend to rotate one of the figure clockwise or anti clock wise mentally. However, in clockwise rotation larger the angel more will be the time taken for instance, it will take longer to rotate 120° than to rotate 40°. Whereas in anti clockwise ration it will be vice versa.