Yes, it can change states of matter from heat. Like ice, ice changes it’s state of matter from the temperature it is in, which causes it to turn into a liquid.(another form of matter) Again with heat, if water is boiled to a certain temperature it converts to a gas. Now with cold. If water is at such a coke temperature it will turn to ice, a form of matter. Now when a gas is cooled down, the atoms comprising the gas have less energy to move around.
Answer:
Explanation:
Given that
Superelation= 0.08ft/ft
Given curve= u•
Curve junction factor= 0.13
DR= 5729.57795
R = 5729.57795/D
R = 5729.57795/4
R = 1432.4ft
c + f = V^2/gG
0.08 + 0.13 = V^2 / (32*1432.4)
V^2 = 9625.728 or V = 98 ft/sec
The designed speed for a project considered is a minimum value which means the highway design elements will meet or exceed the standards for the design speed. The maximum safe speed under normal condition is significantly greater than design speed
Answer:
B. Geosphere
A. Biosphere
A. Atmosphere
Explanation:
Volcanic eruptions occurs within the Geosphere. The geosphere is the rock solid earth make up of rocks that extends into the deep interior.
Magma formed deep within the crust rises to elevated parts and finally erupts as lava on the surface. When they cool, they solidify to form volcanic rocks.
The volcanic eruptions affects the biosphere significantly. The biosphere is the portion of the earth where all life forms exists.
Gases and ash spewed during an eruption into the atmosphere causes severe changes to weather and leads to pollution. The atmosphere is the gaseous envelope round the earth.
I think number 1 is incorrect I believe that answer is D. Number 6 I believe would be B. The rest seem to be correct.
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
