The relation between the refractive index and the optical density of the material is a direct relation.
This means that the more the refractive index is, the more optically dense the material is.
Based on the above, when checking the given choices, the refractive index that represents the most optically dense material would be the largest refractive index which is:
<span>d. 2.65</span>
Answer:
<h2>FOCAL</h2>
Explanation:
<em>The center of a lens is known as its optical center. </em><em>All light rays incident on a particular lens converges at a points a point known as the principal focus or the focal point after reflecting</em><em>. Note that all light incident on a reflecting surface must all converge at this focal point after reflection. </em>
The distance measured from the center of this lens to its principal focus (otherwise known as focal point) is known as the <em>focal length of the lens. </em>
<em>Based on the explanation above, it cam be concluded that the distance from the center of a lens to the location where parallel rays converge or appear to converge is called the</em><em> FOCAL</em><em> length.</em>
Answer:
e) 11 m/s
Explanation:
For a particle under the action of a conservative force, its mechanical energy at point 0 is must be equal to its mechanical energy at point 1:
![K_1+U_1=K_0+U_0\\\\\frac{mv_1^2}{2}+[(8.0\frac{J}{m^2})(x_1)^2+(2.0\frac{J}{m^4})(x_1)^4]=\frac{mv_0^2}{2}+[(8.0\frac{J}{m^2})(x_0)^2+(2.0\frac{J}{m^4})(x_0)^4]](https://tex.z-dn.net/?f=K_1%2BU_1%3DK_0%2BU_0%5C%5C%5C%5C%5Cfrac%7Bmv_1%5E2%7D%7B2%7D%2B%5B%288.0%5Cfrac%7BJ%7D%7Bm%5E2%7D%29%28x_1%29%5E2%2B%282.0%5Cfrac%7BJ%7D%7Bm%5E4%7D%29%28x_1%29%5E4%5D%3D%5Cfrac%7Bmv_0%5E2%7D%7B2%7D%2B%5B%288.0%5Cfrac%7BJ%7D%7Bm%5E2%7D%29%28x_0%29%5E2%2B%282.0%5Cfrac%7BJ%7D%7Bm%5E4%7D%29%28x_0%29%5E4%5D)
In 
the speed is given, so 
and 
. Replacing:
![\frac{mv_1^2}{2}+[(8.0\frac{J}{m^2})(1m)^2+(2.0\frac{J}{m^4})(1m)^4]=\frac{mv_0^2}{2}+[(8.0\frac{J}{m^2})0^2+(2.0\frac{J}{m^4})(0)^4]\\\frac{mv_1^2}{2}+8.0J+2.0J=\frac{mv_0^2}{2}\\\frac{mv_0^2}{2}=\frac{mv_1^2}{2}+10J\\v_0=\sqrt{\frac{2}{m}(\frac{mv_1^2}{2}+10J)}\\v_0=\sqrt{\frac{2}{0.2kg}(\frac{(0.2kg)(5\frac{m}{s})^2}{2}+10J)}\\v_0=11.18\frac{m}{s}](https://tex.z-dn.net/?f=%5Cfrac%7Bmv_1%5E2%7D%7B2%7D%2B%5B%288.0%5Cfrac%7BJ%7D%7Bm%5E2%7D%29%281m%29%5E2%2B%282.0%5Cfrac%7BJ%7D%7Bm%5E4%7D%29%281m%29%5E4%5D%3D%5Cfrac%7Bmv_0%5E2%7D%7B2%7D%2B%5B%288.0%5Cfrac%7BJ%7D%7Bm%5E2%7D%290%5E2%2B%282.0%5Cfrac%7BJ%7D%7Bm%5E4%7D%29%280%29%5E4%5D%5C%5C%5Cfrac%7Bmv_1%5E2%7D%7B2%7D%2B8.0J%2B2.0J%3D%5Cfrac%7Bmv_0%5E2%7D%7B2%7D%5C%5C%5Cfrac%7Bmv_0%5E2%7D%7B2%7D%3D%5Cfrac%7Bmv_1%5E2%7D%7B2%7D%2B10J%5C%5Cv_0%3D%5Csqrt%7B%5Cfrac%7B2%7D%7Bm%7D%28%5Cfrac%7Bmv_1%5E2%7D%7B2%7D%2B10J%29%7D%5C%5Cv_0%3D%5Csqrt%7B%5Cfrac%7B2%7D%7B0.2kg%7D%28%5Cfrac%7B%280.2kg%29%285%5Cfrac%7Bm%7D%7Bs%7D%29%5E2%7D%7B2%7D%2B10J%29%7D%5C%5Cv_0%3D11.18%5Cfrac%7Bm%7D%7Bs%7D)
The analogous formula for magnetic fields is the Ampere's law.
To find the answer, we need to know about the Ampere's law of magnetism.
<h3>What's Ampere's law of magnetism?</h3>
Ampere's law states that the close line integral of magnetic field around a current carrying loop is directly proportional to the current enclosed within it.
<h3>What's is the mathematical expression of Ampere's law?</h3>
Mathematically, Ampere's law is
B•dl= μ₀I
Thus, we can conclude that the analogous formula for gauss law is the Ampere's law in magnetism.
Learn more about the Ampere's law here:
brainly.com/question/17070619
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Answer:
the ball's velocity was approximately 0.66 m/s
Explanation:
Recall that we can study the motion of the baseball rolling off the table in vertical component and horizontal component separately.
Since the velocity at which the ball was rolling is entirely in the horizontal direction, it doesn't affect the vertical motion that can therefore be studied as a free fall, where only the constant acceleration of gravity is affecting the vertical movement.
Then, considering that the ball, as it falls covers a vertical distance of 0.7 meters to the ground, we can set the equation of motion for this, and estimate the time the ball was in the air:
0.7 = (1/2) g t^2
solve for t:
t^2 = 1.4 / g
t = 0.3779 sec
which we can round to about 0.38 seconds
No we use this time in the horizontal motion, which is only determined by the ball's initial velocity (vi) as it takes off:
horizontal distance covered = vi * t
0.25 = vi * (0.38)
solve for vi:
vi = 0.25/0.38 m/s
vi = 0.65798 m/s
Then the ball's velocity was approximately 0.66 m/s
