Potential energy is stored an object , where as kinetic energy is energy in a object during movement . Potential energy is what kinetic energy transfers into

6 0

4 years ago

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HELP!!

QveST [7]

Answer:

Total displacement 155 m

Explanation:

From the question, the formula to apply will be that of cosine rule where:

$c=\sqrt{a^2+b^2-2ab*cosC}$

where

a= 43 m , b=130m and c =? , <C=28° + 90° =118°

Substituting values in the equation as;

$c=\sqrt{43^2+130^2-2*43*130*cos118}$

$c=\sqrt{1849+16900-11180*-0.46947} \\\\c=\sqrt{1849+16900+5249} \\\\c=\sqrt{23998} \\\\\\c=155m$

6 0

3 years ago

Can someone help me please? I am stuck.

Mars2501 [29]

The distance traveled and the time it takes to travel are both going up at a steady rate. The distance goes from lap 1, lap 2, lap 3, lap 4. The time is increasing by a minute each time. When both the x and y scenario are increasing at a STEADY RATE the line will be linear. B displays a linear line.

8 0

4 years ago

Unpolarized light is incident upon two polarization filters that do not have their transmission axes aligned. If 21% of the ligh

Debora [2.8K]

Answer:

49.6°

Explanation:

$I_0$ = Unpolarized light

$I_2$ = Light after passing though second filter = $0.21I_0$

Polarized light passing through first filter

$I_1=\frac{I_0}{2}$

Polarized light passing through second filter

$I_2=\frac{I_0}{2}cos^2\theta\\\Rightarrow 0.21I_0=\frac{I_0}{2}cos^2\theta\\\Rightarrow cos^2\theta=\frac{0.21I_0}{\frac{I_0}{2}}\\\Rightarrow cos\theta=\sqrt{\frac{0.21I_0}{\frac{I_0}{2}}}\\\Rightarrow \theta=cos^{-1}\sqrt{\frac{0.21I_0}{\frac{I_0}{2}}}\\\Rightarrow \theta=cos^{-1}\sqrt{0.21\times 2}\\\Rightarrow \theta=cos^{-1}\sqrt{0.42}\\\Rightarrow \theta=49.6^{\circ}$

The angle between the two filters is 49.6°

5 0

4 years ago