A red apple absorbs all colors of visible light except red, so red light
is the only light left to bounce off of the apple toward our eyes.
(This is a big part of the reason that we call it a "red" apple.)
Here's how the various items on the list make out when they hit the apple:
<span>Red . . . . . reflected
Orange . . absorbed
Yellow . . . </span><span><span>absorbed
</span>Green . </span><span><span>. . absorbed
</span>Blue . . </span><span><span>. . absorbed
</span>Violet .</span><span> . . absorbed</span>
<span>Black . . . no light; not a color
White . . . has all colors in it</span>
Explanation:
Area of ring 
Charge of on ring 
Charge on disk
![\begin{aligned}d v &=\frac{k d q}{\sqrt{x^{2}+a^{2}}} \\&=2 \pi-k \frac{a d a}{\sqrt{x^{2}+a^{2}}} \\v(1) &=2 \pi c k \int_{0}^{R} \frac{a d a}{\sqrt{x^{2}+a^{2}}} \cdot_{2 \varepsilon_{0}}^{2} R \\&=2 \pi \sigma k[\sqrt{x^{2}+a^{2}}]_{0}^{2} \\&=\frac{2 \pi \sigma}{4 \pi \varepsilon_{0}}[\sqrt{z^{2}+R^{2}}-(21)] \\&=\frac{\sigma}{2}(\sqrt{2^{2}+R^{2}}-2)\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7Dd%20v%20%26%3D%5Cfrac%7Bk%20d%20q%7D%7B%5Csqrt%7Bx%5E%7B2%7D%2Ba%5E%7B2%7D%7D%7D%20%5C%5C%26%3D2%20%5Cpi-k%20%5Cfrac%7Ba%20d%20a%7D%7B%5Csqrt%7Bx%5E%7B2%7D%2Ba%5E%7B2%7D%7D%7D%20%5C%5Cv%281%29%20%26%3D2%20%5Cpi%20c%20k%20%5Cint_%7B0%7D%5E%7BR%7D%20%5Cfrac%7Ba%20d%20a%7D%7B%5Csqrt%7Bx%5E%7B2%7D%2Ba%5E%7B2%7D%7D%7D%20%5Ccdot_%7B2%20%5Cvarepsilon_%7B0%7D%7D%5E%7B2%7D%20R%20%5C%5C%26%3D2%20%5Cpi%20%5Csigma%20k%5B%5Csqrt%7Bx%5E%7B2%7D%2Ba%5E%7B2%7D%7D%5D_%7B0%7D%5E%7B2%7D%20%5C%5C%26%3D%5Cfrac%7B2%20%5Cpi%20%5Csigma%7D%7B4%20%5Cpi%20%5Cvarepsilon_%7B0%7D%7D%5B%5Csqrt%7Bz%5E%7B2%7D%2BR%5E%7B2%7D%7D-%2821%29%5D%20%5C%5C%26%3D%5Cfrac%7B%5Csigma%7D%7B2%7D%28%5Csqrt%7B2%5E%7B2%7D%2BR%5E%7B2%7D%7D-2%29%5Cend%7Baligned%7D)
Note: Refer the image attached
I'm happy to know that the diagram shows how it's all set up.
If I could see the diagram, then I could probably do a much
better job with an answer. As it is ... 'flying blind' as it were ...
I'm going to wing it and hope it's somewhat helpful.
If the pulley is movable, then I'm picturing one end of the rope
tied to a hook in the ceiling, then the rope passing down through
the pulley, then back up, and you lifting the free end of the rope.
A very useful rule about movable and combination pulleys is:
the force needed to lift the load is
(the weight of the load)
divided by
(the number of strands of rope supporting the load) .
With the setup as I described it, there are 2 strands of rope
supporting the load ... one on each side of the pulley. So the
force needed to lift the load is
(250 N) / 2 = 125 N .
Answer:B
Explanation:
Given
mass of sledge=82 kg
Force on sledge=160 N
degree
force has two component sin and cos
Normal reaction 
