Answer:
The answer should be C. slanted upward to the right.
Hope this helps. :-)
The concept required to solve this problem is the optical relationship that exists between the apparent depth and actual or actual depth. This is mathematically expressed under the equations.

Where,
Depth of glass
Refraction index of water
Refraction index of glass
Refraction index of air
Depth of water
I enclose a diagram for a better understanding of the problem, in this way we can determine that the apparent depth in the water of the logo would be subject to



Therefore the distance below the upper surface of the water that appears to be the logo is 4.041cm
Answer:
Explanation:
E=(σ/ε0)
As noted by Dirac the field is the same no matter how far you are from the sheet. When your charge covers a conducting plane, as in your case, the field is, D/eo ,(D is charge density). Because the field inside the conductor (no matter how thin) is zero. The only time the field is, D/2eo, is when you have just a sheet of charge, by itself, not on a conducting plane."
Answer:
<em>Because </em><em>of </em><em>the </em><em>given </em><em>stranded</em><em> </em><em>wires </em><em>is </em><em>that </em><em>it's </em><em>thinner </em><em>there </em><em>are </em><em>even </em><em>more </em><em>air </em><em>gaps </em><em>and </em><em>a </em><em>greater </em><em>surface</em><em> </em><em>area </em><em>in </em><em>the </em><em>individual</em><em> </em><em>stranded</em><em> wires</em><em> </em><em>then </em><em>therefore </em><em>it </em><em>carries </em><em>less </em><em>current </em><em>than </em><em>similar </em><em>solid </em><em>wires </em><em>can </em><em>with</em><em> </em><em>each</em><em> </em><em>type </em><em>of </em><em>wire </em><em>,</em><em> insulations</em><em> </em><em>technologies </em><em>can </em><em>greatly</em><em> </em><em>assist </em><em> </em><em>in </em><em>reducing</em><em> </em><em>power </em><em>dissipation</em><em>.</em>
Answer:
2697.75N/m
Explanation:
Step one
This problem bothers on energy stored in a spring.
Step two
Given data
Compression x= 2cm
To meter = 2/100= 0.02m
Mass m= 0.01kg
Height h= 5.5m
K=?
Let us assume g= 9.81m/s²
Step three
According to the principle of conservation of energy
We know that the the energy stored in a spring is
E= 1/2kx²
1/2kx²= mgh
Making k subject of formula we have
kx²= 2mgh
k= 2mgh/x²
k= (2*0.01*9.81*5.5)/0.02²
k= 1.0791/0.0004
k= 2697.75N/m
Hence the spring constant k is 2697.75N/m