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:
m = 105.37 kg
Explanation:
We are given;
Mass of man; m = 113 kg
Length of boat = 6.3m
Now, The position of the center of mass will not change during the motion of the man.
Thus,
X_g,i = X_g,f
So,
[113(6.3) + ma]/(113 + m) = [113(3.26) + m(a +3.26)]/(113 + m)
113 + m will cancel on both sides to give;
113(6.3) + ma = [113(3.26) + m(a +3.26)]
711.9 + ma = 368.38 + ma + 3.26m
ma will cancel out to give;
711.9 - 368.38 = 3.26m
343.52/3.26 = m
m = 105.37 kg
If we go down the list,
It wouldn’t be Primary Producer because Deer don’t really produce anything
Can’t be Carnivore because deer don’t eat meat
Same with tertiary
They do eat plants. So it would have to be the last one
Answer:
ball hit the ground from her feet is 1.83 m far away
Explanation:
given data
speed = 5.3 m/s
angle = 12°
height = 1 m
to find out
how far from her feet ball hit ground
solution
we consider here x is horizontal component and y is vertical component
so in vertical
velocity will be = v sin12
vertical speed u = 5.3 sin 12 = 1.1 m/s downward
and
in horizontal , velocity we know v = 5.3 m/s
so from motion of equation
s = ut + 0.5×a×t²
s is distance t is time a is 9.8
put all value
1 = 1.1 ( t) + 0.5×9.8×t²
solve it we get t
t = 0.353 s
and
horizontal distance is = vcos12 × t
so horizontal distance = 5.3×cos12 × ( 0.353)
horizontal distance = 1.83 m
so ball hit the ground from her feet is 1.83 m far away
