Answer:
The two dogs sitting here are already poor and ignorant
Answer: 1.65m
Explanation:
Refractive index in terms of the depth of liquid is the ratio of the real depth to the apparent depth of the liquid i.e Refractive index =Real depth/apparent depth
Refractive index of water given = 1.33
Real depth is the measure of how deep is the liquid while apparent depth is the depth at the surface of the liquid.
Real depth = 2.2m
Apparent depth =?
Applying the formula above
Apparent depth =Real depth/refractive index
= 2.2/1.33
= 1.65m
Therefore, the circle of light that exits the surface of the water when that light shines in the middle of the night is 1.65m wide
Answer:
8.46E+1
Explanation:
From the question given above, the following data were obtained:
Charge 1 (q₁) = 39 C
Charge 2 (q₂) = –53 C
Force (F) of attraction = 26×10⁸ N
Electrical constant K) = 9×10⁹ Nm²/C²
Distance apart (r) =?
The distance between the two charges can be obtained as follow:
F = Kq₁q₂ / r²
26×10⁸ = 9×10⁹ × 39 × 53 / r²
26×10⁸ = 1.8603×10¹³ / r²
Cross multiply
26×10⁸ × r² = 1.8603×10¹³
Divide both side by 26×10⁸
r² = 1.8603×10¹³ / 26×10⁸
r² = 7155
Take the square root of both side
r = √7155
r = 84.6 m
r = 8.46E+1 m