Answer:
1.7323
Explanation:
To develop this problem, it is necessary to apply the concepts related to refractive indices and Snell's law.
From the data given we have to:
Where n means the index of refraction.
We need to calculate the index of refraction of the liquid, then applying Snell's law we have:
Replacing the values we have:
Therefore the refractive index for the liquid is 1.7323
The correct answer is A) Ipsilateral
Explanation:
The term ipsilateral is commonly used to describe objects or structures that are on the same side of a body or structure. This term is correct to describe the right eye and the right lung because these two organs are on the same side of the body (the right side). This can also be used to describe other organs such as the left humerus and the left hand or the right ear and the right feet because these pairs are also on the same side. According to this, the correct answer is A.
Answer:
This is because motion is intended to occur but at zero acceleration. It means at a constant velocity, henceFor that to happen the pulling force F must exactly equal the frictional force Fk .
Answer:
Vf = 15 m/s
Explanation:
First we consider the upward motion of ball to find the height reached by the ball. Using 3rd equation of motion:
2gh = Vf² - Vi²
where,
g = acceleration due to gravity = -9.8 m/s² (negative sign for upward motion)
h = height =?
Vf = Final Velocity = 0 m/s (Since, ball momentarily stops at highest point)
Vi = Initial Velocity = 15 m/s
Therefore,
2(-9.8 m/s²)h = (0 m/s)² - (15 m/s)²
h = (-225 m²/s²)/(-19.6 m/s²)
h = 11.47 m
Now, we consider downward motion:
2gh = Vf² - Vi²
where,
g = acceleration due to gravity = 9.8 m/s²
h = height = 11.47 m
Vf = Final Velocity = ?
Vi = Initial Velocity = 0 m/s
Therefore,
2(9.8 m/s²)(11.47 m) = Vf² - (0 m/s)²
Vf = √(224.812 m²/s²)
<u>Vf = 15 m/s</u>
Answer:
the maximum allowable current is 7302.967 amperl
Explanation:
The computation of the maximum allowable current is shown below;
Force F = mean ÷ 4π 2 I_1 I_2 ÷d × ΔL
200 N = (10)^-7 (2I × I) ÷ 0.08 × 1.5
200 = 3.75 × 10^-6 I^2
I = √200 ÷ √ 3.75 × 10^-6
= 7302.967 amperl
Hence, the maximum allowable current is 7302.967 amperl
Basically we applied the above formula