The sun not set in the summer and greater than 66 N
I hope that's help !
Answer:
Refractive index of liquid C > Refractive index of liquid B > Refractive index of liquid A
Explanation:
Let the depth of each section is h.
That means the real depth for each section is h.
Apparent depth is liquid A is 7 cm.
Apparent depth in liquid B is 6 cm.
Apparent depth in liquid C is 5 cm.
by the formula of the refractive index
n = real depth / apparent depth
where, n is the refractive index of the liquid.
For liquid A:
.... (1)
For liquid B:
..... (2)
For liquid C:
..... (3)
By comparing all the three equations
nc > nB > nA
Refractive index of liquid C > Refractive index of liquid B > Refractive index of liquid A
(a) 764.4 N
The weight of the astronaut on Earth is given by:

where
m is the astronaut's mass
g is the acceleration due to gravity
Here we have
m = 78.0 kg
g = 9.8 m/s^2 at the Earth's surface
So the weight of the astronaut is

(b) 21.1 N
The spacecraft is located at a distance of

from the center of Earth.
The acceleration due to gravity at a generic distance r from the Earth's center is

where G is the gravitational constant and M is the Earth's mass.
We know that at a distance of r = R (at the Earth's surface) the value of g is 9.8 m/s^2, so we can write:
(1)
the acceleration due to gravity at r=6R instead will be

And substituting (1) into this formula,

So the weight of the astronaut at the spacecratf location is

FBD provides a visual for every force that the object experiences. It can tell you the direction by showing the direction and magnitude of the forces. You can calculate the net force, example being that the applied force is 300N and the force of friction is 100N, the net force is 200N in the direction of the applied force. (Fnet= Fa- Ff).
60.3° from due south and 5.89 m/s For this problem, first calculate a translation that will put John's destination directly on the origin and apply that translation to Mary's destination. Then the vector from the origin to Mary's new destination will be the relative vector of Mary as compared to John. So John is traveling due south at 6.7 m/s. After 1 second, he will be at coordinates (0,-6.7). The translation will be (0,6.7) Mary is traveling 28° West of due south. So her location after 1 second will be (-sin(28)*10.9, -cos(28)*10.9) = (-5.117240034, -9.624128762) After translating that coordinate up by 6.7, you get (-5.117240034, -2.924128762) The tangent of the angle will be 2.924128762/5.117240034 = 0.57142693 The arc tangent is atan(0.57142693) = 29.74481039° Subtract that value from 90 since you want the complement of the angle which is now 60.25518961° So Mary is traveling 60.3° relative to due south as seen from John's point of view. The magnitude of her relative speed is sqrt(-5.117240034^2 + -2.924128762^2) = 5.893783 m/s Rounding the results to 3 significant digits results in 60.3° and 5.89 m/s