The thickness of the glass block in front of a fish tank is 9cm. An insect is present at O in air in front of the glass block. The apparent displacement front point O of the insect to the fish which is observing from the water (refractive index of water = 4/3, glass = 3/2)

1) appears 2cm towards
2) appears 2cm away
3) appears 3cm away
4) appears 4 cm away
5) appears appears 4cm towards

Please show me how you worked it out, along with a brief explanation.​

Answer:

Explanation:

a. The X and Y components of the velocity can be found using trigonometry:

X = V * cos(θ) = 15 m/s * cos(38°) ≈ 11.63 m/s

Y = V * sin(θ) = 15 m/s * sin(38°) ≈ 9.14 m/s

b. The time the ball is in the air can be found using the Y component of the velocity and the acceleration due to gravity:

Y = V * sin(θ) * t - (1/2) * g * t^2

where g = 9.8 m/s^2 is the acceleration due to gravity

Solving for t, we get:

t = 2 * Y / g ≈ 1.87 s

c. The distance the ball travels can be found using the X component of the velocity and the time in the air:

distance = X * time = 11.63 m/s * 1.87 s ≈ 21.78 m

A) The net electric flux through the closed surface S₁ is given by the equation: Net electric flux = q1/4πε0
= 4.45 nC / (4π x 8.85 x 10-12 C2/Nm2)
= 0.541 x 10-3 Nm2/C .

Electricity is a form of energy that exists in nature and is created through the movement of electrons between atoms. It is the force that powers all of the electrical appliances and devices in our homes and offices.

B) The net electric flux through the closed surface S₂ is given by the equation:
Net electric flux = q1/4πε0 + q2/4πε0
= (4.45 nC + (-7.50 nC)) / (4π x 8.85 x 10-12 C2/Nm2)
= -2.05 x 10-3 Nm2/C
C) The net electric flux through the closed surface S₃ is given by the equation:
Net electric flux = q1/4πε0 + q2/4πε0 + q3/4πε0
= (4.45 nC + (-7.50 nC) + 2.15 nC) / (4π x 8.85 x 10-12 C2/Nm2)
= -0.239 x 10-3 Nm2/C
D) The net electric flux through the closed surface S₄ is given by the equation:
Net electric flux = q1/4πε0 + q2/4πε0 + q3/4πε0 + q1/4πε0
= (4.45 nC + (-7.50 nC) + 2.15 nC + 4.45 nC) / (4π x 8.85 x 10-12 C2/Nm2)
= 0.302 x 10-3 Nm2/C

E) The net electric flux through the closed surface S₅ is given by the equation:
Net electric flux = q1/4πε0 + q2/4πε0 + q3/4πε0 + q1/4πε0 + q2/4πε0
= (4.45 nC + (-7.50 nC) + 2.15 nC + 4.45 nC + (-7.50 nC)) / (4π x 8.85 x 10-12 C2/Nm2)
= -1.75 x 10-3 Nm2/C.

To learn more about electric
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A) The net electric flux through the closed surface S₁ is given by the equation: Net electric flux = q1/4πε0

= 4.45 nC / (4π x 8.85 x 10⁻¹² C²/Nm²)

= 0.541 x 10⁻³ Nm²/C .

Electricity is a form of energy that exists in nature and is created through the movement of electrons between atoms. It is the force that powers all of the electrical appliances and devices in our homes and offices.

B) The net electric flux through the closed surface S₂ is given by the equation:

Net electric flux = q1/4πε0 + q2/4πε0

= (4.45 nC + (-7.50 nC)) / (4π x 8.85 x 10⁻¹² C²/Nm²)

= -2.05 x 10⁻³ Nm²/C

C) The net electric flux through the closed surface S₃ is given by the equation:

Net electric flux = q1/4πε0 + q2/4πε0 + q3/4πε0

= (4.45 nC + (-7.50 nC) + 2.15 nC) / (4π x 8.85 x 10⁻¹² C²/Nm²)

= -0.239 x 10⁻³ Nm2/C

D) The net electric flux through the closed surface S₄ is given by the equation:

Net electric flux = q1/4πε0 + q2/4πε0 + q3/4πε0 + q1/4πε0

= (4.45 nC + (-7.50 nC) + 2.15 nC + 4.45 nC) / (4π x 8.85 x 10-12 C²/Nm²)

= 0.302 x 10⁻³ Nm²/C

E) The net electric flux through the closed surface S₅ is given by the equation:

Net electric flux = q1/4πε0 + q2/4πε0 + q3/4πε0 + q1/4πε0 + q2/4πε0

= (4.45 nC + (-7.50 nC) + 2.15 nC + 4.45 nC + (-7.50 nC)) / (4π x 8.85 x 10⁻¹² C²/Nm²)

= -1.75 x 10⁻³ Nm²/C.

To learn more about electricity, visit:

brainly.com/question/1100341

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Answer:

625

Explanation:

To solve this problem, we can use the following kinematic equation:

v^2 = u^2 + 2as

Where:

v = final velocity (0 m/s since the body comes to rest)

u = initial velocity (50 m/s)

a = acceleration (-2 m/s^2 since the body is decelerating)

s = distance

We want to find the distance (s) that the body covers from the time it starts to decelerate to the time it comes to rest. We can rearrange the equation to solve for s:

s = (v^2 - u^2) / (2a)

Substituting the values we have:

s = (0^2 - 50^2) / (2 x (-2)) = 625 meters

Therefore, the body covers a distance of 625 meters from the time it starts to decelerate until it comes to rest.