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3 years ago

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g Light that is incident upon the eye is refracted several times before it reaches the retina. As light passes through the eye,

at which boundary does most of the overall refraction occur?

1 answer:

sertanlavr [38]3 years ago

7 0

Answer

Explanation

:giác mạc

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When a light ray passes from LESS dense water (n = 1.33) into a MORE dense diamond (n = 2.419) at an angle of 45 degrees, its pa

Ivenika [448]

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<u>Explanation:</u>

When light is incident at a transparent surface, the transmitted component of the light changes direction at the interface. Another component of the light is reflected at the surface. When a ray of light passes from water to diamond at an angle 45°, its path is bent towards the normal. This is so because water is less dense than the diamond. The refractive index of water (n = 1.33) is less than the refractive index of diamond (n = 2.419).

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3 years ago

Select the statements that are TRUE. Electrophilic aromatic substitution reactions have energies of activation that are very low

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Answer:

Addition reactions with benzenes lead to the loss of aromaticity.

Benzene and its derivatives undergo a type of substitution reaction in which a hydrogen atom is replaced by a substituent, but the stable aromatic benzene ring is regenerated at the end of the mechanism.

Benzene and its derivatives tend to undergo electrophilic aromatic substitution reactions.

Explanation:

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3 years ago

When the mass is removed, the length of the cable is found to be l0 = 4.66 m. After the mass is added, the length is remeasured

zaharov [31]

Answer:

$\gamma=6.07*10^5\frac{N}{m^2}$

Explanation:

For a linear elastic material Young's modulus is a constant that is given by:

$\gamma=\frac{F/A}{\Delta L/L_0}$

Here, F is the force exerted on an object under tensio, A is the area of the cross-section perpendicular to the applied force, $\Delta L$ is the amount by which the length of the object changes and $L_0$ is the original length of the object. In this case the force is the weight of the mass:

$F=mg\\F=55kg(9.8\frac{m}{s^2})\\F=539N$

Replacing the given values in Young's modulus formula:

$\gamma=\frac{F/\pi r^2}{(L_1-L_0)/L_0}\\\gamma=\frac{539N/\pi(0.045m)^2}{(5.31m-4.66m)/4.66m}\\\gamma=6.07*10^5\frac{N}{m^2}$

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3 years ago

A huge tank of glycerine with a density of 1.260 g/cm3 is vertically stationed on a platform which is 15 m above the ground. The

EleoNora [17]

Answer:

The tank is losing $4.976*10^{-4} m^3/s$

$v_g = 19.81 \ m/s$

Explanation:

According to the Bernoulli’s equation:

$P_1 + 1 \frac{1}{2} \rho v_1^2 + \rho gh_1 = P_2 + \frac{1}{2} \rho v_2^2 + \rho gh_2$

We are being informed that both the tank and the hole is being exposed to air :

∴ P₁ = P₂

Also as the tank is voluminous ; we take the initial volume $v_1$ ≅ 0 ;

then $v_2$ can be determined as: $\sqrt{[2g (h_1- h_2)]$

h₁ = 5 + 15 = 20 m;

h₂ = 15 m

$v_2 = \sqrt{[2*9.81*(20 - 15)]$

$v_2 = \sqrt{[2*9.81*(5)]$

$v_2= 9.9 \ m/s$ as it leaves the hole at the base.

radius r = d/2 = 4/2 = 2.0 mm

(a) From the law of continuity; its equation can be expressed as:

J = $A_1v_2$

J = πr² $v_2$

J = $\pi *(2*10^{-3})^{2}*9.9$

J = $1.244*10^{-4} m^3/s$

b)

How fast is the water from the hole moving just as it reaches the ground?

In order to determine that; we use the relation of the velocity from the equation of motion which says:

v² = u² + 2gh ₂

v² = 9.9² + 2×9.81×15

v² = 392.31

The velocity of how fast the water from the hole is moving just as it reaches the ground is : $v_g = \sqrt{392.31}$

$v_g = 19.81 \ m/s$

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The Answer is 11,121 Kg

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