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3 years ago

Which of the following values represents an index of refraction of an actual material?

1 answer:

3 0

Sadly, we're forced to answer the question without the benefit of the
list of choices, which, for some reason, you decided not to let us see.

Index of refraction of a substance =

(speed of light in vacuum) / (speed of light in the substance).

Any number greater than ' 1 ' can be an index of refraction. A number
less than ' 1 ' can't be . . . that would be saying that the speed of light
in this substance is greater than the speed of light in vacuum.

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Leaves uses_,_and_to make food for the plant

LenaWriter [7]

light, water, carbon dioxide

Explanation:

c02 , h20 and light

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3 years ago

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In the compound MgS, the sulfide ion has 1. lost one electrons. 2. lost two electrons. 3. gained one electron. 4. gained two ele

never [62]

Answer:

4.has gained two electrons

Explanation:

There exist electrovalent bonding the compound MgS . In electrovalent bonding, there is a transfer of electrons from the metal to non-metal.

Magnesium atom has an atomic number 12 and its electron configuration is 2,8,2

Sulfur atom , a non-metal has atomic number of 16 and its electron configuration = 2,8,6

This means that magnesium as a metal needs to loose two electrons from its valence shell to attain its stable structure.Also sulfur requires two more electron to achieve its octet structure.

Hence a transfer of electrons will take place from magnesium atom to sulfur atom, sulfur gaining two electrons.

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3 years ago

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A small ball with mass 1.20 kg is mounted on one end of a rod 0.860 m long and of negligible mass. The system rotates in a horiz

Rufina [12.5K]

Answer:

0.88752 kgm²

0.02236 Nm

Explanation:

m = Mass of ball = 1.2 kg

L = Length of rod = 0.86 m

$\theta$ = Angle = 90°

Rotational inertia is given by

$I=mL^2\\\Rightarrow I=1.2\times 0.86^2\\\Rightarrow I=0.88752\ kgm^2$

The rotational inertia is 0.88752 kgm²

Torque is given by

$\tau=FLsin\theta\\\Rightarrow \tau=2.6\times 10^{-2}\times 0.86sin90\\\Rightarrow \tau=0.02236\ Nm$

The torque is 0.02236 Nm

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3 years ago

A disk of a radius 50 cm rotates at a constant rate of 100 rpm. What distance in meters will a point on the outside rim travel d

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Answer:

the distance in meters traveled by a point outside the rim is 157.1 m

Explanation:

Given;

radius of the disk, r = 50 cm = 0.5 m

angular speed of the disk, ω = 100 rpm

time of motion, t = 30 s

The distance in meters traveled by a point outside the rim is calculated as follows;

$\theta = \omega t\\\\\theta = (100 \frac{rev}{\min} \times \frac{2\pi \ rad}{1 \ rev} \times \frac{1\min}{60 s} ) \times (30 s)\\\\\theta = 100 \pi \ rad\\\\d = \theta r\\\\d = 100\pi \ \times \ 0.5m\\\\d = 50 \pi \ m = 157.1 \ m$

Therefore, the distance in meters traveled by a point outside the rim is 157.1 m

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3 years ago

8750 J of heat are applied to a piece of aluminium causing a 56C increase in its

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Answer:146.8983 grams

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3 years ago