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3 years ago

11

The melting point of a substance occurs at the same temperatures as it’s blank point

2 answers:

azamat3 years ago

4 0

I think it's a.

Explanation:

melting point is boiling

Anna [14]3 years ago

3 0

I agree with them I belive I tea a as well

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What is the prefix for oxygen in As2O5? tri- di- mono- penta-

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The prefix for oxygen in As2O5 is PENTA.
Penta is used to show that the number of oxygen atoms present in the compound is five. Penta means five, while mono mean one, di means two, tri means three and tetra mans four and so on. The chemical name for the given compound is Arsenic pentoxide.

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3 years ago

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Crank

$\\ \rm\Rrightarrow \dfrac{1}{u}+\dfrac{1}{v}=\dfrac{1}{f}$

$\\ \rm\Rrightarrow \dfrac{1}{u}=\dfrac{1}{-10}+\dfrac{1}{38}$

$\\ \rm\Rrightarrow \dfrac{1}{u}=\dfrac{-19+5}{190}$

$\\ \rm\Rrightarrow \dfrac{1}{u}=\dfrac{-14}{190}$

$\\ \rm\Rrightarrow u=\dfrac{190}{-14}$

$\\ \rm\Rrightarrow u=13.6cm$

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2 years ago

At what angle of projectile range of maximum

Agata [3.3K]

Answer:

$45°$

Explanation:

The textbooks say that the maximum range for projectile motion (with no air resistance) is 45 degrees.

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3 years ago

A ray of light traveling through air strikes a piece of diamond at an angle of incidence equal to 56 degrees. Calculate the angu

Montano1993 [528]

Answer:

The angle of separation is $\Delta \theta = 0.93 ^o$

Explanation:

From the question we are told that

The angle of incidence is $\theta _ i = 56^o$

The refractive index of violet light in diamond is $n_v = 2.46$

The refractive index of red light in diamond is $n_r = 2.41$

The wavelength of violet light is $\lambda _v = 400nm = 400*10^{-9}m$

The wavelength of red light is $\lambda _r = 700nm = 700*10^{-9}m$

Snell's Law can be represented mathematically as

$\frac{sin \theta_i}{sin \theta_r} = n$

Where $\theta_r$ is the angle of refraction

=> $sin \theta_r = \frac{sin \theta_i}{n}$

Now considering violet light

$sin \theta_r__{v}} = \frac{sin \theta_i}{n_v}$

substituting values

$sin \theta_r__{v}} = \frac{sin (56)}{2.46}$

$sin \theta_r__{v}} = 0.337$

$\theta_r__{v}} = sin ^{-1} (0.337)$

$\theta_r__{v}} = 19.69^o$

Now considering red light

$sin \theta_r__{R}} = \frac{sin \theta_i}{n_r}$

substituting values

$sin \theta_r__{R}} = \frac{sin (56)}{2.41}$

$sin \theta_r__{R}} = 0.344$

$\theta_r__{R}} = sin ^{-1} (0.344)$

$\theta_r__{R}} = 20.12^o$

The angle of separation between the red light and the violet light is mathematically evaluated as

$\Delta \theta = \theta_r__{R}} - \theta_r__{V}}$

substituting values

$\Delta \theta =20.12 - 19.69$

$\Delta \theta = 0.93 ^o$

6 0

3 years ago