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dlinn [17]
3 years ago
11

Magma forms when solid rock in the crust and upper mantle. True or False

Physics
1 answer:
Shtirlitz [24]3 years ago
7 0

Answer:

True

Explanation:

Magma is known as melted rock deep within the Earth, normally coming from the melting of the upper mantle or crust.  Magma is formed by the partial melting of the mantle and crust and this can occur in different  ways. One way can be  called heat-transfer melting. Rising magma or rock will bring heat with it, and so can melt the surrounding mantle or crustal rock. For example, magmas generated in the mantle tend to be around 1200 degrees Celsius, whereas the more silicate minerals such as quartz and orthoclase feldspar (common in continental crustal rocks) begin to partially melt at around 650-850 degrees Celsius. Therefore, the crustal rock will begin to partially melt due to the introduction of heat from rising magma. A Another way of melting rock is known as decompression melting. During decompression melting, rock from within the mantle is brought to the surface adiabatically (no exchange of heat or energy with its surroundings) and so the lithostatic pressure decreases. This means that the parcel of rising rock crosses the solidus, and so at this point the thermal vibration of the molecules is no longer counteracted by the lithostatic pressure and the rock begins to partially melt.

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Vector A has a magnitude of 16 m and makes an angle of 44° with the positive x axis. Vector B also has a magnitude of 13 m and i
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Answer with explanation:

The given vectors in are reduced to their componednt form as shown

For vector A it can be written as

\overrightarrow{v}_{a}=16cos(44^{o})\widehat{i}+16sin(44^{o})\widehat{j}

Similarly vector B can be written as

\overrightarrow{v}_{b}=-13\widehat{i}

Hence The sum and difference is calculated as

\overrightarrow{v}_{a}+\overrightarrow{v}_{b}=16cos(44^{o})\widehat{i}+16sin(44^{o})\widehat{j}+(-13\widehat{i})\\\\\overrightarrow{v}_{a}+\overrightarrow{v}_{b}=(16cos(44^{o})-13)\widehat{i}+16sin(44^{o})\widehat{j}\\\\\therefore \overrightarrow{v}_{a}+\overrightarrow{v}_{b}=-1.49\widehat{i}+11.11\widehat{j}\\\\\therefore |\overrightarrow{v}_{a}+\overrightarrow{v}_{b}|=\sqrt{(-1.49)^{2}+11.11^{2}}=11.21m

The direction is given by

\theta =tan^{-1}\frac{r_{y}}{r_{x}}\\\\\theta =tan^{-1}\frac{11.11}{-1.49}=97.64^{o}with positive x axis.

Similarly

\overrightarrow{v}_{a}-\overrightarrow{v}_{b}=16cos(44^{o})\widehat{i}+16sin(44^{o})\widehat{j}-(-13\widehat{i})\\\\\overrightarrow{v}_{a}-\overrightarrow{v}_{b}=(16cos(44^{o})+13)\widehat{i}+16sin(44^{o})\widehat{j}\\\\\therefore \overrightarrow{v}_{a}-\overrightarrow{v}_{b}=24.51\widehat{i}+11.11\widehat{j}\\\\\therefore |\overrightarrow{v}_{a}-\overrightarrow{v}_{b}|=\sqrt{(24.51)^{2}+11.11^{2}}=26.91m

The direction is given by

\theta =tan^{-1}\frac{r_{y}}{r_{x}}\\\\\theta =tan^{-1}\frac{11.11}{24.51}=24.38^{o}with positive x axis.

3 0
3 years ago
Although we have discussed single-slit diffraction only for a slit, a similar result holds when light bends around a straight, t
user100 [1]

Answer:

The width of the strand of hair is   1.96 10⁻⁵ m

Explanation:

For this diffraction problem they tell us that it is equivalent to the diffraction of a single slit, which is explained by the equation

<h3>       a sin θ =±  m λ </h3><h3 />

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We can see that the diffraction angle is missing, but we can find it by trigonometry, where L is the distance of the strand of hair to the observation screen and "y" is the perpendicular distance to the first minimum of intensity

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      θ = tan⁻¹ ( 0.0405)

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With this data we can continue analyzing the problem, they indicate that they measure the distance to the first dark strip, thus m = 1

     a = m λ / sin θ

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     a = 1.96 10⁻⁵ m  

     a = 0.0196 mm

The width of the strand of hair is   1.96 10⁻⁵ m

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