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Flauer [41]
3 years ago
14

What happens when a tectonic plate gets subducted

Physics
1 answer:
koban [17]3 years ago
8 0
It starts to melt the minerals in within the earths core.
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ACUTE INFECTION plz help worth 70 pt plz help !!!!! Also brainlest
Paraphin [41]

Answer:

Simple awnser Do it yourself I really would help but I have no clue! Sorry

Explanation:

3 0
3 years ago
A diver shines an underwater searchlight at the surface of a pond (n = 1.33). At what angle (relative to the surface) will the l
allsm [11]

Answer:

41.2°

Explanation:

Total internal reflection is the reflection of the incident ray at the interface between two media in which one of the media has a lower refractive index than the other. It occurs when the angle of incidence in the denser medium exceeds the critical angle.

The critical angle is the angle of incidence in the denser medium when the angle of incidence in the less dense medium is 90°.

Since

n= 1/sin C

C= sin^-(1/n)

C= sin^-(1/1.33)

C= 48.8°

Hence angle of incidence= 90-48.8 = 41.2°

4 0
3 years ago
An apple is placed 20.0 cm in front of a diverging lens of focal length 10.0 cm. Find the image distance and the magnification o
jenyasd209 [6]

Answer:

Image distance of apple=-6.7 cm

Magnification of apple=0.33

Explanation:

We are given that an apple is placed 20.cm in front of a diverging lens.

Object distance=u=-20 cm

Focal length=f=-10 cm

Because focal length of diverging lens is negative.

We have to find the image distance and magnification of the apple.

Lens formula

\frac{1}{f}=\frac{1}{v}-\frac{1}{u}

Substitute the values then we get

-\frac{1}{10}=\frac{1}{v}+\frac{1}{20}

\frac{1}{v}=-\frac{1}{10}-\frac{1}{20}

\frac{1}{v}=\frac{-2-1}{20}=-\frac{3}{20}

v=-\frac{20}{3}=-6.7 cm

Image distance of apple=-6.7 cm

Magnification=m=\frac{v}{u}=\frac{-\frac{20}{3}}{-20}

Magnification of apple=\frac{1}{3}=0.33

Hence, the magnification of apple=0.33

5 0
3 years ago
How do u get rid of a hacker? plz help my friend needs this
Alja [10]
IP address then call the cops
3 0
3 years ago
Read 2 more answers
A cylindrical container with a cross-sectional area of 66.2 cm2 holds a fluid of density 856 kg/m3 . At the bottom of the contai
Ugo [173]

Answer:

A. h = 2.15 m

B. Pb' = 122 KPa

Explanation:

The computation is shown below:

a)  Let us assume the depth be h

As we know that

Pb - Pat = d \times g \times h \\\\ ( 119 - 101) \times 10^3 = 856 \times 9.8 \times h

After solving this,  

h = 2.15 m

Therefore the depth of the fluid is 2.15 m

b)

Given that  

height of the extra fluid is

h' = \frac{2.35 \times 10^{-3}}{ area} \\\\ h' = \frac{2.35 \times 10^{-3}} { 66.2 \times 10^{-4}}

h' = 0.355 m

Now let us assume the pressure at the bottom is Pb'

so, the equation would be

Pb' - Pat = d \times g \times  (h + h')\\\\Pb' = 856 \times 9.8 \times ( 2.15 + 0.355) + 101000

Pb' = 122 KPa

3 0
3 years ago
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