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3 years ago

What type of material is found in the asthenosphere?

Physics

2 answers:

IceJOKER [234]3 years ago

6 0

Answer:

The correct answer is option C.

Explanation:

The layer below the lithosphere , the upper layer of the mantle of the earth with high heat pressure .It is made up of almost solid rocks but also flows not which indicates that it consist of viscoelastic fluid of molten rocks.This is because in some regions there can be presence of molten rocks.

o-na [289]3 years ago

5 0

Answer: A viscoelastic fluid of molten rock is the type of material found in the asthenosphere.

Hope This Helps

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A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the lens strength (a.k.a, lens p

Kipish [7]

Answer:

20.0 cm

Explanation:

Here is the complete question

The normal power for distant vision is 50.0 D. A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?

Solution

Now, the power of a lens, P = 1/f = 1/u + 1/v where f = focal length of lens, u = object distance from eye lens and v = image distance from eye lens.

Given that we require a 10 % increase in the power of the lens to accommodate the image she sees clearly, the new power P' = 50.0 D + 10/100 × 50 = 50.0 D + 5 D = 55.0 D.

Also, since the object is seen clearly, the distance from the eye lens to the retina equals the distance between the image and the eye lens. So, v = 2.00 cm = 0.02 m

Now, P' = 1/u + 1/v

1/u = P'- 1/v

1/u = 55.0 D - 1/0.02 m

1/u = 55.0 m⁻¹ - 1/0.02 m

1/u = 55.0 m⁻¹ - 50.0 m⁻¹

1/u = 5.0 m⁻¹

u = 1/5.0 m⁻¹

u = 0.2 m

u = 20 cm

So, at 55.0 dioptres, the closet object she can see is 20 cm from her eye.

8 0

2 years ago

A radar antenna is tracking a satellite orbiting the earth. At a certain time, the radar screen shows the satellite to be 118 km

ruslelena [56]

Answer:

x component 60.85 m

y component 101.031 m

Explanation:

We have given distance r = 118 km

Angle which makes from ground = 58.9°

(a) X component of distance is given by $r_x=rcos\Theta =118\times cos58.9=118\times 0.5165=118=60.85m$

(b) Y component of distance is given by $r_Y=rcos\Theta =118\times sin58.9=118\times 0.8562=101.0316m$

These are the x and y component of position vector

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3 years ago

Point charges of 21.0 μC and 47.0 μC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el

rewona [7]

Answer:

a) x = 0.200 m

b)E = 3.84*10^{-4} N/C

Explanation:

$q_1 = 21.0\mu C$

$q_1 = 47.0\mu C$

DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m

by relation for electric field we have following relation

$E = \frac{kq}{x}^2$

according to question E = 0

FROM FIGURE

x is the distance from left point charge where electric field is zero

$\frac{k21}{x}^2 = \frac{k47}{0.5-x}^2$

solving for x we get

$\frac{0.5}{x} = 1+ \sqrt{\frac{47}{21}}$

x = 0.200 m

b)electric field at half way mean x =0.25

$E =\frac{k*21*10^{-6}}{0.25^2} -\frac{k*47*10^{-6}}{0.25^2}$

E = 3.84*10^{-4} N/C

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kiruha [24]

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