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Afina-wow [57]
3 years ago
10

Where do subduction zones often occur

Physics
1 answer:
vodka [1.7K]3 years ago
4 0

Subductions occur at convergent boundaries. Its where one tectonic plate moves under another.

-Steel jelly

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PLEASE HELP ASAP
andrew-mc [135]

Answer:0.5

Explanation: i think i may be wrong hope i helped:)

8 0
3 years ago
the focal length of a simple magnifier is 8.50 cmcm . assume the magnifier to be a thin lens placed very close to the eye.
Igoryamba

When the object is at the focal point the angular magnification is 2.94.

Angular magnification:

The ratio of the angle subtended at the eye by the image formed by an optical instrument to that subtended at the eye by the object when not viewed through the instrument.

Here we have to find the angular magnification when the object is at the focal point.

Focal length = 6.00 cm

Formula to calculate angular magnification:

Angular magnification = 25/f

                                            = 25/ 8.5

                                             = 2.94

Therefore the angular magnification of this thin lens is 2.94

To know more about angular magnification refer:: brainly.com/question/28325488

#SPJ4

5 0
1 year ago
An aircraft component is fabricated from an aluminum alloy that has a plane strain fracture toughness of 40 mpa m (36.4 ksi in.
GarryVolchara [31]
We will first determine using the given if an aircraft component will fracture with a given stress level (260 MPa), maximum internal crack length (6.0 mm) and fracture toughness (40 MPa m ), given that fracture occurs for the same component using the same alloy for another stress level and internal crack length. First, it is necessary to solve for the parameter Y, using Equation 8.5, for the conditions under which fracture occurred (i.e., σ = 300 MPa and 2 a = 4.0 mm). Therefore,

 Y = K(Ic)/ sqrt(π a) = 40 MPa( m ) / (300 MPa) sqrt(( π ) ((4 × 10-3 m)/2)) = 1.68 

We will now solve for the product Y σ π a for the other set of conditions, so as to ascertain whether or not this value is greater than the K(Ic) for the alloy. Thus, 

Y sqrt(π a) = (1.68)(260 MPa) sqrt (( π )[(6 × 10^-3 m)/ 2])
                  = 42.4 MPa sqrt (m) (39 ksi in. ) 

Therefore, fracture will occur since this value ( 42.4 MPa sqrt(m)) is greater than the K(Ic) of the material, 40 MPa sqrt(m).
8 0
4 years ago
What is the focal length of concave mirror that magnifies , by a factor of +3.2 , an object that is placed 30cm from the mirror?
guapka [62]

Answer:

23 cm

Explanation:

The formula for magnification is;

Magnification = image distance / object distance

use values as;

3.2 = v / 30   where v is image distance

v =30*3.2

v=96 cm

The relationship of the focal length with image distance and object distance is expressed as;

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

where f is the focal length and u is object distance

use values in the equation as;

\frac{1}{30} +\frac{1}{96 }  =\frac{1}{f}

\frac{1}{f} =\frac{7}{160}

f=160/7

f=22.86

f= 23 cm ----------nearest a cm

7 0
3 years ago
What would the acceleration of a car be from a stoplight if it started at 0 m/s and reached a
trapecia [35]

Answer:

The acceleration of a car would be:  a=12.8 m/s²

Explanation:

Given

Initial velocity = v_1=0 m/s

Final velocity = v_f=100 m/s

Time elapsed = t = 7.8 s

To determine

We need to determine the acceleration of a car.

We know that acceleration is basically the rate of change in velocity over time.

Thus,

We can determine the acceleration using the formula

\:\:\:a=\:\frac{v_f-v_i}{t}

where

  • a is the acceleration
  • v_1 is the initial velocity
  • v_f is the final velocity
  • t is time elapsed

now substituting the values v_1=0,  v_f=100, and t = 7.8 in the formula

\:\:\:a=\:\frac{v_f-v_i}{t}

a=\:\frac{100-0}{7.8}

a=\frac{100}{7.8}

a=12.8 m/s²

Therefore, the acceleration of a car would be:  a=12.8 m/s²

3 0
3 years ago
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