1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Kryger [21]
3 years ago
5

Along boundaries plates are brittle and the crust fractures (breaks and cracks) forming ___

Physics
1 answer:
docker41 [41]3 years ago
5 0

Answer:

Along boundaries plates are brittle and the crust fractures (breaks and cracks) forming Faults

Explanation:

Brittle material are substances that looks hard but can be easily broken by strain. Brittle deformation leads to crack and fracture of the material involve. Brittle crust lacks elastic ability , elasticity in this rocks are almost absent.

Fractures and cracks in rocks form fault. Fault is a discontinuity in rocks with an appreciable displacement. Fault is a displacement in rocks relative to one part. Generally, Fault occur when brittle rock fracture or crack, the fractured rocks move relative to one another.

Differential stress can cause rocks to fracture. If that part of the crust lacks ductility the stress on the rock will cause a brittle deformation. The brittle deformation causes fracture(breaks and cracks) leading to structure like Faults and Joints.    

You might be interested in
A wave transfers:<br> Water<br> particles<br> energy<br> matter
SashulF [63]

Answer:

Particles in a water wave exchange kinetic energy for potential energy. When particles in water become part of a wave, they start to move up or down. This means that kinetic energy (energy of movement) has been transferred to them

Explanation:

hope this helps u ....

<em>pls mark this as the brainliest...</em>

6 0
3 years ago
According to the theory of plate tectonics, which forces cause the movement of plates in the earth's crust?
Sauron [17]
The correct answer is B.
4 0
3 years ago
Read 2 more answers
A batter hits two baseballs with the same force. One hits the ground near third base. The other is a home run out of the park. W
Basile [38]
I think the answer is "<span>The ball that went out of the park shows more work because the distance was greater."</span>
3 0
3 years ago
An insect 5.25 mm tall is placed 25.0 cm to the left of a thin planoconvex lens. The left surface of this lens is flat, the righ
Zigmanuir [339]

Answer:

(A) therefore the image is

  • 63 cm to the right of the lens
  • the image size is -13.22 cm
  • it is real
  • it is inverted

(B) therefore the image is

  • 63 cm to the right of the lens
  • the image size is -13.22 cm
  • it is real
  • it is inverted

Explanation:

height of the insect (h) = 5.25 mm = 0.525 cm

distance of the insect (s) = 25 cm

radius of curvature of the flat left surface (R1) = ∞

radius of curvature of the right surface (R2) = -12.5 cm (because it is a planoconvex lens with the radius in the direction of the incident rays)

index of refraction (n) = 1.7

(A) we can find the location of the image by applying the formula below

\frac{1}{f} =\frac{1}{s'} +\frac{1}{s} where

  • s' = distance of the image
  • f = focal length
  • but we first need to find the focal length before we can apply this formula

\frac{1}{f} =(n-1)(\frac{1}{R1} -\frac{1}{R2} )

\frac{1}{f} =(1.7-1)(\frac{1}{∞} -\frac{1}{-12.5} )

\frac{1}{f} =(0.7)(0 + \frac{1}{12.5} )

\frac{1}{f} =\frac{0.7}{12.5}

f = \frac{12.5}{0.7}

f = 17.9 cm

now that we have the focal length we can apply \frac{1}{f} =\frac{1}{s'} +\frac{1}{s}

\frac{1}{f} - \frac{1}{s} =\frac{1}{s'}

\frac{1}{17.9} - \frac{1}{25} =\frac{1}{s'}

\frac{25 - 17.9}{17.9 x 25} =\frac{1}{s'}

\frac{7.1}{447.5} =\frac{1}{s'}

s' = \frac{447.5}{7.1}[/tex]  = 63 cm to the right of the lens

magnification =\frac{-s'}{s} =\frac{y'}{y}   where y' is the height of the image, therefore

\frac{-s'}{s} =\frac{y'}{y}

\frac{-63}{25} =\frac{y'}{52.5}

y' = \frac{-63}{25} x 0.525 = -13.22 cm

therefore the image is

  • 63 cm to the right of the lens
  • the image size is -13.22 cm
  • it is real
  • it is inverted

(B) if the lens is reversed, the radius of curvatures would be interchanged

radius of curvature of the flat left surface (R1) = ∞

radius of curvature of the right surface (R2) = 12.5 cm

we can find the location of the image by applying the formula below

\frac{1}{f} =\frac{1}{s'} +\frac{1}{s} where

  • s' = distance of the image
  • f = focal length
  • but we first need to find the focal length before we can apply this formula

\frac{1}{f} =(n-1)(\frac{1}{R1} -\frac{1}{R2} )

\frac{1}{f} =(1.7-1)(\frac{1}{12.5} -\frac{1}{∞} )

\frac{1}{f} =(0.7)( \frac{1}{12.5} - 0)

\frac{1}{f} =\frac{0.7}{12.5}

f = \frac{12.5}{0.7}

f = 17.9 cm

now that we have the focal length we can apply \frac{1}{f} =\frac{1}{s'} +\frac{1}{s}

\frac{1}{f} - \frac{1}{s} =\frac{1}{s'}

\frac{1}{17.9} - \frac{1}{25} =\frac{1}{s'}

\frac{25 - 17.9}{17.9 x 25} =\frac{1}{s'}

\frac{7.1}{447.5} =\frac{1}{s'}

s' = \frac{447.5}{7.1}[/tex]  = 63 cm to the right of the lens

magnification =\frac{-s'}{s} =\frac{y'}{y}   where y' is the height of the image, therefore

\frac{-s'}{s} =\frac{y'}{y}

\frac{-63}{25} =\frac{y'}{52.5}

y' = \frac{-63}{25} x 0.525 = -13.22 cm

therefore the image is

  • 63 cm to the right of the lens
  • the image size is -13.22 cm
  • it is real
  • it is inverted

7 0
3 years ago
Two cars, one in front of the other, are traveling down the highway at 25 m/s. the car behind sounds its horn, which has a frequ
netineya [11]
<span>You are given two cars, one in front of the other, that are traveling down the highway at 25 m/s. You are also given a frequency of 500 Hz of the car travelling behind it. You are asked what is the frequency heard by the driver of the lead car. This problem can be solved using the Doppler effect

sound frequency heard by the lead car = [(speed of sound + lead car velocity)/( speed of sound + behind car velocity)] * (sound of frequency of the behind car)
</span>sound frequency heard by the lead car = [(340 m/s + 25 m/s)/(340 m/s - 25 m/s)] * (500 Hz)
sound frequency heard by the lead car = 579 Hz
7 0
3 years ago
Other questions:
  • An approaching storm is moving at 15 km an hour what do you need to know to determine its velocity
    7·1 answer
  • Alexander has a mass of 70 kilograms. his apartment is on the second floor. 5 meters yo from ground level . How much work does h
    14·1 answer
  • A lens for a spotlight is coated so that it does not transmit yellow light. if the light sources is white, what color is the spo
    12·1 answer
  • An object has a velocity of 8 m/s and a kinetic energy of 480 J what is the mass of the object
    14·1 answer
  • Which image represents the force on a positively charged particle caused by an approaching magnet?
    10·1 answer
  • Asap pls answer right will mark brainiest ASAP ...
    14·1 answer
  • Please Help!! When do ionic bonds occur?
    13·1 answer
  • The photoelectric work function of a metal is the minimum energy needed to eject an electron by irradiating the metal with light
    6·1 answer
  • In what part of the earth does convection occur?
    15·1 answer
  • In which situation would a space probe most likely experience centripetal acceleration?
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!