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yaroslaw [1]
4 years ago
13

Jackie studied stresses that affect Earth's crust. How is compression of rock different from shearing?

Physics
1 answer:
Gala2k [10]4 years ago
6 0

Answer:

B

Explanation:

This is because compression makes pressure fore the rockss to squeeze, and shearing pulls them apart forcibly.

Hope this helps! :)

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Will an object with a density of 1.05 g/ml float or sink in water? Explain.
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A sample of an unknown liquid has a volume of 12.0 mL and a mass of 6 g. What is its density? Show your work or explain how you
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A vessel at rest at the origin of an xy coordinate system explodes into three pieces. Just after the explosion, one piece, of ma
ahrayia [7]

Incomplete question as we have not told to find what.So the complete question is here

A vessel at rest at the origin of an xy coordinate system explodes into three pieces. Just after the explosion, one piece, of mass m, moves with velocity (-60 m/s)i and a second piece, also of mass m, moves with velocity (-60 m/s)j. The third piece has mass 3m.Just after the explosion, what are the (a) magnitude and (b) direction of the velocity of the third piece?

Answer:

V_{3}=(20i+20j)m/s

Explanation:

Given data

The vessel at rest

Piece one,of mass m,moves with velocity=(-60 m/s)i

Piece two,of mass m,moves with velocity=(-60 m/s)j

Piece three,of mass 3m

As the linear momentum is conserved in this system,Because the system is closed and no external  force acting on it

So momentum is given as

p_{initial}=p_{final}

As the vessel at rest so the initial momentum is zero

So

m_{1}V_{1}+m_{2}V_{2}+m_{3}V_{3}=0\\m_{3}V_{3}=-m_{1}V_{1}-m_{2}V_{2}\\V_{3}=\frac{-m_{1}V_{1}-m_{2}V_{2}}{m_{3}} \\V_{3}=\frac{-m_{1}(-60m/s)i-m_{2}(-60m/s)j}{3m}\\V_{3}=(20i+20j)m/s

 

5 0
4 years ago
A flat, 179 179 ‑turn, current‑carrying loop is immersed in a uniform magnetic field. The area of the loop is 4.41 cm 2 4.41 cm2
Alecsey [184]

Answer:

The value of the magnetic field is  B =0.1423T

Explanation:

From the question we are told that

              The number of turns is  N = 179

               The area of the loop is A = 4.41cm^2 = \frac{4.41}{10000} = 0.000414m

                 The angle is  \theta  = 59^o

               The torque  is  \tau =2.25 * 10^{- 5} N

                The current is  I = 2.49\ mA

The torque acting on the current carry loop is  mathematically represented as

                     \tau = B * I * N * A * sin \theta

Where is the magnitude of the magnetic filed

Making B the subject

                     B= \frac{\tau}{I * N * A * sin\theta}

Substituting values

                    B = \frac{2.25*10^{-5}}{2.49*10^{-3} * 179 * 0.000414 * sin (59)}

                       =0.1423 T

4 0
4 years ago
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