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r-ruslan [8.4K]
3 years ago
6

Restless tectonic plates move (shift) between one and fifteen centimeters per __________. 1)year 2)month 3)day 4)minute

Physics
2 answers:
Vaselesa [24]3 years ago
8 0

Restless tectonic plates move (shift) between one and fifteen centimeters per <u>"year".</u>



Tectonic  plates refers to a giant chunks of the Earth's outside layer that skim on the asthenosphere, which is the upper part of the mantle of the Earth. The asthenosphere is a semi-liquid layer that twists and moves because of convection flows in the mantle. Plate tectonics is the theory that describes how the movement of tectonic plates and about the results from this movement.  

Tectonic plates move at a moderate rate, around one and fifteen centimeters  per year. The plates move in different ways, once in a while crashing into one another. At different areas they move separated or slide past one another.

Alina [70]3 years ago
4 0
1(year is the answer


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Which one of the following terms is used to describe the bending of waves and subsequent spreading around obstacles or the edges
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Answer:

B)Diffraction

Explanation:

The concept of diffraction in the field of physics is defined as the deflection of a wave when it crosses an opening or hits the edge of an opaque element. Diffraction is a phenomenon that involves all waves: electromagnetic, radio, sound, etc., and it is possible to predict their behavior using different mathematical approaches. There is a method of analysis called the Huygens principle, which allows us to understand diffraction as a wavefront that is seen as a series of emitters capable of redirecting the wave when it oscillates and thus promotes propagation. Although the waves produced by the oscillators are spherical, their interference causes a flat wave that moves in the same direction as the initial one.

6 0
3 years ago
Write the nuclear reaction equation for the beta decay of Iodine-131
den301095 [7]

Answer:

_{53}^{131}I \rightarrow _{54}^{131}Xe + e + \bar{\nu}

Explanation:

In a beta (minus) decay, a neutron in a nucleus turns into a proton, emitting a fast-moving electron (called beta particle) alongside with an antineutrino.

The general equation for a beta decay is:

^A_Z X \rightarrow _{Z+1}^AY+^0_{-1}e+ ^0_0\bar{\nu} (1)

where

X is the original nucleus

Y is the daughter nucleus

e is the electron

\bar{\nu} is the antineutrino

We observe that:

  • The mass number (A), which is the sum of protons and neutrons in the nucleus, remains the same in the decay
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In this problem, the original nucles that we are considering is iodine-131, which is

_{53}^{131}I

where

Z = 53 (atomic number of iodine)

A = 131 (mass number)

Using the rule for the general equation (1), the dauther nucleus must have same mass number (131) and atomic number increased by 1 (54, which corresponds to Xenon, Xe), therefore the equation will be:

_{53}^{131}I \rightarrow _{54}^{131}Xe + e + \bar{\nu}

7 0
3 years ago
Read 2 more answers
A red ball is thrown down with an initial speed of 1.6 m/s from a height of 25 meters above the ground. Then, 0.4 seconds after
Radda [10]

Answer:

1. v =  22.2 m/s

2. t = 2.25 seconds

3. h = 27.05 m

4. t = 1.16 seconds

Explanation:

The questions involve motion under the influence of gravity

1. Using the formula v² = u² + 2gh

where u = 1.6 m/s; g= 9.81 m/s²; h = 25 m; v = ?

v² = (1.6)² + 2 * 9.81 * 25

√v² = √493.06

v =  22.2 m/s

2. Using h = ut + 1/2 gt²

where h = 25 m; u = 0 (since velocity on reaching the ground is zero); g = 9.81 m/s²; t = ?

therefore, h = 1/2 gt²

making t subject of the formula, t = √ (2*h /g)

t = √ (2 * 25 / 9.81)

t = 2.25 seconds

3. Time of travel for the blue ball, t = 2 - 0.4 = 1.6s

using h = ut - gt²

u = 24 m/s; t = 1.6 s; g = 9.81 m/s²

note: since the ball is travelling against gravity, g is negative

h = 24 * 1.6 - 11/2 * 9.81 * 1.6²

h = 38.4 - 12.55 = 25.85 m

since height above the ground is 1.2 m,

total height h = 25.85 m + 1.2 m

h = 27.05 m

4. Let the time of travel of the red ball be t seconds.

So the time of travel of the blue ball = (t - 0.4) seconds.

Both the balls are at the same height :

25 - s = 1.2 + h  where s & h are the displacements of the red & the blue ball respectively.

25 - (ut + 1/2 gt2) = 1.2 + (ut - 1/2 gt2)

25 - (1.6 t + 0.5 * 9.8 t²) = 1.2 + (24(t-0.4) - 0.5*9.8*(t-0.4)²)

solving the equation above for the time after which both the balls are at the same height.

25 - 1.6t - 4.9t² = 1.2 + 24t - 9.6 - 4.9t² + 3.92t - 0.784

collecting like terms

(25 - 1.2 + 9.6 + 0.784) = (24 + 3.92 + 1.6) * t  

t = 34.184 / 33.44

t = 1.16 seconds

6 0
3 years ago
Two masses are connected by a string which passes over a pulley with negligible mass and friction. One mass hangs vertically and
monitta

Answer:

a=2,5m/s^2

Explanation:

From the question we are told that:

Coefficient of kinetic friction \mu= 0.200

Vertical Mass M_v=3kg

Horizontal mass M_h=3.00kg  

Generally the equation for kinetic force F_k is mathematically given by

F_k=\mu N\\F_k=0.2*3\\F_k=0.6

Generally the equation for T is mathematically given by

For M_v=3kg3g-T=3a

For M_h=3kg

T=M_v V+F_k\\T=3.0a+0.6

Therefore substituting

3-3a-0.6=3a\\2.4g=6a

a=2,5m/s^2

3 0
3 years ago
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