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3 years ago

An element's atomic number is 34. How many protons would an atom of this element have?

Physics

1 answer:

Elena-2011 [213]3 years ago

7 0

34. The element would have 34 protons. Also 34 electrons.

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Anon25 [30]

Answer:

D. Principle of original horizontality

B. Principle of faunal succession

A. Uniformitarianism

C. Principle of superposition

Explanation:

Question 1

The principle of original horizontality is one of the foremost relative dating principle that is wide used in stratigraphy.

It states that "sedimentary rocks are laid down flatly on top one another in a sedimentary basin".

Sedimentary rocks will only vary vertically, but laterally, they are uniform and internally homogeneous in space. This is why most sedimentary rocks are stratified and laid layers upon layers just like the pages of a book.

Each layer is called a bed and often times are laterally continuous in space within the same basin.

Whenever we see beds not horizontally continuous, we can conclude that a tectonic event must have disrupted the sequence and it came after the it was formed.

Question 2

Principle of faunal succession succession was proposed by Williams Smith, an English Geologist and a canal worker in the 19th century.

Based on this principle, sedimentary rocks can placed in their proper chronostratigrahic framework based on the fossils they contain in them.

This principle is hinged on theory of evolution.
It is widely accepted that organisms evolved from one another.
Rocks often bear these records in fossil remains and this can help us appropriately fit rocks to the time they were formed.

Question 3

The principle of uniformitarianism was one of the disruptive proposition in earth science.

A Scottish name James Hutton while in the country side made this proposition as he observed how landform in his native changed with each episode of season.

The principle proposes that "the processes occurring today have occurred in the time past at the same rate".

This way, it was much more easier to understand how land changes in pre-historic times have occurred.

Before his theory, the principle of catastrophism was the widely accepted one. This theory suggested that events occurred rapidly and changes to the surface are much more faster.

Question 4

The principle of superposition is one of the relative dating principles. It proposes that "in an undeformed land sequence, the oldest rock is at the bottom and the youngest on top".

The first sediment to get deposited fills the bottom as it aggregates upward. This leaves the youngest lithology to the top of strata.

The principle is correct for undeformed or undisturbed rock strata.

Where the sediments are disturbed, the formation might be overturned and this principle might be difficult to apply.

3 0

3 years ago

What is the relationship between gravity and electromagatism?

Mashcka [7]

Gravitational forece can only attract whereas electomagatism can be attractive and repulsive

5 0

3 years ago

Help me to solve this please

aivan3 [116]

Answer:

2) 7

3) 5

Explanation:

6 0

3 years ago

Read 2 more answers

A small block is attached to an ideal spring and is moving in SHM on a horizontal, frictionless surface. When the amplitude of t

Maslowich

Answer:

a) The time taken to travel from 0.18 m to -0.18m when the amplitude is doubled = 2.76 s

b) The time taken to travel from 0.09 m to -0.09 m when the amplitude is doubled = 0.92 s

Explanation:

a) The period of a simple harmonic motion is given as T = (1/f) = (2π/w)

It is evident that the period doesn't depend on amplitude, that is, it is independent of amplitude.

Hence, the time it would take the block to move from its amplitude point to the negative of the amplitude point (0.09 m to -0.09 m) in the first case will be the same time it will take the block to move from its amplitude point to negative of the amplitude point in the second case (0.18 m to -0.18 m).

Hence, time taken to travel from 0.18 m to -0.18m when the amplitude is doubled is 2.76 s

b) Now that the amplitude has been doubled, the time taken to move from amplitude point to the negative amplitude point in simple harmonic motion, just like with waves, is exactly half of the time period.

The time period is defined as the time taken to complete a whole cycle and a while cycle involves movement from the amplitude to point to the negative amplitude point then fully back to the amplitude point. Hence,

0.5T = 2.76 s

T = 2 × 2.76 = 5.52 s

Note that the displacement of a body undergoing simple harmonic motion from the equilibrium position is given as

y = A cos wt (provided that there's no phase difference, that is, Φ = 0)

A = amplitude = 0.18 m

w = (2π/5.52) = 1.138 rad/s

When y = 0.09 m, the time = t₁₂ = ?

0.09 = 0.18 Cos 1.138t₁ (angles in radians)

Cos 1.138t₁ = 0.5

1.138t₁ = arccos (0.5) = (π/3)

t₁ = π/(3×1.138) = 0.92 s

When y = -0.09 m, the time = t₂ = ?

-0.09 = 0.18 Cos 1.138t₂ (angles in radians)

Cos 1.138t₂ = -0.5

1.138t₂ = arccos (-0.5) = (2π/3)

t₂ = 2π/(3×1.138) = 1.84 s

Time taken to move from y = 0.09 m to y = -0.09 m is then t = t₂ - t₁ = 1.84 - 0.92 = 0.92 s

Hope this Helps!!!

3 0

3 years ago

A uniformly charged ring of radius 10.0 cm has a total charge of 75.0 mC. Find the electric field on the axis of the ring at (a)

wlad13 [49]

Answer:

(a) 6650246.305 N/C

(b) 24150268.34 N/C

(c) 6408227.848 N/C

(d) 665024.6305 N/C

Explanation:

Given:

Radius of the ring (r) = 10.0 cm = 0.10 m [1 cm = 0.01 m]

Total charge of the ring (Q) = 75.0 μC = $75\times 10^{-6}\ \mu C$ [1 μC = 10⁻⁶ C]

Electric field on the axis of the ring of radius 'r' at a distance of 'x' from the center of the ring is given as:

$E_x=\dfrac{kQx}{(x^2+r^2)^\frac{3}{2}}$

Plug in the given values for each point and solve.

(a)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=1.00\ cm=0.01\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.01)}{((0.01)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{6750}{1.015\times 10^{-3}}\\\\E_x=6650246. 305\ N/C$

(b)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=5.00\ cm=0.05\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.05)}{((0.05)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{33750}{1.3975\times 10^{-3}}\\\\E_x=24150268.34\ N/C$

(c)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=30.0\ cm=0.30\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(0.30)}{((0.30)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{202500}{0.0316}\\\\E_x=6408227.848\ N/C$

(d)

Given:

$Q=75\times 10^{-6}\ \mu C$ , $r=0.01\ m, a=100\ cm=1\ m,k=9\times 10^{9}\ Nm^2/C^2$

Electric field is given as:

$E_x=\dfrac{(9\times 10^{9})(75\times 10^{-6})(1)}{((1)^2+(0.1)^2)^\frac{3}{2}}\\\\E_x=\dfrac{675000}{1.015}\\\\E_x=665024.6305\ N/C$

7 0

3 years ago