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alukav5142 [94]
3 years ago
12

The earth's magnetic fleld extends:

Physics
2 answers:
gregori [183]3 years ago
7 0

Answer:

The earth's magnetic field extends from the earth's core to outer space.

I hope it helps and please let me know if it is right or wrong.

Brrunno [24]3 years ago
5 0

Answer:

the earths magnetic field extends from the earths core into outerspace! Hope that helps

Explanation:

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A piece of aluminum (bulk modulus 7.1 x 1010 N/m2) is placed in a vacuum chamber where the air pressure is 0.781 x 105 Pa. The v
Jet001 [13]

Answer:

{\dfrac{\Delta V}{V}}=11\times 10^{-7}

Explanation:

Given that

Bulk modulus ,K = 7.1 x 10¹⁰  Pa

The pressure P₁ = 0.781 x 10⁵ Pa

The final pressure P ₂ = 0 Pa

We know that bulk modulus given as

K=\dfrac{\Delta P}{\dfrac{\Delta V}{V}}

Now by putting the values

K=\dfrac{\Delta P}{\dfrac{\Delta V}{V}}

7.1\times 10^{10}=\dfrac{0.781\times 10^5}{\dfrac{\Delta V}{V}}

{\dfrac{\Delta V}{V}}=\dfrac{0.781\times 10^5}{7.1\times 10^{10}}

{\dfrac{\Delta V}{V}}=11\times 10^{-7}

Therefore fractional change will be 11 x 10⁻⁷.

3 0
4 years ago
Complete the mechanism for the given decarboxylation by adding any missing atoms, bonds, charges, nonbonding electrons, and curv
AlexFokin [52]

The decarboxylation reaction involves the removal and replacement of a carboxyl group with hydrogen.

<h3>What is a decarboxylation reaction?</h3>

A decarboxylation reaction is one in which the carboxyl group in a carboxylic acid is replaced with hydrogen.

In living organisms, decarboxylation reactions are catalyzed by enzymes called decarboxylases or carboxy-lyases.

The decarboxylation of beta-carbonyl esters proceeds through a cyclic transition state giving an enol intermediate which then tautomerises to the carbonyl.

Therefore, a decarboxylation reaction involves the removal and replacement of a carboxyl group with hydrogen.

Leran more about decarboxylation at: brainly.com/question/893601

5 0
2 years ago
A harmonic wave on a string is described by
xxMikexx [17]

\huge{\bold{\orange{\underline{ Solution }}}}

<h3><u>Given </u><u>:</u><u>-</u></h3>

A harmonic wave on a string is described by

\sf{ Y( x, t)  = 0.1 sin(300t + 0.01x + π/3)}

  • x is in cm and t is in seconds

<h3><u>Answer </u><u>1</u><u> </u><u>:</u><u>-</u></h3>

<u>Equation </u><u>for </u><u>travelling </u><u>wave </u><u>:</u><u>-</u>

\sf{ Y( x, t)  = Asin(ωt + kx + Φ)...eq(1)}

<u>Equation</u><u> </u><u>for </u><u>stationary </u><u>wave </u><u>:</u><u>-</u>

\sf{ Y( x, t)  = Acos(ωt - kx )...eq(2)}

<u>Given </u><u>equation </u><u>for </u><u>wave </u><u>:</u><u>-</u>

\sf{ Y( x, t)  = 0.1 \:sin(300t + 0.01x + π/3)...eq(3)}

<u>On </u><u>comparing </u><u>eq(</u><u>1</u><u>)</u><u> </u><u>,</u><u> </u><u>(</u><u>2</u><u>)</u><u> </u><u>and </u><u>(</u><u>3</u><u>)</u>

We can conclude that, Given wave represent travelling wave.

<h3><u>Answer </u><u>2</u><u> </u><u>:</u><u>-</u></h3>

From solution 1 , We can say that,

\sf{ Y( x, t)  = 0.1 \: sin(300t + 0.01x + π/3).}

It is travelling from right to left direction

Hence, The direction of its propagation is right to left that is towards +x direction.

<h3><u>Answer </u><u>3</u><u> </u><u>:</u><u>-</u></h3>

Here, We have to find the wave period

<u>We </u><u>know </u><u>that</u><u>, </u>

Wave period = wavelength / velocity

<u>Wave </u><u>equation</u><u> </u><u>:</u><u>-</u>

\sf{ Y( x, t)  = 0.1 \:sin(300t + 0.01x + π/3).}

  • ω = 300rad/s
  • k = 0.01

<u>We </u><u>know </u><u>that</u><u>, </u>

\sf{v =}{\sf{\dfrac{ ω}{2π}}}{\sf{\: and\:}}{\sf{ λ =}}{\sf{\dfrac{ 2π}{k}}}

<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>

\sf{ wave\: period =}{\sf{\dfrac{ 2π/k}{ω/2π  }}}

\sf{ wave \:period = }{\sf{\dfrac{k}{ω}}}

\sf{ wave\: period =}{\sf{\dfrac{ 0.01}{300}}}

\sf{ wave\: period = 0.000033\: s}

<h3><u>Answer </u><u>4</u><u> </u><u>:</u><u>-</u></h3>

The wavelength of given wave

\bold{ λ = }{\bold{\dfrac{2π}{k}}}

<u>Subsitute </u><u>the </u><u>required </u><u>values</u><u>, </u>

\sf{ λ = }{\sf{\dfrac{2 × 3.14 }{0.01}}}

\sf{ λ = }{\sf{\dfrac{6.28}{0.01}}}

\sf{ λ = 628 \: cm }

<h3><u>Answer </u><u>5</u><u> </u><u>:</u><u>-</u></h3>

We have wave equation

\sf{ Y( x, t)  = 0.1 sin(300t + 0.01x + π/3).}

<u>Travelling </u><u>wave </u><u>equation </u><u>:</u><u>-</u>

\sf{ Y( x, t)  = A\:sin(ωt + kx + Φ)...eq(1)}

<u>Therefore</u><u>, </u>

Amplitude of the wave particle

\sf{ A = 0.1 \: cm}

Hence, The amplitude of the particle is 0.1 cm

8 0
2 years ago
f we apply an electrical stimulus to a muscle cell to cause it to contract, the magnitude of that stimulus must be strong enough
kondor19780726 [428]

Answer:

critical value is known as threshold

Explanation:

solution

when we apply electrical stimulus to the cell of muscle

then magnitude of stimulus will be strong enough for reaching the critical value

and this critical value is threshold because when the size of an stimulus change in requirement for the detection, this is called the threshold

but if change is done by without stimulation then there threshold stimulus is absolute threshold

so correct answer is threshold

4 0
3 years ago
a system that uses reflected radio waves to detect objects and to measure their distance and speed is called
viktelen [127]
Such system is called RADAR (<span>RAdio </span>Detection<span> And Ranging).

The Radar emits radio waves, that are reflected back by the object. Since the speed of the radio waves is known (their speed is equal to the speed of light), by measuring the time the waves take to come back to the source it is possible to infer the distance they covered, and so the distance of the object.</span>
3 0
3 years ago
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