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Natasha_Volkova [10]

4 years ago

13

20 points for an answer Explain how the distance of a planet from the sun affects the motion of the planet? Justify your respons

e in two or more complete sentences.

2 answers:

Nataliya [291]4 years ago

4 0

The sun’s huge mass gives it a strong gravitational pull. Because of this gravitational pull, planets that are closer to the sun tend to have different motion than planets that are further away from the sun, because the gravity becomes stronger the closer you get. I hope this helped!

jekas [21]4 years ago

3 0

The sun’s huge mass gives it a strong gravitational pull. Because of this gravitational pull, planets that are closer to the sun tend to have different motion than planets that are further away from the sun, because the gravity becomes stronger the closer you get. I hope this helped!

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If I move 15ft foward, 15 ft backwards, 15 ft to the right, 15ft to the left where am I?

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Exactly where you started

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4 years ago

Read 2 more answers

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

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2 years ago

Object A has a mass m and a speed v, object B has a mass m/2 and a speed 4v, and object C has a mass 3m and a speed v/3. Rank th

borishaifa [10]

<h2>The rank is : $P_B>P_A=P_C$ .</h2>

We need to rank the objects according to the magnitude of their momentum.

We know momentum , $P = m\times v$ { here m is mass and v is velocity }

Momentum of object A , $P_A = m v.$

Momentum of object B , $P_B= \dfrac{m}{2}\times4v=2\ mv.$

Momentum of object C , $P_C= 3m\times\dfrac{v}{3}=mv.$

Now, we can rank them in order of their magnitude of momentum .

$P_B>P_A=P_C$ .

Hence, this is the required solution.

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