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3 years ago

14

A rabbit runs 4.4 m across a lawn, stops, then runs 2.2 m back in the opposite direction. What is the rabbit’s displacement from

its starting point?

1 answer:

Alla [95]3 years ago

6 0

Answer:

it is 2.2 m

Explanation:

because he goes back 2.2 m so 4.4 minus 2.2 equals 2.2

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What are groups 1,2 and 3 examples of on the periodic table

pishuonlain [190]

<span>The number of the group identifies the column of the standard periodic table in which the element appears.</span>
Group 1 contains the alkali metals ( lithium<span> (</span>Li<span>), </span>sodium<span> (</span>Na<span>), </span>potassium<span> (</span>K<span>), </span>rubidium<span> (</span>Rb<span>), </span>caesium<span> (</span>Cs<span>), and </span>francium(Fr).)<span>
Group 2 contains the alkaline earth metals (</span> beryllium<span> (</span>Be),magnesium<span> (</span>Mg<span>), </span>calcium<span> (</span>Ca<span>), </span>strontium<span> (</span>Sr<span>), </span>barium<span> (</span>Ba<span>) and </span>radium<span> (</span>Ra<span>) )
Group 3: </span><span> Scandium (Sc) and yttrium (Y) </span>

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Which element could be described by the list of properties above?

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Answer:

list the properties

Explanation:

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A supersonic jet, with a mass of 21,000 kg, departs from its home airbase with a velocity of 400 m/s due east. What is the jet’s

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Answer:

Momentum P is 840000kgm/s or 8.4 × 10^6

Explanation:

Data :

Mass = 21000 kg

Velocity = 400 m/s

So momentum is given as

P = mv

P = 21000×400

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Suppose that you measure a galaxy's redshift, and from the redshift you determine that its recession velocity is 30,000 (3×10^4)

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Explanation:

v = Recessional velocity = 30000 km/s[/tex]

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According to Hubble's law

$v=H_0D\\\Rightarrow D=\frac{v}{H_0}\\\Rightarrow D=\frac{3\times 10^{4}}{65}\times 3.2\times 10^6\\\Rightarrow D=1476923076.92307\ ly\\\Rightarrow D=1.4\times 10^9\ ly$

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The gravitational force of a star on an orbiting planet 1 is f1. planet 2, which is three times as massive as planet 1 and orbit

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Let us consider two bodies having masses m and m' respectively.

Let they are separated by a distance of r from each other.

As per the Newtons law of gravitation ,the gravitational force between two bodies is given as - $F = G\frac{mm'}{r^{2} }$ where G is the gravitational force constant.

From the above we see that F ∝ mm' and $F\alpha \frac{1}{r^{2} }$

Let the orbital radius of planet A is $r_{1}$ = r and mass of planet is $m_{1}$ .

Let the mass of central star is m .

Hence the gravitational force for planet A is $f_{1} =G \frac{m_{1}*m }{r^{2} }$

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Hence the gravitational force $f_{2} =G\frac{m m_{2} }{r^{2} }$

$f_{2} =G\frac{m*3m_{1} }{[2r_{1}] ^{2} }$

$= \frac{3}{4} G\frac{mm_{1} }{r_{1} ^{2} }$

Hence the ratio is $\frac{f_{2} }{f_{1} } = \frac{\frac{3}{4}G mm_{1/r_{1} ^2} }{Gmm_{1}/r_{1} ^2 }$

$=\frac{3}{4}$ [ ans]

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