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3 years ago

9

If an object were released in space far away from planted or stars and given an initial momentum, describe what would happen to

that objet

2 answers:

mamaluj [8]3 years ago

7 0

Strange as it may seem, the object would keep moving, in a straight line and at the same speed, until it came near another object. Its momentum and kinetic energy would never change. It might continue like that for a billion years or more.

Have a look at Newton's first law of motion.

-Dominant- [34]3 years ago

4 0

Answer:

The object in motion will stay in motion if there are no other forces affecting it.

Explanation:

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15m/s is how many Newtons

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147.09975 newton meters per second

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What is 541.2 mg in grams

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Answer:

0.5142 grams.

Explanation:

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An aluminum cylinder with a radius of 2.7 cm and a height of 67 cm is used as one leg of a workbench. The workbench pushes down

soldier1979 [14.2K]

Answer:

$1.9\times 10^{-4}$

$1.2\times 10^{-4}\ m$

Explanation:

r = Radius = 2.7 cm

F = Force = $3.2\times 10^4\ N$

A = Area = $\pi r^2$

$\sigma$ = Stress = $\frac{F}{A}$

E = Young's modulus = $7\times 10^{10}\ Pa$

$\epsilon$ = Strain

$L_0$ = Original length = 67 cm

$\Delta L$ = Change in length

Young's modulus is given by

$E=\frac{\sigma}{\epsilon}\\\Rightarrow \epsilon=\frac{\sigma}{E}\\\Rightarrow \epsilon=\frac{\frac{3.2\times 10^4}{\pi 0.027^2}}{7\times 10^{10}}\\\Rightarrow \epsilon=0.0001996=1.9\times 10^{-4}$

Strain is $1.9\times 10^{-4}$

Strain is given by

$\epsilon=\frac{\Delta L}{L_0}\\\Rightarrow \Delta L=\epsilon\times L_0\\\Rightarrow \Delta L=1.9\times 10^{-4}\times 0.67\\\Rightarrow \Delta L=0.0001273\\\Rightarrow \Delta L=1.2\times 10^{-4}\ m$

The cylinder height decreases by $1.2\times 10^{-4}\ m$

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3 years ago

An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) whi

Veronika [31]

Answer:

$s=5.79\ km$

$\theta=47^{\circ}$ east of south

Explanation:

Given:

distance of the person form the initial position, $d'=8.4\ km$

direction of the person from the initial position, $47^{\circ}$ north of east

distance supposed to travel form the initial position, $d=5.3\ km$

direction supposed to travel from the initial position, due North

<u>Now refer the schematic for visualization of situation:</u>

$y=d'.\sin47^{\circ}-d$

$y=8.4\times \sin47-5.3$ ...............(1)

$x=d'.\cos47^{\circ}$

$x=8.4\times \cos47^{\circ}$ .................(2)

<u>Now the direction of the desired position with respect to south:</u>

$\tan\theta=\frac{y}{x}$

$\tan\theta=\frac{8.4\times \sin47}{8.4\times \cos47}$

$\theta=47^{\circ}$ east of south

<u>Now the distance from the current position to the desired position:</u>

$s=\sqrt{x^2+y^2}$

$s=\sqrt{(8.4\times \cos47)^2+(8.4\times \sin47-5.3)^2}$

$s=5.79\ km$

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3 years ago

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Answer:

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Explanation:

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3 years ago