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1 year ago

What is the resultant of the vectors shown?

Physics

1 answer:

ad-work [718]1 year ago

5 0

Answer:

Option B

Explanation:

Option A is the wrong answer because the horizontal vector is in the opposite direction.

Option C is the wrong answer as the horizontal vector is in the opposite direction and all the vectors are connected head to tail [of the arrows] [Triangle law of vector addition]

Option D is the wrong answer as the horizontal vector is in the opposite direction.

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son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

$PE = -\frac{GMm}{R}$

As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

$dKE = -dU$

Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

$v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}$

$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

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2 years ago

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zysi [14]

<span>The support force is 15 N. The net force on the book is zero.</span>

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2 years ago

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kogti [31]

The work is 90 as 5 times 18

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2 years ago

A 1.5 kg ball pushed with a force of 13.5 N accelerates to the left. What is the acceleration of the ball?

GarryVolchara [31]

Explanation:

F= ma

13.5 =1.5a

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2 years ago