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3 years ago

Can you help me please?

Physics

1 answer:

egoroff_w [7]3 years ago

7 0

<h3>Answer</h3>

option B)

19N

<h3>Explanation</h3>

If the object is at equilibrium, then the net force acting upon the object should be 0 N. Thus, if all the forces are added together, horizontal and vertical forces separately, then the resultant force (the vector sum) should be 0 Newton.

As we only need to find the magnitude of x-component of force F

so find all x component/horizontal forces acting on the object.

50cos(40) - 40cos(25) + 30cos(55) + x = 0

38.30 - 36.25 + 17.21 + x + = 0

19.26 + x = 0

x = - 19.26

x ≈ 19 (magnitude only)

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Read 2 more answers

calculate earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the s

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Hello your question is incomplete below is the complete question

Calculate Earths velocity of approach toward the sun when earth in its orbit is at an extremum of the latus rectum through the sun, Take the eccentricity of Earth's orbit to be 1/60 and its Semimajor axis to be 93,000,000

answer : V = 1.624* 10^-5 m/s

Explanation:

First we have to calculate the value of a

a = 93 * 10^6 mile/m * 1609.344 m

= 149.668 * 10^8 m

next we will express the distance between the earth and the sun

$r = \frac{a(1-E^2)}{1+Ecos\beta }$ --------- (1)

a = 149.668 * 10^8

E (eccentricity ) = ( 1/60 )^2

$\beta$ = 90°

input the given values into equation 1 above

r = 149.626 * 10^9 m

next calculate the Earths velocity of approach towards the sun using this equation

$v^2 = \frac{4\pi^2 }{r_{c} }$ ------ (2)

Note :

Rc = 149.626 * 10^9 m

equation 2 becomes

( $V^2 = (\frac{4\pi^{2} }{149.626*10^9})$

therefore : V = 1.624* 10^-5 m/s

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