Who speaks the line "Lord, what fools these mortals be"?

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

IceJOKER [234]

Answer:

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

4 0

3 years ago

A hi-lo lifts an 25 N skid to top of a pallet rack. The pallet rack is 3.6 meters tall. The hi-lo takes 12 seconds to get the sk

mojhsa [17]

Answer:

7.5Watts

Explanation:

Given parameters:

Force of lift = 25N

Height = 3.6m

Time = 12s

Unknown:

Power output = ?

Solution:

Power is the rate at which work is done ;

Power = $\frac{force x height }{time}$

Power = $\frac{25 x 3.6}{12}$ = 7.5Watts

8 0

2 years ago

Three charges, each of magnitude 10 nC, are at separate corners of a square of edge length 3 cm. The two charges at opposite cor

FinnZ [79.3K]

Answer:

The force exerted by three charges on the fourth is $F_{resultant}=2.74\times10^{-5}\ \rm N$

Explanation:

Given:

The magnitude of three identical charges, $q=10\ \rm nC$
Length of the edge of the square a=3 cm
Magnitude of fourth charge ,Q=3 nC

According to coulombs Law the force F between any two charge particles is given by

$F=\dfrac{kQq}{r^2}$

where r is the radial distance between them.

Since the force acting on the charge particle will be in different directions so according to triangle law of vector addition

$F_{resultant}=\sqrt ((\dfrac{kQq}{L^2})^2+(\dfrac{kQq}{L^2} })^2)+\dfrac{kQq}{(\sqrt{2}L)^2}\\F_{resultant}=\dfrac{kQq}{L^2}(\sqrt{2}-\dfrac{1}{2})\\F_{resultant}=\dfrac{9\times10^9\times10\times10^{-10}\times3\times10^{-9}}{0.03^2}(\sqrt{2}-\dfrac{1}{2})\\F_{resultant}=2.74\times 10^{-5}\ \rm N$