The most energy-efficient way to send a spacecraft to the moon is to boost its speed while it is in circular orbit about the earth such that its new orbit is an ellipse. the boost point is the perigee of the ellipse, and the point of arrival at the moon is the apogee. calculate the percentage increase in speed required to achieve such an orbit. assume that the spacecraft is initially in a low-flying circular orbit about earth. the distance between earth and the moon is approximately 60 re, where re is the radius of earth

98.33 %

Explanation:

On an elliptical orbit, angular momentum will be conserved .

Angular momentum = I ω = mvR

So mv₁R₁ = mv₂R₂

= v₁R₁ = v₂R₂

where v₁ is velocity and R₁ radius in low orbit (perigee)and v₂ and R₂ is velocity and radius in high orbit ( apogee ).

Here R₁ = Radius of the earth , R₂ is distance between moon and earth.

R₁ / R₂ = 1/60

v₁ /v₂ = R₂ / R₁  = 60

v₂ / v₁ = 1 / 60

1 - (v₂ / v₁ ) = 1 -( 1 / 60)

(v₁ -v₂)/v₁ = ( 60-1 )/60

(v₁ -v₂)/v₁ x 100 = 5900/60 = 98.33 %

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz