Acomet goes around the sun in an elliptical orbit. at its farthest point, 490 million miles from the sun, it is traveling with a speed of 15,470 mi/h. how fast is it traveling at its closest approach to the sun, at a distance of 124 million miles?

V = 61,131.45 mi/h

Explanation:

In this case, we need to use the expression for linear momentum which is:

L = mVd

Where:

m = mass of the object

V = speed of the object

d = distance of the object or radius.

We also need to know that the law of conservation of energy is applied here, so:

L1 = L2

We can say that L1 would be the linear momentum of the comet at it's farthest point and L2 would be the linear momentum of the comet at it's closest point. Then, the expressions for linear momentum for both cases are:

m*V1*d1 = m*V2*d2

And now, all we need to do is calculate V2. So solving for V2:

V2 = m*V1*d1 / m*d2

V2 = V1*d1 / d2

Replacing all the data we have:

V2 = 490,000,000 * 15,470 / 124,000,000

V2 = 61,131.45 mi/h

  the grams     of oxygen gas   that can be produced     is 20.6   grams of o₂

calculation

2liclo₃ →   2 licl + 3 o₂

step 1:   find the moles of liclo₃

moles = mass÷ molar mass

from periodic table the molar mass of liclo₃ =6.9 +35.5   +(16 x3) =90.4 g/mol

moles=38.7 g÷90.4 g/mol=0.4281   moles

step 2:   use the   mole ratio   to calculate the moles of o2

from equation above, liclo₃: o₂   is 2: 3  

therefore the moles of   o₂=0.4281 moles x3/2 =0.6422   moles

step 3: find mass   of o₂

mass= moles x molar mass

from periodic table the   molar mass of   o₂ = 16 x2= 32 g/mol

mass=0.6422   moles   x   32 g/mol = 20.6 grams of o₂

c) up is the direction the magnetic force is acting on the charge.