Solution :
When the spacecraft is at halfway point, the distance from the Earth as well as Mars are same. We have to account the masses of the planets. The gravitational force that is exerted by the Earth is greater because of its combined mass with the space probe.
The mass of Earth is greater than the mass of Mars. Therefore, the force of Earth is more than Mars.
Answer:
The correct answer is 
Explanation:
The formula for the electron drift speed is given as follows,
where n is the number of of electrons per unit m³, q is the charge on an electron and A is the cross-sectional area of the copper wire and I is the current. We see that we already have A , q and I. The only thing left to calculate is the electron density n that is the number of electrons per unit volume.
Using the information provided in the question we can see that the number of moles of copper atoms in a cm³ of volume of the conductor is 
. Converting this number to m³ using very elementary unit conversion we get 
. If we multiply this number by the Avagardo number which is the number of atoms per mol of any gas , we get the number of atoms per m³ which in this case is equal to the number of electron per m³ because one electron per atom of copper contribute to the current. So we get,
if we convert the area from mm³ to m³ we get 
.So now that we have n, we plug in all the values of A ,I ,q and n into the main equation to obtain,
which is our final answer.
Answer
D. move a small magnet back and forth within a section of the coiled wire.
Explanation:
i put that for the test and i got it right
Calculating acceleration involves dividing velocity by time — or in terms of units, dividing meters per second [m/s] by second [s]. Dividing distance by time twice is the same as dividing distance by the square of time. Thus the SI unit of acceleration is the meter per second squared.
This I what it showed when I looked it up so if your writing this down try and put some in your own words
In addition to acceleration of gravity we experience centrifugal acceleration away from the axis of rotation of the earth. this additional acceleration has value ac = r w^2 where w = angular velocity and r is distance from your spot on earth to the earth's axis of rotation so r = R cos(l) where l = 60 deg is the lattitude and R the earth's radius and w = 1 / (24hr x 3600sec/hr)
<span>now you look up R and calculate ac then you combine the centrifugal acc. vector ac with the gravitational acceleration vector ag = G Me/R^2 to get effective ag' = ag -</span>
