A soft metal core made into a magnet by the passage of electric current through a coil surrounding it.
9,100 is the answer if you do the work so i failed for yall you welcome
a. 46 m/s east
The jet here is moving with a uniform accelerated motion, so we can use the following suvat equation to find its velocity:

where
v is the velocity calculated at time t
u is the initial velocity
a is the acceleration
The jet in the problem has, taking east as positive direction:
u = +16 m/s is the initial velocity
is the acceleration
Substituting t = 10 s, we find the final velocity of the jet:
And since the result is positive, the direction is east.
b. 310 m
The displacement of the jet can be found using another suvat equation
where
s is the displacement
u is the initial velocity
a is the acceleration
t is the time
For the jet in this problem,
u = +16 m/s is the initial velocity
is the acceleration
t = 10 s is the time
Substituting into the equation,

Answer:
No, it is not necessary for them to have same mass.
Explanation:
Let both bodies have a density d1 and d2 respectively.
Since their volumes are equal V1 = V2
we know that, https://tex.z-dn.net/?f=%5Cfrac%7Bmass%7D%7Bvolume%7D
Hence, d1 = and d2 =
Taking the ratio of densities,we get
This implies that unless the bodies have same densities, the mass of the two bodies will not be same.
Without counting wind resistance, They will both reach the ground at the same time. If we apply the concept of kinematics, such as the equation vf^2=vi^2 + 2ad. This equation doesn't count how big or how heavy the mass is, it only focuses on how fast where they in the start and how far are both of them from the ground. So if they both have the same distance and same initial veloctity, then they will reach the ground at the same time.
For example, Try dropping a pen and a paper(Vertically) at the same height, you'll see they'll reach the ground at the same time.
If you count wind resistance, the heavier ball will hit the ground faster, because the air molecules will resist the lighter ball compared to the heavier ball.