Answer:
93.125 × 10^(19)
Explanation:
We are told the asteroid has acquired a net negative charge of 149 C.
Thus;
Q = -149 C
charge on electron has a value of:
e = -1.6 × 10^(-19) C
Now, for us to determine the excess electrons on the asteroid, we will just divide the net charge in excess on the asteroid by the charge of a single electron.
Thus;
n = Q/e
n = -149/(-1.6 × 10^(-19))
n = 93.125 × 10^(19)
Thus, it has 93.125 × 10^(19) more electrons than protons
We/Wm = ge/gm = 120N/1.2N
or
gm = ge/100 = 0.1 m/s^2
density = mass/volume = 3M/(4pir^3)
Re-arranging this equation, we get
M/r^2 = (4/3)×pi×(density)×r
From Newton's universal law of gravitation, the acceleration due to gravity on the moon gm is
gm = G(M/r^2) = G×(4/3)×pi×(density)×r
Solving for density, we get the expression
density = 3gm/(4×pi×G×r)
= 3(0.1)/(4×3.14×6.67×10^-11×2.74×10^6)
= 130.6 kg/m^3
Answer:
after it has hit the ground
Statement :- We assume the orthagonal sequence
in Hilbert space, now
, the Fourier coefficients are given by:

Then Bessel's inequality give us:

Proof :- We assume the following equation is true

So that,
is projection of
onto the surface by the first
of the
. For any event, 
Now, by Pythagoras theorem:


Now, we can deduce that from the above equation that;

For
, we have

Hence, Proved
Answer:
t = 4 s
Explanation:
As we know that the particle A starts from Rest with constant acceleration
So the distance moved by the particle in given time "t"



Now we know that B moves with constant speed so in the same time B will move to another distance

now we know that B is already 349 cm down the track
so if A and B will meet after time "t"
then in that case


on solving above kinematics equation we have
