Ek = (m*V^2) / 2 where m is mass and V is speed, then we can take this equation and manipulate it a little to isolate the speed.
Ek = mv^2 / 2 — multiply both sides by 2
2Ek = mv^2 — divide both sides by m
2Ek / m = V^2 — switch sides
V^2 = 2Ek / m — plug in values
V^2 = 2*30J / 34kg
V^2 = 60J/34kg
V^2 = 1.76 m/s — sqrt of both sides
V = sqrt(1.76)
V = 1.32m/s (roughly)
Answer:
Explanation:
A charge located at a point will experience a zero electrostatic force if the resultant electric field on it due to any other charge(s) is zero.
is located at the origin. The net force on it will only be zero if the resultant electric field intensity due to 
and 
at the origin is equal to zero. Therefore we can perform this solution without necessarily needing the value of .
Let the electric field intensity due to be +
and that due to be -
since the charge is negative. Hence at the origin;
From equation (1) above, we obtain the following;
From Coulomb's law the following relationship holds;
where 
is the distance of from the origin, 
is the distance of from the origin and k is the electrostatic constant.
It therefore means that from equation (2) we can write the following;
k can cancel out from both side of equation (3), so that we finally obtain the following;
Given;
Substituting these values into equation (4); we obtain the following;
Answer:
0.11 kg
Explanation:
Ft = MV
Ft = momentum 5.22kg m/s
M = mass
V = velocity 48.3m/s
Therefore
5.22 = M x 48.3
Divide both sides by 48.3
5.22/48.3 = M x 48.3/48.3
0.11 = M
M = 0.11kg
Answer:
a. the core will spin faster.
Explanation:
By law of conservation of angular momentum
(mvR)i= (mvR)f
m= mass of star
v= speed of star
R= radius of star
i= initial
f= final
since, size(R) of the star is reduced by factor of 10,000 and mass remains the same, the velocity must increase by the same factor to keep the angular momentum conserved.
Hence, a. the core will spin faster.
I would say that insofar as the two stars temperatures are presumably closely related to their luminosity, that the blue star at 156,100 k compared to 3000k for the red star then the blue star has a luminosity of 52 times that of the red star.
