<span>First, we use the kinetic energy equation to create a formula:
Ka = 2Kb
1/2(ma*Va^2) = 2(1/2(mb*Vb^2))
The 1/2 of the right gets cancelled by the 2 left of the bracket so:
1/2(ma*Va^2) = mb*Vb^2 (1)
By the definiton of momentum we can say:
ma*Va = mb*Vb
And with some algebra:
Vb = (ma*Va)/mb (2)
Substituting (2) into (1), we have:
1/2(ma*Va^2) = mb*((ma*Va)/mb)^2
Then:
1/2(ma*Va^2) = mb*(ma^2*Va^2)/mb^2
We cancel the Va^2 in both sides and cancel the mb at the numerator, leving the denominator of the right side with exponent 1:
1/2(ma) = (ma^2)/mb
Cancel the ma of the left, leaving the right one with exponent 1:
1/2 = ma/mb
And finally we have that:
mb/2 = ma
mb = 2ma</span>
Rw^2 = GmM/r^2
<span> Leads to
</span><span> w^2 r^3 = GM
</span><span> (2pi /T) ^2 r^3 = GM
</span><span> 4pi^2 r^3 = GM T^2
</span><span> r^3 = GM T^2 / 4pi^2
</span><span> Work out r^3 then r.
</span> T = 125 min = 125(60) = 7500 s
<span> R = 6.38E6 m
</span><span> m = 5.97E24 kg
</span><span> G = 6.673E-11
</span> r=<span>
8279791.78</span><span> m
Since r = radius R of Earth + height above urface,h
</span><span> h = r - R = </span><span>
8279791.78 - </span>6.38E6 = <span>
<span>1899791.78 m
h=</span></span><span>
<span>1899.79178 Km</span></span>
Four electrons are placed at the corner of a square
So we will first find the electrostatic potential at the center of the square
So here it is given as
here
r = distance of corner of the square from it center
now the net potential is given as
now potential energy of alpha particle at this position
Now at the mid point of one of the side
Electrostatic potential is given as
here we know that
now potential is given as
now final potential energy is given as
Now work done in this process is given as
Answer:
5.843 m
Explanation:
suppose that the arrow leave the bow with a horizontal speed , towards he bull's eye.
lets consider that horizontal motion
distance = speed * time
time = 40/ 37 = 1.081 s
arrow doesnot have a initial vertical velocity component. but it has a vertical motion due to gravity , which may cause a miss of the target.
applying motion equation
(assume g = 10 m/s²)
Arrow misses the target by 5.843m ig the arrow us split horizontally
