we have myrrh and would like to obtain address for Eurpa and asia companies dealing with myrrh.
Go to the embasssies in your country of the nations you wish to import to and ask for export assistance.
I am creating a nativity scence for some kids and need things that would represent or symbolise the gifts of Gold, Frankincense & Myrrh… Any ideas?
Potpourri, baby oil, and spray paint a brick metallic gold.
No
Santa is just a sort of nic name for Saint Nicolas.
Like Xmas is being used for Christmas.
And Saint Nicolas’s bishops robes were transfigured into the red and white Santa costume by the marketing department od Coca Cola. I asume the traditional catholic bisops robes didn’t mean a lot to most kids in predominatly protestant North America. Hence the need for a fashion change. Same goes for the term "Saint".
If you go to some European countries you will actually find that Santa is still called Saint Nicolas there and visits the kids in his traditional bishops robes bearing a bag of presents and sweets.
Men traditionally wear curved daggers known as Khanjar.
Oman
Wouldn’t something from Mothercare ( or that days equivelent )have been better ?
jana 11. Facts ?????
If these guys hadn`t brought along the incence, the church wouldn`t have an excuss to go round swinging their smoldering handbags!!
I am doing a clay art project and want to use the houses that the men built as the creation. The one built the house on the rock and one on the sand? Weren’t there three houses? This seems like such a good idea, can you help?
No, there were just two. The wise man built his house upon the rock and the foolish man built his house upon the sand. "And the rains came tumbling down…"
I think you are thinking of the pigs who built three houses…sticks, hay and brick.
Your spacecraft (mass m) has to travel to a far away (distance d) planet. Your engine can deliver a useful energy E. You must start and finish at rest.
What is the minimum travel time given classical mechanics?
What is the minimum travel time (proper and earth) given relativistic mechanics?
I think that the time-optimal approach is to accelerate till the maximum speed (given by the condition T = E/2) as soon as possible, then keep the speed constant until we are close enough to start decelerating in a similar manner.
If we assume that the distance travelled during the first and last stage is negligible to d, the optimal time needed is
t = d / v_max.
The two descriptions of mechanics differ in the expression of v_max.
In classical mechanics,
1/2*m*v_max^2 = E/2
v_max = sqrt(E/m)
t1 = d*sqrt(m/E).
In special relativity,
m / sqrt(1-v_max^2) = m + E/2
1 – v_max^2 = m^2 / (m + E/2)^2
v_max^2 = ((E/2)^2 + mE) / (m + E/2)^2
t2 = d * (m + E/2) / sqrt((E/2)^2 + mE)
(this is the Earth time),
for the proper time,
t3 = τ = ∫ (dτ/dt) dt
However, dτ/dt = sqrt(1-v^2) is constant for most of the interval, leaving a limit of
t3 = sqrt(1-v_max^2) * t2 = m / (m + E/2) * t2 = d * m / sqrt((E/2)^2 + mE).
It can be quickly seen that
t3/t1 = sqrt(mE / (mE + (E/2)^2)) < 1,
from which it follows that
t3 < t1.
Similarly, expanding t2^2 – t1^2, we find that
t1 < t2.
In summary, the special relativity causes a longer travelling time when observed from Earth but shorter proper time for the travellers.
Appendix: the derivation of the condition used in the beginning
For the time-wise optimal scenario, we will use variational calculus. The function to be optimized will be the dependency of consumed energy on the position, E(x). For symmetry reasons, we will study only the acceleration part (we would need some absolute values for the second part, the power is apparently not <0 when decelerating).
Our fixed boundary conditions will be E(0) = 0 and E(d/2) = E/2 and we will restrict the solution on nondecreasing functions.
The cost function is defined as the total time,
Φ[E(x)] = ∫ dt = ∫ (dt/dx) dx = ∫ [0 to d/2] 1/v dx,
where v is expressed in terms of E(x). Apparently, there is no dependence on E’(x), so the Euler-Lagrange equations are unapplicable and there is no regular optimum. However, the message is clear: when there is no restriction on the behaviour of E’(x), the closer we get v(E(x)) to its potential maximum, the better will be the cost. For both mechanics, v is monotoneously increasing in E, so the solution is the better the closer is E(x) to the constant function E(d/2) = E/2 in L1.
I know that we call it the star of Bethlehem, but what was it really?
Considering the fact that it apparently stayed in roughly the same position around the Earth for multiple days worth of time, it would almost have to have been a comet/meteoriod that got caught by Earth’s gravity in such a way that it almost became a geosynchronous natural satellite.
Three Kings Day is Epiphany (Jan 6th), and as such is globally celebrated
Title and annotations says all.