The North Star, or Polaris, is the brightest star in the constellation Ursa Minor, the little bear (also known as the Little Dipper). As viewed by observers in the Northern Hemisphere, Polaris occupies a special place
Answer:
0.0000076 grams
Explanation:
We're given the half life of Tritium to be 12.3 years. In order to find out the amount of substabce remaining:
Let's first find how many 'half lives' are in 250 years.

Now what is half life? It means the time taken for a given quantity of an element to lose half it's mass.
So in 12.3 years we can find that The amount of 250 g of Tritium will be 250/2 = 125 g. In 24.6 years we'll have 125/2 = 62.5 g
So now we can devise a formula:

Where m is the remaining amount and n is th number of half lives in the time given.
Using this formula we can calculate:

Doing this calculation we get:

As we can see a very small value remains.
Answer: E = 
Explanation: The formulae for intensity of an electric field of a solid metal sphere relative to a point is given below
r
where
,
, r = 0.1m r = is the position vector of the charge.
it has been stated in the question that the charge is placed at the center thus it has no position vector.

The value of spring constant and the oscillator's damping constant is
K= 6605.667008, b= 0.002884387
Explanation:
For Weakly damping spring oscillator
K/m = W_0^2 (at resonance)
K= mW_0^2
=0.206 * ( 2π * 28.5) ^2
=0.206 * (2π)^2 * (28.5)^2
K= 6605.667008
F = - bV
b= -F/V = -F/ -W_0 * m
=F/W_0 * m
= 0.438N / 2π * 28.5 * 0.848
b= 0.002884387
Answer:
4948020
Explanation:
(6.9*10-6)(770*102)
<u>Multiply 6.9 by 10
.
</u>
(
69
−
6
)
(
770
⋅
102
)
<u>Subtract 6 from 69
.
</u>
63
(
770
⋅
102
)
<u>Multiply 770 by 102
.
</u>
63
⋅
78540
<u>Multiply 63 by 78540
.
</u>
4948020