Answer:
The last option is correct
ΔT1 = 4 sec ΔT2 = 4 sec ΔT3 = 4 sec
S = V0 t + 1/2 a t^2
S1 = 1/2 a t^2 = 8 a where V0 is the speed at the start of the interval
During any interval (of 4 sec) the particle travels 1/2 a t^2 = 8 a due to its acceleration - and you need to include the speed at the start of the interval
S1 = 8a
S2 = 8 a + 8 a = 16 a
S3 = 16 a + 8 a = 24 a
Note: V2 = V1 + a t for any interval
V2 - V1 = V1 + a t - V1 = a t
and a = (V2 - V1) / t = a the speed increase is constant during the interval
<h3><u>Answer;</u></h3>
Carbon-14 levels in a sample are undetectable after approximately 9 half lives
<h3><u>Explanation;</u></h3>
- <em><u>The half life of Carbon-14 is 5,730 years . Half life is the time taken by a radioactive material to decay by half of its original mass. Therefore, it would take a time of 5730 years for a sample of 100 g of carbon-14 to decay to 50 grams</u></em>
- <em><u>A period of 50,000 years, is equivalent to; </u></em>
<em><u> 50,000÷5,730 </u></em>
<em><u>= 8.73 half lives</u></em>
<em>Which is approximately equal to 9 half lives.</em>
- Therefore, if the age of an object older than 50,000 years cannot be determined by radiocarbon dating, then <em><u>Carbon-14 levels in a sample are undetectable after approximately 9 half lives</u></em>.
Answer:
Esto se debe a que la soda está más caliente que el hielo, lo que hace que pase por su punto de fusión.
Explanation:
Me enteré de esto durante la clase de ciencias hace un par de años. jajaja
Answer:
D. Newton is a derived unit.
Answer:
Explanation:
One charge is situated at x = 1.95 m . Second charge is situated at y = 1.00 m
These two charges are situated outside sphere as it has radius of .365 m with center at origin. So charge inside sphere = zero.
Applying Gauss's theorem
Flux through spherical surface = charge inside sphere / ε₀
= 0 / ε₀
= 0 Ans .