1. Mitosis and Meiosis are both ways cells duplicate DNA
2. They both go through PMAT ( prophase, metaphase, anaphase and telophase)
<h2><em>So there is two truths given. After an amount of time Ttotal (lets call it ‘t’):
</em></h2><h2><em>
</em></h2><h2><em>The car’s speed is 25m/s
</em></h2><h2><em>The distance travelled is 75m
</em></h2><h2><em>Then we have the formulas for speed and distance:
</em></h2><h2><em>
</em></h2><h2><em>v = a x t -> 25 = a x t
</em></h2><h2><em>s = 0.5 x a x t^2 -> 75 = 0.5 x a x t^2
</em></h2><h2><em>Now, we know that both acceleration and time equal for both truths. So we can say:
</em></h2><h2><em>
</em></h2><h2><em>t = 25 / a
</em></h2><h2><em>t^2 = 75 / (0.5 x a) = 150 / a
</em></h2><h2><em>Since we don’t want to use square root at 2) we go squared for 1):
</em></h2><h2><em>
</em></h2><h2><em>t^2 = (25 / a) ^2 = 625 / a^2
</em></h2><h2><em>t^2 = 150 / a
</em></h2><h2><em>Since t has the same value for both truths we can say:
</em></h2><h2><em>
</em></h2><h2><em>625 / a^2 = 150 / a
</em></h2><h2><em>
</em></h2><h2><em>Thus multiply both sides with a^2:
</em></h2><h2><em>
</em></h2><h2><em>625 = 150 x a, so a = 625 / 150 = 4.17
</em></h2><h2><em>
</em></h2><h2><em>We can now calculate t as well t = 25 * 150 / 625 = 6</em></h2>
Answer:
8. 2.75·10^-4 s^-1
9. No, too much of the carbon-14 would have decayed for radiation to be detected.
Explanation:
8. The half-life of 42 minutes is 2520 seconds, so you have ...
1/2 = e^(-λt) = e^(-(2520 s)λ)
ln(1/2) = -(2520 s)λ
-ln(1/2)/(2520 s) = λ ≈ 2.75×10^-4 s^-1
___
9. Reference material on carbon-14 dating suggests the method is not useful for time periods greater than about 50,000 years. The half-life of C-14 is about 5730 years, so at 65 million years, about ...
6.5·10^7/5.73·10^3 ≈ 11344
half-lives will have passed. Whatever carbon 14 may have existed at the time will have decayed completely to nothing after that many half-lives.
Answer:
Parallel universe, or alternate reality, is a hypothetical self-contained plane of existence, co-existing with one's own
Answer:
Q=185.84C
Explanation:
We have to take into account the integral
In this case we have a superficial density in coordinate system.
Hence, we have for R: x2 + y2 ≤ 4
but, for symmetry:
![Q=4\int_0^2\int_0^{\sqrt{4-x^2}}\rho dydx\\\\Q=4\int_0^2\int_0^{\sqrt{4-x^2}}(4x+4y+4x^2+4y^2) dydx\\\\Q=4\int_0^{2}[4x\sqrt{4-x^2}+2(4-x^2)+4x^2\sqrt{4-x^2}+\frac{4}{3}(4-x^2)^{3/2}]dx\\\\Q=4[46.46]=185.84C](https://tex.z-dn.net/?f=Q%3D4%5Cint_0%5E2%5Cint_0%5E%7B%5Csqrt%7B4-x%5E2%7D%7D%5Crho%20dydx%5C%5C%5C%5CQ%3D4%5Cint_0%5E2%5Cint_0%5E%7B%5Csqrt%7B4-x%5E2%7D%7D%284x%2B4y%2B4x%5E2%2B4y%5E2%29%20dydx%5C%5C%5C%5CQ%3D4%5Cint_0%5E%7B2%7D%5B4x%5Csqrt%7B4-x%5E2%7D%2B2%284-x%5E2%29%2B4x%5E2%5Csqrt%7B4-x%5E2%7D%2B%5Cfrac%7B4%7D%7B3%7D%284-x%5E2%29%5E%7B3%2F2%7D%5Ddx%5C%5C%5C%5CQ%3D4%5B46.46%5D%3D185.84C)
HOPE THIS HELPS!!
