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SOVA2 [1]
3 years ago
13

What do Ice core samples with lower ratios of O-18 to O-16 isotopes tell scientist about past climates

Physics
1 answer:
Andrews [41]3 years ago
6 0

Answer:

Ocean-floor sediments can also be used to determine past climate. ... Ice cores contain more 16O than ocean water, so ice cores have a lower 18O/ 16O ratio than ocean water or ocean-floor sediments. Water containing the lighter isotope 16O evaporates more readily than 18O in the warmer subtropical regions

Explanation:

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zubka84 [21]

Answer:at 21.6 min they were separated by 12 km

Explanation:

We can consider the next diagram

B2------15km/h------->Dock

|

|

B1 at 20km/h

|

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V

So by the time B1 leaves, being B2 traveling at constant 15km/h and getting to the dock one hour later means it was at 15km from the dock, the other boat, B1 is at a distance at a given time, considering constant speed of 20km/h*t going south, where t is in hours, meanwhile from the dock the B2 is at a distance of (15km-15km/h*t), t=0, when it is 8pm.

Then we have a right triangle and the distance from boat B1 to boat B2, can be measured as the square root of (15-15*t)^2 +(20*t)^2. We are looking for a minimum, then we have to find the derivative with respect to t. This is 5*(25*t-9)/(sqrt(25*t^2-18*t+9)), this derivative is zero at t=9/25=0,36 h = 21.6 min, now to be sure it is a minimum we apply the second derivative criteria that states that if the second derivative at the given critical point is positive it means here we have a minimum, and by calculating the second derivative we find it is 720/(25 t^2 - 18 t + 9)^(3/2) that is positive at t=9/25, then we have our answer. And besides replacing the value of t we get the distance is 12 km.

3 0
3 years ago
A uranium ion and an iron ion are separated by a distance of =61.10 nm. The uranium atom is singly ionized; the iron atom is dou
scoray [572]

Answer:

Explanation:

Charge on uranium ion = charge of a single electron

= 1.6 x 10⁻¹⁹ C

charge on doubly ionised iron atom = charge of 2 electron

= 2 x 1.6 x 10⁻¹⁹ C = 3.2 x 10⁻¹⁹ C

Let the required distance from uranium ion be d .

force on electron at distance d from uranium ion

= 9 x 10⁹ x 1.6 x 10⁻¹⁹ / r²

force on electron at distance 61.10 x 10⁻⁹ - r from iron  ion

= 9 x 10⁹ x 3.2 x 10⁻¹⁹ / (61.10 x 10⁻⁹ - r )²

For equilibrium ,

9 x 10⁹ x 1.6 x 10⁻¹⁹ / r² = 9 x 10⁹ x 3.2 x 10⁻¹⁹ / (61.10 x 10⁻⁹ - r )²

2 d² = (61.10 x 10⁻⁹ - r )²

1.414 r = 61.10 x 10⁻⁹ - r

2.414 r = 61.10 x 10⁻⁹

r = 25.31 nm .

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