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3 years ago

Based on Planet Z's size, orbital distance, and rotation rate, which of the following properties is it likely to have?

Physics

1 answer:

d1i1m1o1n [39]3 years ago

3 0

Answer:

Planet Z will have the following properties;

Active Volcanoes

Active Tectonics

An Atmosphere produced by outgassing

Explanation:

The little terrestrial worlds have heat shorter than the much bigger terrestrial worlds, so the bigger worlds tend to have active volcanism and tectonics. These active volcanism and tectonics are likely to erase ancient craters. The active volcanism and tectonics would create an atmosphere by producing gases.

It is know that the Terrestrial worlds that are not far from the star have higher surface temperature.

Fast rate of rotation can cause winds and strong storms but here it is slower compared to earth. Also, a tilt of axis causes seasons.

The properties the star have are active volcanoes, active tectonics and an atmosphere produced by outgassing.

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A tree falls in a forest. How many years must pass before the 14C activity in 1.03 g of the tree's carbon drops to 1.02 decay pe

Illusion [34]

Answer:

t = 5.59x10⁴ y

Explanation:

To calculate the time for the ¹⁴C drops to 1.02 decays/h, we need to use the next equation:

$A_{t} = A_{0}\cdot e^{- \lambda t}$ (1)

where $A_{t}$ : is the number of decays with time, A₀: is the initial activity, λ: is the decay constant and t: is the time.

To find A₀ we can use the following equation:

$A_{0} = N_{0} \lambda$ (2)

where N₀: is the initial number of particles of ¹⁴C in the 1.03g of the trees carbon

From equation (2), the N₀ of the ¹⁴C in the trees carbon can be calculated as follows:

$N_{0} = \frac{m_{T} \cdot N_{A} \cdot abundance}{m_{^{12}C}}$

where $m_{T}$ : is the tree's carbon mass, $N_{A}$ : is the Avogadro's number and $m_{^{12}C}$ : is the ¹²C mass.

$N_{0} = \frac{1.03g \cdot 6.022\cdot 10^{23} \cdot 1.3\cdot 10^{-12}}{12} = 6.72 \cdot 10^{10} atoms ^{14}C$

Similarly, from equation (2) λ is:

$\lambda = \frac{Ln(2)}{t_{1/2}}$

where t 1/2: is the half-life of ¹⁴C= 5700 years

$\lambda = \frac{Ln(2)}{5700y} = 1.22 \cdot 10^{-4} y^{-1}$

So, the initial activity A₀ is:

$A_{0} = 6.72 \cdot 10^{10} \cdot 1.22 \cdot 10^{-4} = 8.20 \cdot 10^{6} decays/y$

Finally, we can calculate the time from equation (1):

$t = - \frac{Ln(A_{t}/A_{0})}{\lambda} = - \frac {Ln(\frac{1.02decays \cdot 24h \cdot 365d}{1h\cdot 1d \cdot 1y \cdot 8.20 \cdot 10^{6} decays/y})}{1.22 \cdot 10^{-4} y^{-1}} = 5.59 \cdot 10^{4} y$

I hope it helps you!

4 0

3 years ago

If earth's axis was tilted 45 degrees at what latitudes

givi [52]

23.5 DEGREES is the answer.

7 0

3 years ago

What is the density of the solid given that its mass is 200

alina1380 [7]

Answer:

a) 2cm³

b) 100 g/cm³

Explanation:

a- 9-7= 2cm³

b- 200 divided by 2= 100 g/cm³

Hope this helps... correct me if i'm wrong

7 0

3 years ago

The measure of a spring’s resistance to being compressed or stretched is the

larisa86 [58]

Answer;

Spring constant 

Explanation;

The measure of a spring’s resistance to being compressed or stretched is the spring constant.

The symbol of spring constant is K, since it is a constant. From the Hooke's law,for a helical spring or any elastic material, the extension force is directly proportional to the extension provided the elastic limit is not exceeded.

Therefore; the spring constant = Force/extension. That is; K = F/e; where k is the spring constant, F is the extension force and e is the extension.

Spring constant depicts the resistance of the spring to compressional and stretching forces.

7 0

3 years ago

Read 2 more answers

You have covered a grounded metal surface with a layer of photoconductor. Working in the dark, you sprinkle negative charge onto

Keith_Richards [23]

Answer:

A. the left half becomes neutral while the right half remains negatively charged

Explanation:

This is because wherever light strikes the photoconductor, it transforms from an insulator into a conductor. The charge will then migrate through it and leaves its surface. By exposing the left half of the photoconductor to light, you allow its local charge to leave and it becomes neutral.

6 0

3 years ago