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3 years ago

How do data collected from tree rings provide evidence for climate

Physics

1 answer:

Nimfa-mama [501]3 years ago

7 0

Answer:

Climate scientists compare the tree growth records to local weather records. For locations where a good statistical match exists between tree growth and temperature or precipitation during the period of overlap, the ring widths can be used to estimate past temperature or precipitation over the lifetime of the tree.

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The change in momentum that occurs when a 1. 0 kg ball traveling at 4. 0 m/s strikes a wall and bounces back at 2. 0 m/s is.

Doss [256]

Answer:

The change is momentum is given by ∆p=p(inital) - p(final) =4-2=2 kg.m/s

Explanation:

momentum is the product of mass and velocity (speed)

So it's initial momentum would be:

p=mv=(1)(4)=4 kg.m/s

It's final momentum is given by:

p=mv=(1)(2)=2 kg.m/s

7 0

2 years ago

In an experiment, a disk is set into motion such that it rotates with a constant angular speed. As the disk spins, a small spher

boyakko [2]

Answer:

L₀ = L_f , K_f < K₀

Explanation:

For this exercise we start as the angular momentum, with the friction force they are negligible and if we define the system as formed by the disk and the clay sphere, the forces during the collision are internal and therefore the angular momentum is conserved.

This means that the angular momentum before and after the collision changes.

Initial instant. Before the crash

L₀ = I₀ w₀

Final moment. Right after the crash

L_f = (I₀ + mr²) w

we treat the clay sphere as a point particle

how the angular momentum is conserved

L₀ = L_f

I₀ w₀ = (I₀ + mr²) w

w = $\frac{I_o}{I_o + m r^2}$ w₀

having the angular velocities we can calculate the kinetic energy

starting point. Before the crash

K₀ = ½ I₀ w₀²

final point. After the crash

K_f = ½ (I₀ + mr²) w²

sustitute

K_f = ½ (I₀ + mr²) ( $\frac{I_o}{I_o + m r^2}$ w₀)²

Kf = ½ $\frac{I_o^2}{ I_o + m r^2}$ w₀²

we look for the relationship between the kinetic energy

$\frac{K_f}{K_o}$ = $\frac{I_o}{I_o + m r^2}$

$\frac{K_f}{K_o } < 1$

K_f < K₀

we see that the kinetic energy is not constant in the process, this implies that part of the energy is transformed into potential energy during the collision

6 0

3 years ago

A car moves 50km in the first 30 minutes and 50km in the second 30 minutes and continues in this way which motion is this ------

egoroff_w [7]

Answer:

Uniform rectilinear movement (m. r. u.)

Explanation:

It is a continuous movement without acceleration, that is, it moves at a constant speed. The speed does not change over time, for this reason, there is no change in acceleration.

Vf = final velocity = 50/30 [km/s] = 1.67 [km/s]

Vo = initial velocity = 50/30 [km/s] = 1.67 [km/s]

3 0

3 years ago

What are the three rules for writing the electron configuration of elements

astraxan [27]

Answer:

The aufbau principle

The Pauli exclusion principle

Hund's rule of maximum multiplicity

Explanation:

The aufbau principle:

The fundamental electronic configuration is achieved by placing the electrons one by one in the different orbitals available for the atom, which are arranged in increasing order of energy.

The Pauli exclusion principle:

Two electrons of the same atom cannot have their four equal quantum numbers. Because each orbital is defined by the quantum numbers n, l, and m, there are only two possibilities ms = -1/2 and ms = +1/2, which physically reflects that each orbital can contain a maximum of two electrons, having opposite spins

Hund's rule of maximum multiplicity:

This rule says that when there are several electrons occupying degenerate orbitals, of equal energy, they will do so in different orbitals and with parallel spins, whenever this is possible. Because electrons repel each other, the minimum energy configuration is one that has electrons as far away as possible from each other, and that is why they are distributed separately before two electrons occupy the same orbital.

4 0

3 years ago

A satellite that goes around the earth once every 24 hours (86,400 s) is called a geosynchronous satellite. If a geosynchronous

Olegator [25]

Answer:

42244138.951 m

Explanation:

G = Gravitational constant = 6.667 × 10⁻¹¹ m³/kgs²

r = Radius of orbit from center of earth

M = Mass of Earth = 5.98 × 10²⁴ kg

m = Mass of Satellite

The satellite revolves around the Earth at a constant speed

Speed = Distance / Time

The distance is the perimeter of the orbit

$v=\frac{2\pi \times r}{24\times 3600}$

The Centripetal force of the satellite is balanced by the universal gravitational force

$m\frac{v^2}{r}=\frac{GMm}{r^2}\\\Rightarrow \frac{\left(\frac{2\pi \times r}{24\times 3600}\right)^2}{r}=\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}}{r^2}\\\Rightarrow \left(\frac{2\pi \times r}{24\times 3600}\right)^2=6.667\times 10^{-11}\times 5.98\times 10^{24}\\\Rightarrow r^3=\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}\times (24\times 3600)^2}{(2\pi)^2}\\\Rightarrow r=\left(\frac{6.667\times 10^{-11}\times 5.98\times 10^{24}\times (24\times 3600)^2}{(2\pi)^2}\right)^{\frac{1}{3}}\\\Rightarrow r=42244138.951\ m$

The radius as measured from the center of the Earth) of the orbit of a geosynchronous satellite that circles the earth is 42244138.951 m

6 0

3 years ago