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Dennis_Churaev [7]
4 years ago
14

How does volcanic ash help scientists to determine the age of a rock sequence?

Physics
1 answer:
xeze [42]4 years ago
7 0

Answer:

Volcanic ash helps scientists to determine the age of a rock sequence by using the process of RADIOMETRIC DATING

Explanation:

When scientist want to determine the age of a rock sequence, they try to find it they is a volcanic ash present. This volcanic ash tend to contains radio isotopes or radioactive elements that can be used to determine the age of the rock sequence. This process is called a Radiometric dating.

Radiometric dating can be defined as the process of dating rocks or other materials based or the known rate of decay of radioactive isotopes.

Radiometric dating has to involve parent and daughter radioactive isotopes therefore the following must be observed before carrying out radiometric dating.

a. The half life of the Parent radioactive isotopes must be known

b. The daughter radioactive isotopes must be available in substantial amounts and it must be differentiated from the parent radioactive isotopes

c. The parent radioactive isotope must have a longer half-life present in the volcanic ash as at the time of radiometric dating.

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Water absorbs and releases large amount of energy before changing temperature, a characteristic known as
Anton [14]
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3 0
3 years ago
a car accelerates from rest at 3 m / s^2 along a straight road. how far has the car travelled after 4 s
evablogger [386]

Answer:

24 m

Explanation:

The motion of the car is a uniformly accelerated motion (=at constant acceleration), therefore we  can find the distance covered by using the following suvat equation:

s=ut+\frac{1}{2}at^2

where

s is the distance covered

u is the initial velocity

t is the time elapsed

a is the acceleration

For the car in this problem:

u = 0, since the car starts from rest

a=3 m/s^2 is the acceleration

t = 4 s is the time elapsed

Therefore, the distance covered is:

s=0+\frac{1}{2}(3)(4)^2=24 m

4 0
3 years ago
Explain why the wave model of light cannot explain the energy emissions from a blackbody
Cerrena [4.2K]

Answer:

As the temperature decreases, the peak of the black-body radiation curve moves to lower intensities and longer wavelengths. The black-body radiation graph is also compared with the classical model of Rayleigh and Jeans.

So as you see the wavelengths are in the x axis so all wavelengths are covered.

Black-body radiation provides insight into the thermodynamic equilibrium state of cavity radiation. If each Fourier mode of the equilibrium radiation in an otherwise empty cavity with perfectly reflective walls is considered as a degree of freedom capable of exchanging energy, then, according to the equipartition theorem of classical physics, there would be an equal amount of energy in each mode. Since there are an infinite number of modes this implies infinite heat capacity (infinite energy at any non-zero temperature), as well as an unphysical spectrum of emitted radiation that grows without bound with increasing frequency, a problem known as the ultraviolet catastrophe. Instead, in quantum theory the occupation numbers of the modes are quantized, cutting off the spectrum at high frequency in agreement with experimental observation and resolving the catastrophe. The study of the laws of black bodies and the failure of classical physics to describe them helped establish the foundations of quantum mechanics.

The above explains why the classical assumptions lead to a wrong spectrum.

Explanation:

i don't know if It helps you..parang Ang layo naman Ng sagot ko sa tanong mo

5 0
4 years ago
Need help ASAP
devlian [24]

Answer:

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Explanation:

5 0
3 years ago
Read 2 more answers
⦁ A 68 kg crate is dragged across a floor by pulling on a rope attached to the crate and inclined 15° above the horizontal. (a)
GREYUIT [131]

Answer:

303.29N and 1.44m/s^2

Explanation:

Make sure to label each vector with none, mg, fk, a, FN or T

Given

Mass m = 68.0 kg

Angle θ = 15.0°

g = 9.8m/s^2

Coefficient of static friction μs = 0.50

Coefficient of kinetic friction μk =0.35

Solution

Vertically

N = mg - Fsinθ

Horizontally

Fs = F cos θ

μsN = Fcos θ

μs( mg- Fsinθ) = Fcos θ

μsmg - μsFsinθ = Fcos θ

μsmg = Fcos θ + μsFsinθ

F = μsmg/ cos θ + μs sinθ

F = 0.5×68×9.8/cos 15×0.5×sin15

F = 332.2/0.9659+0.5×0.2588

F =332.2/1.0953

F = 303.29N

Fnet = F - Fk

ma = F - μkN

a = F - μk( mg - Fsinθ)

a = 303.29 - 0.35(68.0 * 9.8- 303.29*sin15)/68.0

303.29-0.35( 666.4 - 303.29*0.2588)/68.0

303.29-0.35(666.4-78.491)/68.0

303.29-0.35(587.90)/68.0

(303.29-205.45)/68.0

97.83/68.0

a = 1.438m/s^2

a = 1.44m/s^2

7 0
4 years ago
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