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3 years ago

Scientists can find the absolute age of a rock by using

Physics

1 answer:

yaroslaw [1]3 years ago

6 0

Explanation:

Carbon-14 is normally used to date living organisms (all living things we know today contain mostly carbon). It can be used for samples up to about 60000 years old. Carbon-12 is the non-radioactive form that we encounter daily. Relative dating determines dates relative to each other, such as ordering of period or events.

Uranium 235 dating can determine date samples a million years old and up, so is more suited for rocks.

You might be interested in

Which statement about nuclear fusion is correct? A. Two hydrogen electrons become protons during fusion. B. Helium nuclei can fu

velikii [3]

Answer:

A is the answer

Explanation:

Yes, Two hydrogen electrons become protons during fusion.

By far, the most common fusion reaction in nature combines two hydrogen atoms to make a helium atom.

8 0

2 years ago

Read 2 more answers

The graph in the accompanying figure (Figure 1) shows the magnitude of the force exerted by a given spring as a function of the

serious [3.7K]

The elastic potential energy stored in the stretched spring is 1 J.

<h3>What is Hooke's law?</h3>

Hooke's law states that; provided the elastic limit is not exceeded, the extension of the spring is directly proportional to the force on the spring.

Given that;

Force on the spring = 350 Newton

Distance stretched = 7 centimeters or 0.07 m

Hence;

F = ke

k = F/e = 350 Newton/0.07 m = 5000 N/m

Work done in stretching a spring = 1/2ke^2

= 0.5 × 5000 × (2 × 10^-2)^2 =1 J

Learn more about elastic potential energy: brainly.com/question/156316

4 0

2 years ago

A player is positioned 35 m[40 degrees W of S] of the net. He shoot the puck 25 m [E] to a teammate. What second displacement do

Lubov Fominskaja [6]

Answer:

x=22.57 m

Explanation:

Given that

35 m in W of S

angle = 40 degrees

25 m in east

From the diagram

The angle

$\theta=90-40=50^o$

From the triangle OAB

$cos40^o=\frac{35^2+25^2-x^2}{2\times 35\times 25}$

$1340.57=35^2+25^2-x^2$

x=22.57 m

Therefore the answer of the above problem will be 22.57 m

4 0

3 years ago

Some beetles break down the remains of dead animals. Some mushrooms breakdown the remains of dead trees. How do these actions mo

Y_Kistochka [10]

Answer:

A By returning nutrients to the soil

Explanation:

Here are the options

A By returning nutrients to the soil

B By releasing oxygen into the air

C By making space for new animals

D By decreasing the population of herbivores

The process by which mushrooms act on dead trees provide nutrients that are necessary for the process of photosynthesis.

Types of decomposers

Fungi - mushroom is an example of a fungi
Bacteria
Insects - beetle is an example of an insect
Earthworms

Functions of a decomposer

A decomposer recycles dead plants and animals to release nutrients into the soil.
They are ecological cleaners. they carry out this function by disintegrating dead plants and animal remains.

3 0

3 years ago

An object is propelled straight up from ground level with an initial velocity of 48 feet per second. Its height at time t is mod

Ket [755]

Answer:

Explanation:

For a. its max height and when it occurs. First the max height. That's a y-dimension thing, and in the y-dimension we have this info:

v₀ = 48 ft/s

a = -32 ft/s/s

v = 0 (the max height of an object occurs when the final velocity of the object is 0). Use the following equation for this part of the problem:

v² = v₀² + 2aΔx and filling in:

$0=48^2+2(-32)$ Δx and

0 = 2300 - 64Δx and

-2300 = -64Δs so

Δx = 36 feet.

Now for the time it takes to get to this max height. Final velocity is still 0 here, but the equation is a different one for this part of the problem. Use:

v = v₀ + at and filling in:

0 = 48 - 32t and

-48 = -32t so

t = 1.5 sec. That's part a. Onto part b:

The object hits the ground when its displacement, Δx, is 0. Use this equation for this problem:

Δx = v₀t + $\frac{1}{2}at^2$ and filling in:

$0=48t+\frac{1}{2}(-32)t^2$ and

$0=48t-16t^2$ and

0 = 16t(3 - t) so

t = 0 and t = 3. t = 0 is before the object is propelled, so it makes sense that at 0 seconds, the object was still on the ground, right? Then at 3 seconds, it's back on the ground. (Isn't math just perfectly, beautifully sensible!?) Now onto part c:

We are looking for the time interval when the object is >32 feet. So we use the same equation we just used, but with an inequality instead of an equals sign:

$48t+\frac{1}{2}(-32)t^2 >32$ and get everything on one side and factor it again:

$-16t^2+48t-32>0$ and we find that

1 < t < 2 so the time interval is between 1 and 2 seconds that the object is over 32 feet in the air.

8 0

3 years ago