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3 years ago

Which isotope is use to date ancient artifacts such as fossils?

Physics

1 answer:

meriva3 years ago

5 0

the isotope they use is carbon-14

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In the following equation , which compounds are the reactants? NaCL+AgNo>NaNo+AgCl

NeX [460]

The first two are always the reactants the products come after so they are last

3 0

3 years ago

Use the data provided to calculate the amount of heat generated for each cylinder height. Round your answers to the nearest tent

luda_lava [24]

100m: 4.9 kJ

200m: 9.8 kJ

1000m: 49.1 kJ

8 0

3 years ago

Read 2 more answers

sasho [114]

Answer:

The answer is 40 cm.

Explanation:

7 0

3 years ago

An object at rest on a flat, horizontal surface explodes into two fragments, one seven times as massive as the other. The heavie

leva [86]

To solve the problem it is necessary to apply conservation of the moment and conservation of energy.

By conservation of the moment we know that

$MV=mv$

Where

M=Heavier mass

V = Velocity of heavier mass

m = lighter mass

v = velocity of lighter mass

That equation in function of the velocity of heavier mass is

$V = \frac{mv}{M}$

Also we have that $m/M = 1/7 times$

On the other hand we have from law of conservation of energy that

$W_f = KE$

Where,

W_f = Work made by friction

KE = Kinetic Force

Applying this equation in heavier object.

$F_f*S = \frac{1}{2}MV^2$

$\mu M*g*S = \frac{1}{2}MV^2$

$\mu g*S = \frac{1}{2}( \frac{mv}{M})^2$

$\mu = \frac{1}{2} (\frac{1}{7}v)^2$

$\mu = \frac{1}{98}v^2$

$\mu = \frac{1}{g(98)(5.1)}v^2$

Here we can apply the law of conservation of energy for light mass, then

$\mu mgs = \frac{1}{2} mv^2$

Replacing the value of $\mu$

$\frac{1}{g(98)(5.1)}v^2 mgs = \frac{1}{2}mv^2$

Deleting constants,

$s= \frac{(98*5.1)}{2}$

$s = 249.9m$

7 0

3 years ago

You throw a 3.00 N rock vertically into the air from ground level. You observe that when it is 16.0 m above the ground, it is tr

Morgarella [4.7K]

Answer:

The final speed of the stone as it lift the ground is 23.86 m/s.

Explanation:

Given that,

Force acting on the rock, F = 3 N

Distance, d = 16 m

Initial speed of the stone, u = 22 m/s

We need to find the rock's speed just as it left the ground. It can be calculated using work energy theorem as :

$W=\Delta E\\\\W=\dfrac{1}{2}m(v^2-u^2)\\\\Fd=\dfrac{1}{2}m(v^2-u^2)\\\\v^2=\dfrac{2Fd}{m}+u^2\\\\v^2=\dfrac{2mgd}{m}+u^2\\\\v^2=2\times 9.8\times 16+(16)^2\\\\v=23.86\ m/s$

So, the final speed of the stone as it lift the ground is 23.86 m/s.

4 0

3 years ago