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Advocard [28]
3 years ago
5

The closure temperature represents the point when ________. A. a magma cools to the point where minerals begin to crystallize B.

radioactive isotopes begin to decay C. isotopes are no longer free to move out of a crystal lattice D. the last remaining magma crystallizes
Physics
1 answer:
melomori [17]3 years ago
5 0

The closure temperature represents the point when isotopes are no longer free to move out of a crystal lattice.

Answer: Option C

<u>Explanation:</u>

The closure temperature can also be termed as blocking temperature. It is mostly used in radiometric dating. As the temperature decreases, below a certain point the isotopes may get freeze in their lattice positions. And there may be slowing of diffusion.

At the closure temperature, that rate of diffusion will be zero as the isotopes will be no longer free to move out of crystal lattice. So, this is termed as closure or blocking temperature. As the isotopes loose their ability to move, their concentration will remain fixed in their position leading to measurement of radiation dating.

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Calculate the period of a ball tied to a string of length 0.468 m making 3.8 revolutions every second. Answer in units of s. You
Bogdan [553]

Answer:

0.26315 s

Explanation:

The frequency of the ball tied to a string system is 3.8 rev/s.

That means in one second the ball will complete 3.8 revolutions.

The time period will be the reciprocal of this frequency

T=\dfrac{1}{f}\\\Rightarrow T=\dfrac{1}{3.8}\\\Rightarrow T=0.26315\ s

The time period is 0.26315 s

It can be also solved in the following way

1\ s=3.8\ rev\\\Rightarrow 1\ rev=\dfrac{1}{3.8}\ s\\\Rightarrow 1\ rev=0.26315\ s

The time period is 0.26315 s

7 0
3 years ago
Suppose you are helping Galileo measure the acceleration due to gravity by dropping a canon ball from the tower of Pisa, You mea
Anni [7]

It is given that the height of the tower is

h=183 ft.

The uncertainty the measurement of this height is

\Delta h=0.2 ft

Drop time is measured as:

t=3.5s

The uncertainty in measurement of time is:

\Delta t=0.5 s

Using the equation of motion: h=ut+\frac{1}{2} at^2 where, h is the distance covered, u is the initial velocity, a is the acceleration and t is the time.

u=0 (because canon ball is in free fall). we need to calculate the value of a=g.

\Rightarrow h=\frac{1}{2}gt^2

\Rightarrow g=\frac{2h}{t^2}\\ \Rightarrow g=\frac{2\times 183ft}{(3.5s)^2}=29.87 ft/s^2

The uncertainty in this value is given by:

\Delta g=g\sqrt{(\frac{\Delta h}{h})^2+(\frac{2\Delta t}{t})^2}

Substitute the values:

\Delta g=29.87\sqrt{(\frac{0.2 }{183})^2+(\frac{2\times 0.5}{3.5})^2}=29.87\sqrt{1.19\times 10^{-6}+0.08}=29.87\times \sqrt{0.08}=29.87\times 0.28=8.44 ft/s^2



5 0
4 years ago
When you stifle a sneeze, you can damage delicate tissuesbecause the pressure of the air that is not allowed to escape mayrise b
anzhelika [568]

Answer:

2.49379624842 N

Explanation:

P = Pressure = 45 kPa

d = Diameter of eardrum = 8.44 mm

A = Area = \pi\dfrac{d^2}{4}

F = Force

Pressure is given by

P=\dfrac{F}{A}

\Rightarrow F=PA

\Rightarrow F=45000\times \pi\dfrac{(8.4\times 10^{-3})^2}{4}

\Rightarrow F=2.49379624842\ N

The outward force is 2.49379624842 N

5 0
4 years ago
What happens to the sound as frequency decreases? *
lisov135 [29]
I did a project on this in 8th grade I would try go with b because when the sound decrease the pitch is not affected but when frequency decreases the pitch is affected
6 0
3 years ago
During the filming of a certain movie scene, the director wants a small car (mass 795 kg) traveling due east at 35.0 m/s to coll
frosja888 [35]

Answer:

The trucker should drive at a speed of 11 m/s m/s

Explanation:

mass of the small car, m₁ = 795 kg

initial velocity of the small car, u₁ = 35 m/s(due east)

Mass of the truck, m₂ = 3360 kg

Initial velocity of the truck, u₂ = ?

Final velocity, V = u₁ sin37 ( the line between the camera and the collision makes an angle 37° with respect to the north)

According to the law of momentum conservation:

For the x -component

m₁u₁  = (m₁ + m₂)Vsinθ

m₁u₁  = (m₁ + m₂)Vsin37

(795*35) = (795 + 3360)* Vsin37

27825 = 2500.54V

V = 27825/2500.54

V = 11.13 m/s

For the y - component

m₂u₂  = (m₁ + m₂)Vcos37

3360*u₂ = (795+3360)*11.13*cos37

3360*u₂ =36933.02

u₂ = 36933.02/3360

u₂ = 11 m/s

7 0
3 years ago
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