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Whitepunk [10]
3 years ago
15

Select all of the answers that apply.

Physics
2 answers:
kolezko [41]3 years ago
7 0

Answer:

-law of original horizontality

-law of superposition

-law of original lateral continuity

-law of cross-cutting or intrusive relationships

Explanation:

Relative dating is the science that many scientist use to set an order, or to figure out the order in which past events happened, this is without actually invetigating the natural age of the elements, and from stratigraphy and paleontology the laws of relative dating are:

  1.    Uniformitarianism
  2.    Intrusive relationships
  3.    Cross-cutting relationships
  4.    Inclusions and components
  5.    Original horizontality
  6.    Superposition
  7.    Faunal succession
  8.    Lateral continuity
postnew [5]3 years ago
3 0
Here are the answers to the given question above.
<span>Relative dating uses laws or principles of stratigraphy and paleontology. These laws of relative dating are: 
-</span><span>law of original horizontality 
-</span><span>law of superposition 
-</span><span>law of original lateral continuity 
-</span><span>law of cross-cutting or intrusive relationships 
Hope these are the answers that you are looking for.</span>
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aleksandrvk [35]

Answer:

25

Explanation:

Because u don't know the gravity so u take that force and just put it in the gravity spot

5 0
3 years ago
a car traveling at 24 m/s starts to decelerate steadily. It comes to a complete stop in 6 seconds what is its acceleration?
Irina18 [472]
We can solve for the acceleration by using a kinematic equation. First we should identify what we know so we can choose the  correct equation.

We are given an original velocity of 24 m/s, a final velocity of 0 m/s, and a time  of 6 s. We and looking for acceleration (a) in m/s^2.

The following equation has everything we need:

v_f=v_i + at

So plug in the known values and solve for a:

0 = 24 + 6a

-24 = 6a

a = -4 m/s^2
8 0
4 years ago
A 10 g particle undergoes SHM with an amplitude of 2.0 mm and a maximum acceleration of magnitude 8.0 multiplied by 103 m/s2, an
Nat2105 [25]

Answer:

a)T=0.0031416s

b)v_{max}=6.283\frac{m}{s}

c) E=0.1974J

d)F=80N

e)F=40N

Explanation:

1) Important concepts

Simple harmonic motion is defined as "the motion of a mass on a spring when it is subject to the linear elastic restoring force given by Hooke's Law (F=-Kx). The motion experimented by the particle is sinusoidal in time and demonstrates a single resonant frequency".

2) Part a

The equation that describes the simple armonic motion is given by X=Acos(\omega t +\phi)    (1)

And taking the first and second derivate of the equation (1) we obtain the velocity and acceleration function respectively.

For the velocity:

\frac{dX}{dt}=v(t)=-A\omega sin(\omega t +\phi)   (2)

For the acceleration

\frac{d^2 X}{dt}=a(t)=-A\omega^2 cos(\omega t+\phi)   (3)

As we can see in equation (3) the acceleration would be maximum when the cosine term would be -1 and on this case:

A\omega^2=8x10^{3}\frac{m}{s^2}

Since we know the amplitude A=0.002m  we can solve for \omega like this:

\omega =\sqrt{\frac{8000\frac{m}{s^2}}{0.002m}}=2000\frac{rad}{s}

And we with this value we can find the period with the following formula

T=\frac{2\pi}{\omega}=\frac{2 \pi}{2000\frac{rad}{s}}=0.0031416s

3) Part b

From equation (2) we see that the maximum velocity occurs when the sine function is euqal to -1 and on this case we have that:

v_{max}=A\omega =0.002mx2000\frac{rad}{s}=4\frac{m rad}{s}=4\frac{m}{s}

4) Part c

In order to find the total mechanical energy of the oscillator we can use this formula:

E=\frac{1}{2}mv^2_{max}=\frac{1}{2}(0.01kg)(6.283\frac{m}{s})^2=0.1974J

5) Part d

When we want to find the force from the 2nd Law of Newton we know that F=ma.

At the maximum displacement we know that X=A, and in order to that happens cos(\omega t +\phi)=1, and we also know that the maximum acceleration is given by::

|\frac{d^2X}{dt^2}|=A\omega^2

So then we have that:

F=ma=mA\omega^2

And since we have everything we can find the force

F=ma=0.01Kg(0.002m)(2000\frac{rad}{s})^2 =80N

6) Part e

When the mass it's at the half of it's maximum displacement the term cos(\omega t +\phi)=1/2 and on this case the acceleration would be given by;

|\frac{d^2X}{dt^2}|=A\omega^2 cos(\omega t +\phi)=A\omega^2 \frac{1}{2}

And the force would be given by:

F=ma=\frac{1}{2}mA\omega^2

And replacing we have:

F=\frac{1}{2}(0.01Kg)(0.002m)(2000\frac{rad}{s})^2 =40N

8 0
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Vilka [71]

I might be wrong but I'm pretty sure It's B

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Temka [501]

Answer:

C

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