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3 years ago

14

What rock type does the principle of superposition apply to and how is the principle of superposition used to study the history

of life on Earth?

2 answers:

azamat3 years ago

8 0

Your answer would be sedimentary I believe

Strike441 [17]3 years ago

3 0

Answer: Sedimentary rocks

The principle of superposition is the basic principle of geology, which states that on examining the sequence of the strata, the stratum that is younger lies as the top whereas the older one rests at the bottom. This principle applies to the sedimentary rocks. As, the sediments are formed and piled one over the other, the compaction of the sediments occurs due to heat and pressure will result in formation of the sedimentary rocks.

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The moment of inertia of a uniform-density disk rotating about an axle through its center can be shown to be . This result is ob

Naddik [55]

(a) $0.2888 kg m^2$

The moment of inertia of a uniform-density disk is given by

$I=\frac{1}{2}MR^2$

where

M is the mass of the disk

R is its radius

In this problem,

M = 16 kg is the mass of the disk

R = 0.19 m is the radius

Substituting into the equation, we find

$I=\frac{1}{2}(16 kg)(0.19 m)^2=0.2888 kg m^2$

(b) 142.5 J

The rotational kinetic energy of the disk is given by

$K=\frac{1}{2}I\omega^2$

where

I is the moment of inertia

$\omega$ is the angular velocity

We know that the disk makes one complete rotation in T=0.2 s (so, this is the period). Therefore, its angular velocity is

$\omega=\frac{2\pi}{T}=\frac{2\pi}{0.2 s}=31.4 rad/s$

And so, the rotational kinetic energy is

$K=\frac{1}{2}(0.2888 kg m^2)(31.4 rad/s)^2=142.5 J$

(c) $9.07 kg m^2 /s$

The rotational angular momentum of the disk is given by

$L=I\omega$

where

I is the moment of inertia

$\omega$ is the angular velocity

Substituting the values found in the previous parts of the problem, we find

$L=(0.2888 kg m^2)(31.4 rad/s)=9.07 kg m^2 /s$

8 0

3 years ago

An electron travels through a particular point in an experimental apparatus with a velocity of 0.949 \times 10^6×10 6 m/s an

Rzqust [24]

Answer:

The magnitude of the acceleration of the electron at this point is $3.94\times10^{14}\ m/s^2$ .

Explanation:

Given that,

Velocity $v= 0.949\times10^{6}\ m/s$

Angle = 61.5°

Magnetic field = 0.01 T

We need to calculate the magnetic force

Using formula of magnetic force

$F=Bev\cos\theta$

Where, B = magnetic field

v = velocity

e = charge of electron

Put the value into the formula

$F=0.01\times1.6\times10^{-19}\times0.949\times10^{6}\cos61.5$

$F=3.59\times10^{-16}\ N$

We need to calculate the acceleration

Using newton's second law

$F= ma$

$a = \dfrac{F}{m}$

Put the value into the formula

$a=\dfrac{3.59\times10^{-16}}{9.1\times10^{-31}}$

$a=3.94\times10^{14}\ m/s^2$

Hence, The magnitude of the acceleration of the electron at this point is $3.94\times10^{14}\ m/s^2$ .

5 0

3 years ago

An atom with 6 protons and 5 electrons would be what kind of atom?

dybincka [34]

cation

anion is negatively charged, so more electrons then protons,

cation is positively charged, so more protons thatn electrons

4 0

3 years ago

Read 2 more answers

PLEASE HELP ON MORE SCIENCE! :)

VLD [36.1K]

1 is standing wave
2 is Constructive Interference
<span>3 is longitudinal
4 is first choice
5 is first choice</span>

8 0

3 years ago

Read 2 more answers

If the work required to speed up a car from 11 km/h to 21 km/h is 6.0×103 J , what would be the work required to increase the ca

goblinko [34]

Explanation:

We need convert the velocities first to m/s and we get the following:

v2 = 21 km/hr = 5.8 m/s

v1 = 11 km/hr = 3.1 m/s

We need to find the mass of the car also for later use do using the work-energy theorem:

$delta \: w = \frac{1}{2} m(v \frac{2}{2} - v\frac{2}{1} )$

6.0x10^3 J = (0.5) m [(5.8)^2 - (3.1)^2]

or

m = 499.4 kg

Now we determine work needed delta W to change its velocity from 21 km/hr to 33 km/hr

v2 = 33 km/hr = 9.2 m/s

v1 = 21 km/hr = 5.8 m/s

delta W = (0.5)(499.4)[(9.2)^2 - (5.8)^2]

= 1.3 x 10^4 J

6 0

3 years ago