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3 years ago

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Visualize five horizontal sedimentary strata exposed in a cliff or canyon wall identified by consecutive numbers, 1 being the lo

west bed and 5 being the highest. Which of the following statements concerning the strata are true?
a. Bed 4 is older than bed 2.
b. Bed 3 is older than beds 2 and 4.
c. Bed 5 is the oldest.
d. Beds l and 3 are older than bed 4.

2 answers:

Grace [21]3 years ago

6 0

Answer:

Only the 4 th statement is true that is bed 1 and 3 are older than 4.

Explanation:

The 5 beds are numbered from 1 to 5 , 1 being the lowest and 5 being the topmost bed.

We are given 4 statements and we have to find out which all are true.

(<u>a)Bed 4 is older than bed 2 </u>

This is wrong because the lower beds are older than beds that are higher.

<u>(b)Bed 3 is older than beds 2 and 4</u>

This is also wrong because the 2 is older than 3

(<u>c)Bed 5 is the oldest </u>

This is wrong because bed 1 is the oldest

(<u>d)Beds 1 and 3 are older than 4</u>

This is true as lower beds are older

<u />

Vlad1618 [11]3 years ago

4 0

Answer:

bed 2 was deposited before bed 3

Explanation:

Which of the following conclusions can be made about the sedimentary layers?

bed 2 was deposited after bed 3

bed 1 is the youngest

bed 3 was deposited before bed 1

bed 2 was deposited before bed 3

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Answer:

a) a geostationary satellite is that it is always at the same point with respect to the planet,

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c) d) w = 1.745 10⁻⁴ rad / s

Explanation:

a) The definition of a geostationary satellite is that it is always at the same point with respect to the planet, that is, its period of revolutions is the same as the period of the planet

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b) the period the frequency are related

T = 1 / f

f = 1 / T

f = 1 / 3.6 104

f = 2.7777 10⁻⁵ Hz

c) the distance traveled by the satellite in 1 day

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d = 2pi (R + r)

d = 2pi (69 911 103 + 120 106)

d = 1193.24 m

d) the angular velocity is the angle traveled between the time used.

.w = 2pi /t

w = 2pi / 3.6 10⁴

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how fast is

v = w r

v = 1.75 10-4 (69.911 106 + 120 106)

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Answer:

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Explanation:

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<h3>What are these properties an example of?</h3>

These are all examples of the physical properties of a sample. Since we cannot see the sample that David is using, it would be safest to assume that he would have no trouble providing evidence as to the physical properties of the soil, the:

Color
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3 years ago

Read 2 more answers

A package of mass m is released from rest at a warehouse loading dock and slides down a 3.0-m-high frictionless chute to a waiti

LuckyWell [14K]

Answer:

The speed of the package of mass m right before the collision $= 7.668\ ms^-1$

Their common speed after the collision $= 2.56\ ms^-1$

Height achieved by the package of mass m when it rebounds $= 0.33\ m$

Explanation:

Have a look to the diagrams attached below.

a.To find the speed of the package of mass m right before collision we have to use law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

where $K$ is Kinetic energy and $U$ is Potential energy.

$K= \frac{mv^2}{2}$ and $U= mgh$

Considering the fact $K_{initial} = 0\ and U_{final} =0$ we will plug out he values of the given terms.

So $V_{1}{(initial)} =\sqrt{2gh} = \sqrt{2\times9.8\times3} = 7.668\ ms^-1$

Keypoints:

Sum of energies and momentum are conserved in all collisions.
Sum of KE and PE is also known as Mechanical energy.
Only KE is conserved for elastic collision.
for elastic collison we have $e=1$ that is co-efficient of restitution.

<u>KE = Kinetic Energy and PE = Potential Energy</u>

b.Now when the package stick together there momentum is conserved.

Using law of conservation of momentum.

$m_1V_1(i) = (m_1+m_2)V_f$ where $V_1{i} =7.668\ ms^-1$ .

Plugging the values we have

$m\times 7.668 = (3m)\times V_{f}$

Cancelling m from both sides and dividing 3 on both sides.

$V_f = 2.56\ ms^-1$

Law of conservation of energy will be followed over here.

c.Now the collision is perfectly elastic $e=1$

We have to find the value of $V_{f}$ for m mass.

As here $V_{f}=-2.56\ ms^-1$ we can use that if both are moving in right ward with $2.56$ then there is a $-2.56$ velocity when they have to move leftward.

The best option is to use the formulas given in third slide to calculate final velocity of object $1$ .

So

$V_{1f} = \frac{m_1-m_2}{m_1+m_2} \times V_{1i}= \frac{m-2m}{3m} \times7.668=\frac{-7.668}{3} = -2.56\ ms^-1$

Now using law of conservation of energy.

$K_{initial} + U_{initial} = K_{final}+U_{final}$

$\frac{m\times V(f1)^2}{2} + 0 = 0 +mgh$

$\frac{v(f1)^2}{2g} = h$

$h= \frac{(-2.56)^2}{9.8\times 3} =0.33\ m$

The linear momentum is conserved before and after this perfectly elastic collision.

So for part a we have the speed $=7.668\ ms^-1$ for part b we have their common speed $=2.56\ ms^-1$ and for part c we have the rebound height $=0.33\ m$ .

3 0

3 years ago