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Briefly describe the characteristics of each soil horizon from the top layer to the bottom layer

inna [77]

<span>There is six horizen. 1. O Horizon - The top, organic layer of soil, 2. A Horizon - The layer called topsoil; 3. E Horizon - This layer is beneath the A Horizon and above the B Horizon. It is made up mostly of sand. 4. B Horizon - Also called the subsoil - this layer is beneath the E Horizon and above the C Horizon. 5. C Horizon - it's called regolith: the layer beneath the B Horizon and above the R Horizon. 6 R Horizon - this is last and the unweathered rock layer that is beneath all the other layers.</span>

7 0

4 years ago

In which of the following elevator situations would the acceleration be positive. Select TWO answers

kap26 [50]

Both options 5 and 6

Explanation:

Let us consider option 5,

In option 5 body is moving up with initial velocity lower than that of final velocity which gets accelerated. Therefore the acceleration is positive in this case.

Let us consider option 6,

In option 6 body is moving down with initial velocity lower than that of final velocity which gets accelerated. Therefore the acceleration is positive in this case.

4 0

3 years ago

Planet 1 orbits Star 1 and Planet 2 orbits Star 2 in circular orbits of the same radius. However, the orbital period of Planet 1

hichkok12 [17]

Answer:

The mass of Star 2 is Greater than the mass of Start 1. (This, if we suppose the masses of the planets are much smaller than the masses of the stars)

Explanation:

First of all, let's draw a free body diagram of a planet orbiting a star. (See attached picture).

From the free body diagram we can build an equation with the sum of forces between the start and the planet.

$\sum F=ma$

We know that the force between two bodies due to gravity is given by the following equation:

$F_{g} = G\frac{m_{1}m_{2}}{r^{2}}$

in this case we will call:

M= mass of the star

m= mass of the planet

r = distance between the star and the planet

G= constant of gravitation.

so:

$F_{g} =G\frac{Mm}{r^{2}}$

Also, if the planet describes a circular orbit, the centripetal force is given by the following equation:

$F_{c}=ma_{c}$

where the centripetal acceleration is given by:

$a_{c}=\omega ^{2}r$

where

$\omega = \frac{2\pi}{T}$

Where T is the period, and $\omega$ is the angular speed of the planet, so:

$a_{c} = ( \frac{2\pi}{T})^{2}r$

or:

$a_{c}=\frac{4\pi^{2}r}{T^{2}}$

so:

$F_{c}=m(\frac{4\pi^{2}r}{T^{2}})$

so now we can do the sum of forces:

$\sum F=ma$

$F_{g}=ma_{c}$

$G\frac{Mm}{r^{2}}=m(\frac{4\pi^{2}r}{T^{2}})$

in this case we can get rid of the mass of the planet, so we get:

$G\frac{M}{r^{2}}=(\frac{4\pi^{2}r}{T^{2}})$

we can now solve this for $T^{2}$ so we get:

$T^{2} = \frac{4\pi ^{2}r^{3}}{GM}$

We could take the square root to both sides of the equation but that would not be necessary. Now, the problem tells us that the period of planet 1 is longer than the period of planet 2, so we can build the following inequality:

$T_{1}^{2}>T_{2}^{2}$

So let's see what's going on there, we'll call:

$M_{1}$ = mass of Star 1

$M_{2}$ = mass of Star 2

So:

$\frac{4\pi^{2}r^{3}}{GM_{1}}>\frac{4\pi^{2}r^{3}}{GM_{2}}$

we can get rid of all the constants so we end up with:

$\frac{1}{M_{1}}>\frac{1}{M_{2}}$

and let's flip the inequality, so we get:

$M_{2}>M_{1}$

This means that for the period of planet 1 to be longer than the period of planet 2, we need the mass of star 2 to be greater than the mass of star 1. This makes sense because the greater the mass of the star is, the greater the force it applies on the planet is. The greater the force, the faster the planet should go so it stays in orbit. The faster the planet moves, the smaller the period is. In this case, planet 2 is moving faster, therefore it's period is shorter.

6 0

3 years ago

An ac power generator produces 73.37 a (rms) at 4623 v. the voltage is stepped up to 105,033 v by an ideal transformer, and the

evablogger [386]

Step up transformer is a device which is used to step up the voltage which is input with some value.

This is based upon the principle of mutual inductance and in this the voltage input and voltage output is different because of number of turns.

Here if ideal transformer is given then power input and power output of the transformer must be same as there is no power loss in ideal transformer.

So we can write

$i_pV_p = i_sV_s$

here

$i_p$ = 73.37 A

$V_p$ = 4623 V

$V_s$ = 105033 A

now using above equation we will have

$73.37*4623 = 105033*i_s$

solving above we will have

$i_s = 3.23 A$

7 0

3 years ago

Air bags can be dangerous or fatal under certain circumstances. you should not sit closer than ______ inches from the steering w

pishuonlain [190]

<span>10 inches
You are at risk of serious injury if you sit less than 10 inches away from the steering wheel, because of the speed and force the airbag deploys at. This is also part of the reason why driving instructors now instruct you to hold the steering wheel from the lower parts, rather than the top, which can cause your thumbs to break if the air bag deploys.</span>

5 0

4 years ago

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