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3 years ago

A soil sample contains 55 percent sand, 25 percent clay, and 20 percent silt. What type of soil is present in the sample?

Physics

1 answer:

Step2247 [10]3 years ago

7 0

Answer: D (sandy clay loam)

Explanation:

It is because it is 55 percent sand, and 25 percent clay, so it has to be more sandy than silty. Much of it is by sand, so it is sandy. Loam is a fertile soil of clay and sand, so it is D.

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An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m. Part A How m

r-ruslan [8.4K]

PART A)

Electrostatic potential at the position of origin is given by

$V = \frac{kq_1}{r_1} + \frac{kq_2}{r_2}$

here we have

$q_1 = 1.6 \times 10^{-19} C$

$q_2 = -1.6 \times 10^{-19} C$

$r_1 = r_2 = 1 m$

now we have

$V = \frac{Ke}{r} - \frac{Ke}{r}$

$V = 0$

Now work done to move another charge from infinite to origin is given by

$W = q(V_f - V_i)$

here we will have

$W = e(0 - 0) = 0$

so there is no work required to move an electron from infinite to origin

PART B)

Initial potential energy of electron

$U = \frac{Kq_1e}{r_1} + \frac{kq_2e}{r_2}$

$U = \frac{9\times 10^9(-1.6\times 10^{-19}(-1.6 \times 10^{-19})}{19} + \frac{9\times 10^9(1.6\times 10^{-19}(-1.6 \times 10^{-19})}{21}$

$U = (2.3\times 10^{-28})(\frac{1}{19} - \frac{1}{21})$

$U = 1.15\times 10^{-30}$

Now we know

$KE = \frac{1}{2}mv^2$

$KE = \frac{1}{2}(9.1\times 10^{-31}(100)^2$

$KE = 4.55 \times 10^{-27} kg$

now by energy conservation we will have

So here initial total energy is sufficient high to reach the origin

PART C)

It will reach the origin

4 0

3 years ago

Write a short description of how the motion of the racers might change from the start of the race to the finish line

Ad libitum [116K]

The motion of the racers might change from the start because the pressure goes up so all the racer wants is to speed up and win, so when the racer first starts he or she is calm because he's not driving yet and when he or she is on his/hers way to he finish line he/she just wants to win and gets under pressure so he speeds up even more and drifts. Your welcome

6 0

3 years ago

In preparation for the final exam, our astronomy study group has reconvened to discuss Mercury's unique orbital properties. They

grigory [225]

Answer:

The only incorrect statement is from student B

Explanation:

The planet mercury has a period of revolution of 58.7 Earth days and a rotation period around the sun of 87 days 23 ha, approximately 88 Earth days.

Let's examine student claims using these rotation periods

Student A. The time for 4 turns around the sun is

t = 4 88

t = 352 / 58.7 Earth days

In this time I make as many rotations on itself each one with a time to = 58.7 Earth days

#_rotaciones = t / to

#_rotations = 352 / 58.7

#_rotations = 6

therefore this statement is TRUE

student B. the planet rotates 6 times around the Sun

t = 6 88

t = 528 s

The number of rotations on itself is

#_rotaciones = t / to

#_rotations = 528 / 58.7

#_rotations = 9

False, turn 9 times

Student C. 8 turns around the sun

t = 8 88

t = 704 days

the number of turns on itself is

#_rotaciones = t / to

#_rotations = 704 / 58.7

#_rotations = 12

True

The only incorrect statement is from student B

6 0

3 years ago

El ancho de la uña de su dedo meñique se acerca a una unidad metrica comun ¿cual?

BartSMP [9]

Answer:

I have no clue what's really going on I'm just here to get answer maybe I will just try to get an answer but I have no clue I'm sorry I am confused and dint really know what to do here.

3 0

3 years ago

Find the resultant of an easterly force of 100 N and a southeast force of 80 N acting at 65 degrees to the 100 N force

saveliy_v [14]

Answer:

Resultant is 152 N at 28.5 degrees south to the 100 N force

Explanation:

7 0

4 years ago