Ask question

Ask question

All categories

2 years ago

5

Which of these is most likely a step in the formation of soil? a. erosion of soil b. crystallization of rocks c. animals digging

rocks d. solidification of organic matter

2 answers:

Lyrx [107]2 years ago

7 0

<span>C. Measure the soil's volume and then the volume of water that the soil can absorb.</span>

elixir [45]2 years ago

7 0

It's B! I already took the test

You might be interested in

A ball rolls from 10 m to -25 m in 2.5 seconds. What was<br> its average velocity?<br> (Units = m/s)

marissa [1.9K]

Answer:

v = -14 m/s

Explanation:

Given that,

Initial location of the ball, X₁ = 10 m

Final position of the ball, X₂ = -25 m

Time taken to travel is, t = 2.5 s

The average velocity of the ball is given by the formula,

V = X₂ - X₁ / t m/s

Substituting the values in the above equation,

V = -25 - 10 / 2.5

= -14 m/s

The negative sign in the velocity indicates that ball rolls in the opposite direction.

Hence, the average velocity of the ball is v = -14 m/s

8 0

3 years ago

What is the pressure drop due to the Bernoulli Effect as water goes into a 3.00-cm-diameter nozzle from a 9.00-cm-diameter fire

Marianna [84]

Answer:

The pressure and maximum height are $1.58\times10^{6}\ N/m^2$ and 161.22 m respectively.

Explanation:

Given that,

Diameter = 3.00 cm

Exit diameter = 9.00 cm

Flow = 40.0 L/s²

We need to calculate the pressure

Using Bernoulli effect

$P_{1}+\dfrac{1}{2}\rho v_{1}^2+\rho g h_{1}=P_{2}+\dfrac{1}{2}\rho v_{2}^2+\rho g h_{2}$

When two point are at same height so ,

$P_{1}+\dfrac{1}{2}\rho v_{1}^2=P_{2}+\dfrac{1}{2}\rho v_{2}^2$ ....(I)

Firstly we need to calculate the velocity

Using continuity equation

For input velocity,

$Q=A_{1}v_{1}$

$v_{1}=\dfrac{Q}{A_{1}}$

$v_{1}=\dfrac{40.0\times10^{-3}}{\pi\times(1.5\times10^{-2})^2}$

$v_{1}=56.58\ m/s$

For output velocity,

$v_{2}=\dfrac{40.0\times10^{-3}}{\pi\times(4.5\times10^{-2})^2}$

$v_{2}=6.28\ m/s$

Put the value into the formula

$P_{1}-P_{2}=\dfrac{1}{2}\rho(v_{1}^2-v_{2}^2)$

$\Delta P=\dfrac{1}{2}\times1000\times(56.58^2-6.28^2)$

$\Delta P=1.58\times10^{6}\ N/m^2$

(b). We need to calculate the maximum height

Using formula of height

$\Delta P=\rho g h$

Put the value into the formula

$1.58\times10^{6}=1000\times9.8\times h$

$h=\dfrac{1.58\times10^{6}}{1000\times9.8}$

$h=161.22\ m$

Hence, The pressure and maximum height are $1.58\times10^{6}\ N/m^2$ and 161.22 m respectively.

8 0

3 years ago

Compressing the air by squeezing the bottle was accompanied by a(n) ________ in the temperature of air inside the bottle

Georgia [21]

Increase in temprature

3 0

3 years ago

A particle moves at a constant speed of 34 m/s in a circular path of radius 6.3 m. From this information, what can be calculated

Yakvenalex [24]

Answer:

Centripetal acceleration

Explanation:

The centripetal acceleration is the motion inwards towards the center of a circular path.

<em><u>Centripetal acceleration is given by; the square of the velocity, divided by the radius of the circular path. </u></em>

That is;

ac = v²/r

Where; ac = acceleration, centripetal, m/s², v is the velocity, m/s and r is the radius, m

3 0

3 years ago

Find the time t2 that it would take the charge of the capacitor to reach 99.99% of its maximum value given that r=12.0ω and c=50

defon

Answer:

Explanation:

Given that, .

R = 12 ohms

C = 500μf.

Time t =? When the charge reaches 99.99% of maximum

The charge on a RC circuit is given as

A discharging circuit

Q = Qo•exp(-t/RC)

Where RC is the time constant

τ = RC = 12 × 500 ×10^-6

τ = 0.006 sec

The maximum charge is Qo,

Therefore Q = 99.99% of Qo

Then, Q = 99.99/100 × Qo

Q = 0.9999Qo

So, substituting this into the equation above

Q = Qo•exp(-t/RC)

0.9999Qo = Qo•exp(-t / 0.006)

Divide both side by Qo

0.9999 = exp(-t / 0.006)

Take In of both sodes

In(0.9999) = In(exp(-t / 0.006))

-1 × 10^-4 = -t / 0.006

t = -1 × 10^-4 × - 0.006

t = 6 × 10^-7 second

So it will take 6 × 10^-7 a for charge to reached 99.99% of it's maximum charge

8 0

3 years ago