What is the equation that defines work

Suppose you were digging a well into saturated sediments. Why is the sediment’s permeability an important factor in deciding whe

zloy xaker [14]

Answer:

The importance of the sediments permeability is that if it is permeable, water will flow easily through the sediment and thereby produce a very good supply of water for the well.

Explanation:

When digging a well into saturated sediments, the possibility of the sediment with either little saturation or full saturation being able to provide steady water supply will be limited by how permeable it is. Now, the importance of the sediments permeability is that if it is permeable, water will flow easily through the sediment and thereby produce a very good supply of water for the well.

4 0

3 years ago

Hello everyone. This is a question about Dimensional Analysis and I came across this question but I am unable to wrap my head ar

omeli [17]

Answer:

2. [B] = [L]/[T] and [C] = [L]/[T]

Explanation:

I assume you mean this:

A = B² + 2B⁴/C²

Since you can't add numbers with different units (for example, you can't add seconds to meters), each term in the sum must have the same units as A.

B² = [L]²/[T]²

B = [L]/[T]

B⁴/C² = [L]²/[T]²

C²/B⁴ = [T]²/[L]²

C² = B⁴ [T]²/[L]²

C² = ([L]/[T])⁴ [T]²/[L]²

C² = [L]²/[T]²

C = [L]/[T]

Notice we ignore the 2 coefficient, which is unitless.

7 0

3 years ago

A mass weighing 14 pounds stretches a spring 2 feet. The mass is attached to a dashpot device that offers a damping force numeri

Elodia [21]

Answer:

The motion is over-damped when λ^2 - w^2 > 0 or when $b^{2}$ > 0.86

The motion is critically when λ^2 - w^2 = 0 or when $b^{2}$ = 0.86

The motion is under-damped when λ^2 - w^2 < 0 or when $b^{2}$ < 0.86

Explanation:

Using the newton second law

k is the spring constante

b positive damping constant

m mass attached

$m\frac{d^{2} x}{dt^{2}} = - kx - b\frac{dx}{dt}$

x(t) is the displacement from the equilibrium position

$\frac{d^{2} x}{dt^{2}} +\frac{b}{m}\frac{dx}{dt} + \frac{k}{m}x = 0$

Converting units of weights in units of mass (equation of motion)

$m = \frac{W}{g} = \frac{14}{32} = 0.43 slug$

From hook's law we can calculate the spring constant k

$k = \frac{W}{s} = \frac{14}{2} = 7 lb/ft$

If we put m and k into the DE, we get

$\frac{d^{2} x}{dt^{2}} +\frac{b}{0.43}\frac{dx}{dt} + 16.28x = 0$

Denoting the constants

2λ = $\frac{b}{m}$ = $\frac{b}{0.43}$

λ = b/0.215

$w^{2} = \frac{k}{m} = 16.28$

λ^2 - w^2 = $\frac{b^{2} }{0.046} - 16.28$

This way,

The motion is over-damped when λ^2 - w^2 > 0 or when $b^{2}$ > 0.86

The motion is critically when λ^2 - w^2 = 0 or when $b^{2}$ = 0.86

The motion is under-damped when λ^2 - w^2 < 0 or when $b^{2}$ < 0.86

3 0

3 years ago

Anna Litical analyzes the force between a planet and its moon, varying the mass of

Travka [436]

Answer:

Trial 1 is the largest, trial 3 is the smallest

Explanation:

Given:

Trial 1

M₁ = 6·10²² kg

d₁ = 3 500 km = 3.5·10⁶ м

Trial 2

M₂ = 6·10²² kg

d₂ = 7 000 km = 7·10⁶ м

Trial 3

M₃ = 3·10²² kg

d₃ = 7 000 km = 7·10⁶ м

___________

F - ?

Gravitational force:

F₁ = G·m·M₁ / d₁² = m·6.67·10⁻¹¹·6·10²² / (3.5·10⁶)² = 0.37·m (N)

F₂ = G·m·M₂ / d₂² = m·6.67·10⁻¹¹·6·10²² / (7·10⁶)² = 0.08·m (N)

F₃ = G·m·M₃ / d₃² = m·6.67·10⁻¹¹·3·10²² / (7·10⁶)² = 0.04·m (N)

Trial 1 is the largest, trial 3 is the smallest

5 0

1 year ago

Consider a projectile of mass 20 kg launched with a speed 9 m/s at an elevation angle of 45 degrees. Taking the launch point as

viva [34]

Answer:

a) L=0. b) L = 262 k ^ Kg m²/s and c) L = 1020.7 k^ kg m²/s

Explanation:

It is angular momentum given by

L = r x p

Bold are vectors; where L is the angular momentum, r the position of the particle and p its linear momentum

One of the easiest ways to make this vector product is with the use of determinants

${array}\right] \left[\begin{array}{ccc}i&j&k\\x&y&z\\px&py&pz\end{array}\right]$

Let's apply this relationship to our case

Let's start by breaking down the speed

v₀ₓ = v₀ cosn 45

voy =v₀ sin 45

v₀ₓ = 9 cos 45

voy = 9 without 45

v₀ₓ = 6.36 m / s

voy = 6.36 m / s

a) at launch point r = 0 whereby L = 0

. b) let's find the position for maximum height, we can use kinematics, at this point the vertical speed is zero

vfy² = voy²- 2 g y

y = voy² / 2g

y = (6.36)²/2 9.8

y = 2.06 m

Let's calculate the angular momentum

L= $\left[\begin{array}{ccc}i&j&k\\x&y&0\\px&0&0\end{array}\right]$

L = -px y k ^

L = - (m vox) (2.06) k ^

L = - 20 6.36 2.06 k ^

L = 262 k ^ Kg m² / s

The angular momentum is on the z axis

c) At the point of impact, at this point the height is zero and the position on the x-axis is the range

R = vo² sin 2θ / g

R = 9² sin (2 45) /9.8

R = 8.26 m

L = $\left[\begin{array}{ccc}i&j&k\\x&0&0\\px&py&0\end{array}\right]$

L = - x py k ^

L = - x m voy

L = - 8.26 20 6.36 k ^

L = 1020.7 k^ kg m² /s

5 0

3 years ago