All categories

3 years ago

Which "spheres" are interacting when water evaporates from plants

2 answers:

8 0

The plant grows in the solid part of earth, the lithosphere. When water evaporates from the plant, it enters the hydrosphere, the portion if earth on kand and in the air that contains water. The atmosphere is part of the hydrosphere.

Volgvan3 years ago

4 0

Answer:

Explanation:

You might be interested in

the attendance qt a seminar in a year 1 was 500 in year 2 the attendance changed by 100 what was the percent change in attendanc

TiliK225 [7]

20% change from year one to year two.

5 0

4 years ago

Please help

inysia [295]

Explanation:

Start with what you know and list your knowns and unknowns

F = ma

F= 3N

m = 6kg

a =?

3N = 6kg x a

solve for a

3N / 6kg = a

3 0

3 years ago

Two gratings A and B have slit separations dA and dB, respectively. They are used with the same light and the same observation s

77julia77 [94]

Answer:

a) dB / dA = 2 ,

b) Network B Network A

2 1

4 2

6 3

Explanation:

a) The expression for grating diffraction is

d sin θ = m λ

where d the distance between two slits, λ the wavelength and m an integer that represents the diffraction range

In this exercise we are told that the two spectra are in the same position, let's write the expression for each network

Network A

m = 1

sin θ = 1 λ / dA

Network B

m = 2

sin θ = 2 λ / dB

they ask us for the relationship between the distances, we match the equations

λ / dA) = 2 λ / dB

dB / dA = 2

b) let's write the equation of the networks

sin θ = m_A λ / dA

sin θ = m_B λ / dB

we equalize

m_A λ/ dA = m_B λ / dB

we use that

dB / dA = 2

m_A 2 = m_B

therefore the overlapping orders are

Network B Network A

2 1

4 2

6 3

4 0

3 years ago

A(n) 28.3 g bullet is shot into a(n) 5004 g wooden block standing on a frictionless surface. The block, with the bullet in it, a

Natali [406]

Answer:

the speed of the bullet before striking the block is 302.3 m/s.

Explanation:

Given;

mass of the bullet, m₁ = 28.3 g = 0.0283 kg

mass of the wooden block, m₂ = 5004 g = 5.004 kg

initial velocity of the block, u₂ = 0

final velocity of the bullet-wood system, v = 1.7 m/s

let the initial velocity of the bullet before striking the block = u₁

Apply the principle of conservation of linear momentum to determine the initial velocity of the bullet.

m₁u₁ + m₂u₂ = v(m₁ + m₂)

0.0283u₁ + 5.004 x 0 = 1.7(0.0283 + 5.004)

0.0283u₁ = 8.5549

u₁ = 8.5549 / 0.0283

u₁ = 302.3 m/s

Therefore, the speed of the bullet before striking the block is 302.3 m/s.

4 0

3 years ago

The inner cylinder of a long, cylindrical capacitor has radius r and linear charge density +λ. It is surrounded by a coaxial cyl

Ulleksa [173]

Hi there!

We can begin by using the equation for energy density.

$U = \frac{1}{2}\epsilon_0 E^2$

U = Energy (J)

ε₀ = permittivity of free space

E = electric field (V/m)

First, derive the equation for the electric field using Gauss's Law:
$\Phi _E = \oint E \cdot dA = \frac{Q_{encl}}{\epsilon_0}$

Creating a Gaussian surface being the lateral surface area of a cylinder:
$A = 2\pi rL\\\\E \cdot 2\pi rL = \frac{Q_{encl}}{\epsilon_0}\\\\Q = \lambda L\\\\E \cdot 2\pi rL = \frac{\lambda L}{\epsilon_0}\\\\E = \frac{\lambda }{2\pi r \epsilon_0}$

Now, we can calculate the energy density using the equation:
$U = \frac{1}{2} \epsilon_0 E^2$

Plug in the expression for the electric field and solve.

$U = \frac{1}{2}\epsilon_0 (\frac{\lambda}{2\pi r \epsilon_0})^2\\\\U = \frac{\lambda^2}{8\pi^2r^2\epsilon_0}$

Now, we can integrate over the volume with respect to the radius.

Recall:
$V = \pi r^2L \\\\dV = 2\pi rLdr$

Now, we can take the integral of the above expression. Let:
$r_i$ = inner cylinder radius

$r_o$ = outer cylindrical shell inner radius

Total energy-field energy:

$U = \int\limits^{r_o}_{r_i} {U_D} \, dV = \int\limits^{r_o}_{r_i} {2\pi rL *U_D} \, dr$

Plug in the equation for the electric field energy density and solve.

$U = \int\limits^{r_o}_{r_i} {2\pi rL *\frac{\lambda^2}{8\pi^2r^2\epsilon_0}} \, dr\\\\U = \int\limits^{r_o}_{r_i} { L *\frac{\lambda^2}{4\pi r\epsilon_0}} \, dr\\$

Bring constants in front and integrate. Recall the following integration rule:
$\int {\frac{1}{x}} \, dx = ln(x) + C$

Now, we can solve!

$U = \frac{\lambda^2 L}{4\pi \epsilon_0}\int\limits^{r_o}_{r_i} { \frac{1}{r}} \, dr\\\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(r)\left \| {{r_o} \atop {r_i}} \right. \\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} (ln(r_o) - ln(r_i))\\\\U = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})$

To find the total electric field energy per unit length, we can simply divide by the length, 'L'.

$\frac{U}{L} = \frac{\lambda^2 L}{4\pi \epsilon_0} ln(\frac{r_o}{r_i})\frac{1}{L} \\\\\frac{U}{L} = \boxed{\frac{\lambda^2 }{4\pi \epsilon_0} ln(\frac{r_o}{r_i})}$

And here's our equation!

3 0

2 years ago