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3 years ago

The waste product of photosynthesis is

Physics

1 answer:

Vsevolod [243]3 years ago

5 0

Answer:

oxygen

Plants need to excrete excess carbon dioxide and oxygen. Carbon dioxide is a waste product of aerobic respiration in plant cells. Oxygen is a waste product of photosynthesis .

Explanation:

yw!

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An electric field is described by the vector e=yj -xi. what is the electric flux through a rectangular box in the x-z plane that

Varvara68 [4.7K]

The flux of $\vec E = -x\,\vec\imath + y\,\vec\jmath$ is given by the surface integral

$\displaystyle \iint_S \vec E \cdot d\vec\sigma$

where $S$ is the given square region, which we can parameterize by

$\vec s(x, z) = x\,\vec\imath + z\,\vec k$

with $0\le x\le 1$ and $0\le z\le 1$ . The area element is

$d\vec\sigma = \vec n \, dx\,dz$

where $\vec n$ is the normal vector to $S$ . Depending on the orientation of $S$ , this vector could be

$\vec n = \dfrac{\partial\vec s}{\partial x} \times \dfrac{\partial\vec s}{\partial z} = -\vec\jmath$

or $-\vec n = \vec \jmath$ ; either way, the integral reduces to

$\displaystyle \iint_S \vec E \cdot d\,\vec\sigma = \int_0^1 \int_0^1 (-x\,\vec\imath + z\,\vec k) \cdot (\pm\vec\jmath) \, dx\,dz = \boxed{0}$

3 0

2 years ago

Plzzz help its for my civics class what answer is it????????????

ahrayia [7]

Answer:

the correct answer is The English Bill of Rights

Explanation:

This is the answer because i have taken the test.

7 0

3 years ago

Read 2 more answers

A thin spherical spherical shell of radius R which carried a uniform surface charge density σ. Write an expression for the volum

ozzi

Answer:

Explanation:

From the given information:

We know that the thin spherical shell is on a uniform surface which implies that both the inside and outside the charge of the sphere are equal, Then

The volume charge distribution relates to the radial direction at r = R

∴

$\rho (r) \ \alpha \ \delta (r -R)$

$\rho (r) = k \ \delta (r -R) \ \ at \ \ (r = R)$

$\rho (r) = 0\ \ since \ r< R \ \ or \ \ r>R---- (1)$

To find the constant k, we examine the total charge Q which is:

$Q = \int \rho (r) \ dV = \int \sigma \times dA$

$Q = \int \rho (r) \ dV = \sigma \times4 \pi R^2$

∴

$\int ^{2 \pi}_{0} \int ^{\pi}_{0} \int ^{R}_{0} \rho (r) r^2sin \theta \ dr \ d\theta \ d\phi = \sigma \times 4 \pi R^2$

$\int^{2 \pi}_{0} d \phi* \int ^{\pi}_{0} \ sin \theta d \theta * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

$(2 \pi)(2) * \int ^{R}_{0} k \delta (r -R) * r^2dr = \sigma \times 4 \pi R^2$

Thus;

$k * 4 \pi \int ^{R}_{0} \delta (r -R) * r^2dr = \sigma \times R^2$

$k * \int ^{R}_{0} \delta (r -R) r^2dr = \sigma \times R^2$

$k * R^2= \sigma \times R^2$

$k = R^2 --- (2)$

Hence, from equation (1), if k = $\sigma$

$\mathbf{\rho (r) = \delta* \delta (r -R) \ \ at \ \ (r=R)}$

$\mathbf{\rho (r) =0 \ \ at \ \ rR}$

To verify the units:

$\mathbf{\rho (r) =\sigma \ * \ \delta (r-R)}$

↓ ↓ ↓

c/m³ c/m³ × 1/m

Thus, the units are verified.

The integrated charge Q

$Q = \int \rho (r) \ dV \\ \\ Q = \int ^{2 \ \pi}_{0} \int ^{\pi}_{0} \int ^R_0 \rho (r) \ \ r^2 \ \ sin \theta \ dr \ d\theta \ d \phi \\ \\ Q = \int ^{2 \pi}_{0} \ d \phi \int ^{\pi}_{0} \ sin \theta \int ^R_{0} \rho (r) r^2 \ dr$