Ask question

Ask question

All categories

3 years ago

14

From the poem Black woman what does it mean by Savannah stretching to clear horizons, Savannah shuddering beneath the East wind'

s eager caresses.

1 answer:

My name is Ann [436]3 years ago

8 0

Answer:

From the poem Black woman what does it mean by Savannah stretching to clear horizons, Savannah shuddering beneath the East wind's eager caresses.

Explanation:

You might be interested in

Two capacitors with capacitances of 1.0 m F and 0.50 m F, respectively, are connected in series. The system is connected to a 10

forsale [732]

Answer:

Therefore energy is stored in the 1.0 mF capacitor is 5.56×10⁻⁹ J

Explanation:

Series capacitor: The ending point of a capacitor is the starting point of other capacitor.

If C₁ and C₂ are connected in series then the equivalent capacitance is C.

where $\frac{1}{C} =\frac{1}{C_1}+\frac{1}{C_2}$

Given that,

C₁ = 1.0 mF=1.0×10⁻³F and C₂ = 0.50mF=0.50×10⁻³F

If C is equivalent capacitance.

Then $\frac{1}{C} =\frac{1}{1.0}+\frac{1}{0.5}$

$\Rightarrow \frac{1}{C} =\frac{3}{1}$

$\Rightarrow C=\frac{1}{3}$ mF

Again given that the system is connected to a 100-v battery.

We know that

q=Cv

q= charge

C= capacitor

v= potential difference

Therefore

$q=(\frac{1}{3} \times 10^{-3}\times10) C$

$=\frac{10^{-2}}{3} C$

The electrical potential energy stored in a capacitor can be expressed

$U=\frac{q^2C}{2}$

q= charge

c=capacitance of a capacitor

Therefore energy is stored in the 1.0 mF capacitor is

$U=\frac{q^2C_1}{2}$

$\Rightarrow U=\frac{(\frac{10^{-2}}{3})^2\times 10^{-3} }{2}$

$\Rightarrow U= 5.56\times 10^{-9}$ J

8 0

3 years ago

ty sm i got a better score than the first time i had a 25.6 =v= but i got a 85.4 bc your guys helpped ty

mezya [45]

Answer:

Im glad you got a good score, love!

Explanation:

3 0

3 years ago

Read 2 more answers

Equations 22-8 and 22-9 are approximations of the magnitude of the electric field of an electric dipole, at points along the dip

OleMash [197]

On axis of the dipole the electric field is given by

$E = \frac{kq}{(r - \frac{d}{2})^2} - \frac{kq}{(r + \frac{d}{2})^2}$

$E = \frac{kq(r + d/2)^2 - (r - d/2)^2}{(r^2 - \frac{d^2}{4})^2}$

$E_{act} = \frac{kq*2rd}{(r^2 - \frac{d^2}{4})^2}$

now if we approximate above equation

$E_{eppr} = \frac{kq*2d}{r^3}$

now we will find the ratio of these two

$\frac{E_{eppr}}{E_{act}} = \frac{(r^2 - \frac{d^2}{4})^2}{r^4}$

put r = 3d

$\frac{E_{eppr}}{E_{act}} = \frac{(9d^2 - \frac{d^2}{4})^2}{81d^4}$

$\frac{E_{eppr}}{E_{act}} = \frac{(9- \frac{1}{4})^2}{81}$

$\frac{E_{eppr}}{E_{act}} = 0.945$

<em>so above is the ratio of two field</em>

6 0

4 years ago

Read 2 more answers

Consider two identical containers. Container A is filled with water to the top. Container B has a block of wood floating in it,

Elodia [21]

Answer:

<em>Container A will weigh more</em>

<em></em>

Explanation:

Both containers are identical, so we assume that they weigh the same.

They both have the same volume, and will contain an equal volume of a material.

Since they both contain water to the top, this means that their volume is fully occupied. But container B contain a block of wood <em><u>floating</u></em> in it.

The fact that the block of wood floats in the water in container B shows that it is less dense than the water around it, and in the container A, this same space is completely filled with water.

What we derive from this is that the portion of space contained by the block of wood in container B is occupied by water in container A, but, in container B, the density of this space is lesser now, since the wood block floats.

<em>Since density is mass per unit volume, and weight is proportional to mass, then we can see that the weight of this volume portion in container B is lesser than that of container A. The consequence is that container A will weigh more than container B because of this extra weight.</em>

4 0

3 years ago

Think of something from everyday life that follows a two-dimensional path. It could be a kicked football, a bus that's turning a

OLEGan [10]

Answer:

Let us consider the case of a bus turning around a corner with a constant velocity, as the bus approaches the corner, the velocity at say point A is Va, and is tangential to the curve with direction pointing away from the curve. Also, the velocity at another point say point B is Vb and is also tangential to the curve with direction pointing away from the curve.<em> </em><em>Although the velocity at point A and the velocity at point B have the same magnitude, their directions are different (velocity is a vector quantity), and hence we have a change in velocity. By definition, an acceleration occurs when we have a change in velocity, so the bus experiences an acceleration at the corner whose direction is away from the center of the corner</em>.

The acceleration is not aligned with the direction of travel because<em> the change in velocity is at a tangent (directed away) to the direction of travel of the bus.</em>

4 0

3 years ago