When was hawaii volcanoes national park established

The work-energy theorem states that the work done on an object is equal to a change in which quantity?

Fynjy0 [20]

The work-energy theorem states that the net work done by the forces on an object equals the change in its kinetic energy.

8 0

2 years ago

A "spherical capacitor" is constructed of two thin, concentric spherical shells of conducting material. Let a be the radius of t

Shalnov [3]

Answer:

$C=\frac{ab}{k(b-a)}$

Explanation:

We can assume this problem as two concentric spherical metals with opposite charges.

We have also to take into account the formulas for the electric field and the capacitance. Hence we have

$C=\frac{Q}{V}\\\\E=k\frac{Q}{r^2}\\$

Where k is the Coulomb's constant. Furthermore, by taking into account the expression for the potential and by integrating

$dV=Edr\\\\V=\int_{R_1}^{R_2}Edr=-\int_{R_1}^{R_2}\frac{kQ}{r^2}dr\\\\V=kQ[\frac{1}{R_2}-\frac{1}{R_1}]$

Hence, the capacitance is

$C=\frac{1}{k[\frac{1}{R_2}-\frac{1}{R_1}]}$

but R1=a and R2=b

$C=\frac{ab}{k(b-a)}$

HOPE THIS HELPS!!

3 0

3 years ago

Read 2 more answers

Ecologically speaking, which is bigger, a population or a community? Explain.

exis [7]

Here are a few of the fundamental words in ecology, which are simple, but may be easy to mix up because they are so similar. It is, however, quite important to be clear of what they mean. I will here try to explain how they differ by defining them and giving a few examples to illustrate how they could be applied.

A habitat is basically the site where an organism or a group lives. It may be anything from a stone in a lake, on which algae grows, to a forest containing all sorts of creatures. Note that groups within a habitat do not need to be of the same species. However, one usually speaks of habitats of individuals, species, or larger groups. For instance, the habitat of the algae would be the stone in the lake, and the forest could be the habitat of a single bear – regardless of what other organisms live there and how they are geographically distributed; here we are interested in the bear, so we define the habitat as its home range, and all that falls within it will arbitrarily be a apart of its habitat. hope this helps

3 0

2 years ago

Read 2 more answers

Could I get help on this question . I’m scared I’m wrong

goldfiish [28.3K]

Answer:

It is correct!

Explanation:

5 0

2 years ago

Read 2 more answers

A charged paint is spread in a very thin uniform layer over the surface of a plastic sphere of diameter 18.0 cm , giving it a ch

vladimir1956 [14]

A) The electric field inside the paint layer is zero

B) The electric field just outside the paint layer is $3.2\cdot 10^7 N/C$ (radially inward)

C) The electric field at 6.00 cm from the surface is $1.2\cdot 10^7 N/C$ (radially inward)

Explanation:

A)

We can solve the problem by applying Gauss Law, which states that the electric flux through a Gaussian surface must be equal to the charge contained in the surface divided by the vacuum permittivity:

$\int EdS = \frac{q}{\epsilon_0}$

where

E is the magnitude of the electric field

dS is the element of the surface

q is the charge contained within the surface

$\epsilon_0$ is the vacuum permittivity

By taking a sphere centered in the origin,

$\int E dS = E \cdot 4\pi r^2$

where $4\pi r^2$ is the surface of the Gaussian sphere of radius r.

In this problem, we want to find the electric field just inside the paint layer, so we take a value of r smaller than

$R=9.0 cm = 0.09 m$ (radius of the plastic sphere is half of the diameter)

Since the charge is all distributed over the plastic sphere, the charge contained within the Gaussian sphere is zero:

$q=0$

And therefore,

$E4\pi r^2 = 0\\\rightarrow E = 0$

So, the electric field inside the plastic sphere is zero.

B)

Here we apply again Gauss Law:

$E\cdot 4 \pi r^2 = \frac{q}{\epsilon_0}$

In this case, we want to calculate the electric field just outside the paint layer: this means that we take r as the radius of the plastic sphere, so

$r=R=0.18 m$

The charge contained within the Gaussian sphere is therefore

$q=-29.0 \mu C = -29.0\cdot 10^{-6}C$

Therefore, the electric field is

$E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-29.0\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.09)^2}=-3.2\cdot 10^7 N/C$

And the negative sign indicates that the direction of the field is radially inward (because the charge that generates the field is negative). However, the text of the question says "Enter positive value if the field is directed radially inward and negative value if the field is directed radially outward", so the answer to this part is

$E=3.2\cdot 10^7 N/C$

C)

For this part again, we apply Gauss Law:

$E\cdot 4 \pi r^2 = \frac{q}{\epsilon_0}$

In this case, we want to calculate the field at a point 6.00 cm outside the surface of the paint layer; this means that the radius of the Gaussian sphere must be

r = 9 cm + 6 cm = 15 cm = 0.15 m

While the charge contained within the sphere is again

$q=-29.0 \mu C = -29.0\cdot 10^{-6}C$

Therefore, the electric field in this case is

$E=\frac{q}{4\pi \epsilon_0 R^2}=\frac{-29.0\cdot 10^{-6}}{4\pi (8.85\cdot 10^{-12})(0.15)^2}=-1.2\cdot 10^7 N/C$