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3 years ago

Which of the following is an example of a way that an individual can help the environment?

Physics

2 answers:

7nadin3 [17]3 years ago

8 0

Answer:

it is A

Explanation:

just took the quiz

wolverine [178]3 years ago

5 0

We certainly don’t want to increase waste, so B and C are incorrect. We wouldn’t be helping the environment by sending more waste into landfills, so D is incorrect as well. A is the correct answer, because by increasing the recycling of waste, we can help the environment.

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What were Niels Bohr's experiments?

Travka [436]

Niel depicted that the atom was a small, positively-charged nucleus surrounded by negatively-charged electrons that travel in circular orbits around the nucleus

5 0

3 years ago

A thin, 100 g disk with a diameter of 8.0 cm rotates about an axis through its center with 0.15 J of kinetic energy. What is the

butalik [34]

Answer:

v=2.45 m/s

Explanation:

The rotational kinetic energy is calculated with:

$K=\frac{I\omega^2}{2}$

The moment of inertia of a disk is:

$I=\frac{mr^2}{2}$

A point on the rim of a disk will have a speed of:

$v=r\omega$

Putting all together:

$v=r\sqrt{\frac{2K}{I}}=r\sqrt{\frac{2(2K)}{mr^2}}=2\sqrt{\frac{K}{m}}$

Which for our values is:

$v=2\sqrt{\frac{0.15J}{0.1Kg}}=2.45m/s$

3 0

4 years ago

Read 2 more answers

If you observe an impaired driver, a safe behavior for you would be ____________.

timama [110]

Get someone else to drive me

6 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'.

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