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4 years ago

I NEED HELP PLEASE, THANKS! :)

Physics

1 answer:

Neko [114]4 years ago

4 0

Answer:

Take a look at the attachment below.

Explanation:

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It is interesting to speculate on the properties of a universe with different values for the fundamental constants.

qwelly [4]

Answer:

Part a)

$\lambda = 0.345 m$

Part b)

$\Delta x = 0.274 m$

Part c)

$r = 2.8 \times 10^{11} m$

Explanation:

Part a)

De broglie wavelength is given as

$\lambda = \frac{h}{mv}$

$\lambda = \frac{1}{(0.145)(20)}$

$\lambda = 0.345 m$

Part b)

By principle of uncertainty we know that

$\Delta x \times \Delta P = \frac{h}{4\pi}$

$\Delta x \times (0.145)(21 - 19) = \frac{1}{4\pi}$

$\Delta x = 0.274 m$

Part c)

As we know that

$\frac{kq_1q_2}{r^2} = \frac{mv^2}{r}$

also we know

$mvr = \frac{nh}{2\pi}$

$v = \frac{h}{2\pi mr}$

now we have

$\frac{ke^2}{r} = \frac{mh^2}{4\pi^2m^2 r^2}$

$r = \frac{h^2}{4\pi^2mke^2}$

$r = 2.8 \times 10^{11} m$

5 0

3 years ago

An infinite long straight wire is uniformly charged, the charge density is a. Use Coulomb's law to calculate the electric field

bixtya [17]

Answer:

$\vec{E} = \frac{a}{2\pi \epsilon_0 R}\^R$

Explanation:

Since the wire is infinitely long, we will use Gauss' Law:

$\int\vec{E}d\vec{a} = \frac{Q_{enc}}{\epsilon_0}$

We will draw an imaginary cylindrical surface with height h around the wire. The electric flux through the imaginary surface will be equal to the net charge inside the surface.

In that case, the net charge inside the imaginary surface will be the portion of wire with height h. Then the charge of that portion will be equal to

$Q_{enc} = ah$

The left-hand side of the Gauss' Law is the flux through the imaginary surface. Since we choose our surface as a cylinder, of which we know the area, we do not have to take the surface integral.

$\int\vec{E}d\vec{a} = E2\pi R h$

where R is the radius of the imaginary cylinder.

Finally, Gauss' Law gives

$E2\pi Rh = \frac{ah}{\epsilon_0}\\E = \frac{a}{2\pi \epsilon_0 R}$

The vector expression is

$\vec{E} = \frac{a}{2\pi \epsilon_0 R}\^R$

As you can see, the electric field is independent from the height h, since that is merely an imaginary cylinder to apply Gauss' Law. In the end, what matters is the charge density of the wire and the distance from the wire.

4 0

4 years ago

Consider two thin disks, of negligible thickness, of radiusR oriented perpendicular to thex axis such that the x axis runs throu

Veseljchak [2.6K]

Complete question

The complete question is shown on the first and second uploaded image

Answer:∈

Answer to first question is shown on the second uploaded image.

Part B the Answer is:

The ratio $\frac{R}{a}$ is evaluated to be 49.99

Explanation:

The explanation is shown on the third ,fourth and fifth image.

7 0

4 years ago

Oceanic water particles mainly move in circles; is this movement greater on the ocean's surface or below the surface? Explain yo

goblinko [34]

I think that the oceanic water particles mainly move in circles greater in the oceans surface because of how big the waves can be and how wind and air impact the motion. The water particles move more on the surface because of the other factors that impact it such as people, wind, air, etc...

5 0

3 years ago

*Brainliest if answered with in 5 min*

levacccp [35]

Your answer is going to be 0 degrees Kelvin!

8 0

3 years ago

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