Classify the following matter

There are 2 concentric cylinders. These cylinders are very long with length L. The inner cylinder has a radius R1 and is a solid

OLga [1]

To get the charge along the inner cylinder, we use Gauss Law
E = d R1/2εo
For the outer cylinder the charge can be calculated using
E = d R2^2/2εoR1
where d is the charge density
Use these two equations to get the charge in between the cylinders and the capacitance between them.

5 0

3 years ago

A cat dozes on a stationary merry-go-round, at a radius of 4.4 m from the center of the ride. The operator turns on the ride and

monitta

Answer:

The coefficient of static friction is 0.29

Explanation:

Given that,

Radius of the merry-go-round, r = 4.4 m

The operator turns on the ride and brings it up to its proper turning rate of one complete rotation every 7.7 s.

We need to find the least coefficient of static friction between the cat and the merry-go-round that will allow the cat to stay in place, without sliding. For this the centripetal force is balanced by the frictional force.

$\mu mg=\dfrac{mv^2}{r}$

v is the speed of cat, $v=\dfrac{2\pi r}{t}$

$\mu=\dfrac{4\pi^2r}{gt^2}\\\\\mu=\dfrac{4\pi^2\times 4.4}{9.8\times (7.7)^2}\\\\\mu=0.29$

So, the least coefficient of static friction between the cat and the merry-go-round is 0.29.

4 0

3 years ago

1. A baseball is thrown vertically at 16.7 m/s. What is the maimum height of the baseball?

Anvisha [2.4K]

Answer:

14.2 m

Explanation:

Using conservation of energy:

PE at top = KE at bottom

mgh = ½ mv²

h = v² / (2g)

h = (16.7 m/s)² / (2 × 9.8 m/s²)

h = 14.2 m

Using kinematics:

Given:

v₀ = 16.7 m/s

v = 0 m/s

a = -9.8 m/s²

Find: Δy

v² = v₀² + 2aΔy

(0 m/s)² = (16.7 m/s)² + 2 (-9.8 m/s²) Δy

Δy = 14.2 m

7 0

3 years ago

Calculate curls of the following vector functions (a) AG) 4x3 - 2x2-yy + xz2 2

aleksandr82 [10.1K]

Answer:

The curl is $0 \hat x -z^2 \hat y -4xy \hat z$

Explanation:

Given the vector function

$\vec A (\vec r) =4x^3 \hat{x}-2x^2y \hat y+xz^2 \hat z$

We can calculate the curl using the definition

$\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\A_x&X_y&A_z\end{array}\right|$

Thus for the exercise we will have

$\nabla \times \vec A (\vec r ) = \left|\begin{array}{ccc}\hat x&\hat y&\hat z\\\partial/\partial x&\partial/\partial y&\partial/\partial z\\4x^3&-2x^2y&xz^2\end{array}\right|$

So we will get

$\nabla \times \vec A (\vec r )= \left( \cfrac{\partial}{\partial y}(xz^2)-\cfrac{\partial}{\partial z}(-2x^2y)\right) \hat x - \left(\cfrac{\partial}{\partial x}(xz^2)-\cfrac{\partial}{\partial z}(4x^3) \right) \hat y + \left(\cfrac{\partial}{\partial x}(-2x^2y)-\cfrac{\partial}{\partial y}(4x^3) \right) \hat z$

Working with the partial derivatives we get the curl

$\nabla \times \vec A (\vec r )=0 \hat x -z^2 \hat y -4xy \hat z$

6 0

3 years ago

Two wires each carry 10.0 A of current (in opposite directions) and are 2.50 mm apart. What is the magnetic field 37.0 cm away a

lyudmila [28]

Answer:

see answer below

Explanation:

Before we do any kind of calculation, we need to convert the proper units of the exercise. All the units of distance must be in meters, so, let's change distance of the wire, and the magnetic field to meters:

Separation between the wires are 2.5 mm:

2.5 mm * (1 m / 1000 mm) = 0.0025 m

The distance of P from the bottom of the wires is 37 cm:

37 cm * (1 m/100 cm) = 0.37 m

The distance of P from the top of the wires is just the sum of the two distances:

R = 0.37 + 0.0025 = 0.3725 m

Now that we have the distance, we can determine the magnetic field, using the following expression:

B = B(bottom) - B(top) or just B₂ - B₁

And B = μ₀ I / 2πR

Replacing in the above expression we have:

B = μ₀ I / 2π ( 1/R₂ - 1/R₁)

Now we can determine the magnetic field:

B = (4πx10⁻⁷ * 10 / 2π) (1/0.37 - 1/0.3725)

Which means that the magnetic field is out of the page.

Hope this helps

4 0

3 years ago