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3 years ago

What type of energy requires the motion of an object?

Physics

1 answer:

hichkok12 [17]3 years ago

8 0

Answer:

Mechanical Energy.

Explanation:

This can occur as either kinetic or potential energy.

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9966 [12]

Answer: de?

Explanation:

7 0

3 years ago

Two coaxial conducting cylindrical shells have equal and opposite charges. The inner shell has charge +q and an outer radius a,

Leviafan [203]

Answer:

$\Delta V = \frac{q ln(\frac{b}{a})}{2\pi \epsilon_0 L}$

Explanation:

As we know that the charge per unit length of the long cylinder is given as

$\lambda = \frac{q}{L}$

here we know that the electric field between two cylinders is given by

$E = \frac{2k\lambda}{r}$

now we know that electric potential and electric field is related to each other as

$\Delta V = - \int E.dr$

$\Delta V = -\int_a^b (\frac{2k\lambda}{r})dr$

$\Delta V = -2k \lambda ln(\frac{b}{a})$

$\Delta V = \frac{\lambda ln(\frac{b}{a})}{2\pi \epsilon_0}$

$\Delta V = \frac{q ln(\frac{b}{a})}{2\pi \epsilon_0 L}$

7 0

3 years ago

Two paths lead to the top of a big hill. One is steep and direct, while the other is twice as long but less steep. How much more

umka2103 [35]

Answer:

None.

Explanation:

Gravitational potential energy, depends only of the mass on which the gravity is doing work, and the displacement produced by this force.
As displacement depends only on the final and initial positions (in this case, the height of the hill), if we choose as our zero reference level the bottom of the hill, the change in gravitational potential energy will be as follows:

$\Delta U = U_{f} -U_{0} = m*g*h - 0 = m*g*h$

As we can see, the only value of distance involved is the height of the hill , so it is independent of the distance travelled.

6 0

4 years ago

In refraction, when a wave travels from one medium to another, it

kirill115 [55]

In refraction, when a wave travels from one medium to another it changes speed.

Correct option is A.

3 0

3 years ago

Read 2 more answers

Normalize the equations

tatyana61 [14]

Answer:

Solution is in explanation

Explanation:

part a)

For normalization we have

$\int_{0}^{\infty }f(x)dx=1\\\\\therefore \int_{0}^{\infty }ae^{-kx}dx=1\\\\\Rightarrow a\int_{0}^{\infty }e^{-kx}dx=1\\\\\frac{a}{-k}[\frac{1}{e^{kx}}]_{0}^{\infty }=1\\\\\frac{a}{-k}[0-1]=1\\\\\therefore a=k$

Part b)

$\int_{0}^{L }f(x)dx=1\\\\\therefore Re(\int_{0}^{L }ae^{-ikx}dx)=1\\\\\Rightarrow Re(a\int_{0}^{L }e^{-ikx}dx)=1\\\\\therefore Re(\frac{a}{-ik}[\frac{1}{e^{ikx}}]_{0}^{L})=1\\\\\Rightarrow Re(\frac{a}{-ik}(e^{-ikL}-1))=1\\\\\frac{a}{k}Re(\frac{1}{-i}(cos(-kL)+isin(-kL)-1))=1$

$\frac{a}{k}Re(\frac{1}{-i}(cos(-kL)+isin(-kL)-1))=1\\\\\frac{a}{k}Re(icos(-kL)+sin(kL)+\frac{1}{i})=1\\\\\frac{a}{k}sin(kL)=1\\\\a=\frac{k}{sin(kL)}$

7 0

4 years ago