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2 years ago

The gravitational potential energy of an object is due to

Physics

1 answer:

ExtremeBDS [4]2 years ago

8 0

d. all of these

look at the picture to know from where we get this law with application on it

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Hi, Solve for λ<br> E=hc/λ

Paul [167]

Answer:

λ=hc/E

Explanation:

E=hc/λ

Eλ=hc

λ=hc/E

4 0

3 years ago

A uniformly charged, one-dimensional rod of length L has total positive charge Q. Itsleft end is located at x = ????L and its ri

GREYUIT [131]

Answer:

$|\vec{F}| = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))$

Explanation:

The force on the point charge q exerted by the rod can be found by Coulomb's Law.

$\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}\^r$

Unfortunately, Coulomb's Law is valid for points charges only, and the rod is not a point charge.

In this case, we have to choose an infinitesimal portion on the rod, which is basically a point, and calculate the force exerted by this point, then integrate this small force (dF) over the entire rod.

We will choose an infinitesimal portion from a distance 'x' from the origin, and the length of this portion will be denoted as 'dx'. The charge of this small portion will be 'dq'.

Applying Coulomb's Law:

$d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qdq}{x + x_0}(\^x)$

The direction of the force on 'q' is to the right, since both charges are positive, and they repel each other.

Now, we have to write 'dq' in term of the known quantities.

$\frac{Q}{L} = \frac{dq}{dx}\\dq = \frac{Qdx}{L}$

Now, substitute this into 'dF':

$d\vec{F} = \frac{1}{4\pi\epsilon_0}\frac{qQdx}{L(x+x_0)}(\^x)$

Now we can integrate dF over the rod.

$\vec{F} = \int{d\vec{F}} = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}\int\limits^{L}_0 {\frac{1}{x+x_0}} \, dx = \frac{1}{4\pi\epsilon_0}\frac{qQ}{L}(\ln(L+x_0)-\ln(x_0))(\^x)$

4 0

2 years ago

Type of force that holds the nucleus of an atom together

Kamila [148]

Answer:

Nuclear Forces

Explanation:

The type of force that holds the nucleus of an atom together is called Nuclear Forces.

7 0

3 years ago

Except for the nodes on a standing wave, what is the frequency f of the points executing simple harmonic motion?

Katen [24]

Take into account that in a standing wave, the frequency f of the points executing simple harmonic motion, is simply a multiple of the fundamental harmonic fo, that is:

f = n·fo

where n is an integer and fo is the first harmonic or fundamental.

fo is given by the length L of a string, in the following way:

fo = v/λ = v/(L/2) = 2v/L

becasue in the fundamental harmonic, the length of th string coincides with one hal of the wavelength of the wave.

6 0

1 year ago

Imagine that asteroid A that has an escape velocity of 50 m/s. If asteroid B has twice the mass and twice the radius, it would h

padilas [110]

Answer:

The same as the escape velocity of asteorid A (50m/s)

Explanation:

The escape velocity is described as follows:

$v=\sqrt{\frac{2GM}{R}}$

where $G$ is the universal gravitational constant, $M$ is the mass of the asteroid and $R$ is the radius

and since the scape velocity is 50m/s:

$50m/s=\sqrt{\frac{2GM}{R}}$

Now, if the astroid B has twice mass and twice the radius, we have that tha mass is: $2M$

and the radius is: $2R$

inserting these values into the formula for escape velocity:

$v=\sqrt{\frac{2G(2M)}{2R} } =\sqrt{\frac{4GM}{2R} } =\sqrt{\frac{2GM}{R} }$

and we have found that $50m/s=\sqrt{\frac{2GM}{R}}$ , so the two asteroids have the same escape velocity.

We found that the expression for escape velocity remains the same as for asteroid A, this because both quantities (radius and mass) doubled, so it does not affect the equation.

The answer is

Asteroid B would have an escape velocity the same as the escape velocity of asteroid A

7 0

3 years ago