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3 years ago

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When a high-energy proton or pion traveling near the speed of light collides with a nucleus, it may travel 2.5 x 10-15 m before

interacting. From this information, find the time interval required for the strong interaction to occur.

1 answer:

Tasya [4]3 years ago

8 0

Explanation:

It is given that,

When a high-energy proton or pion traveling near the speed of light collides with a nucleus, $d=2.5\times 10^{-15}\ m$

Speed of light, $c=3\times 10^{8}\ m\s$

Let t is the time interval required for the strong interaction to occur. The speed is given by :

$c=\dfrac{d}{t}$

$t=\dfrac{d}{c}$

$t=\dfrac{2.5\times 10^{-15}\ m}{3\times 10^{8}\ m/s}$

$t=8.33\times 10^{-24}\ s$

So, the time interval required for the strong interaction to occur is $8.33\times 10^{-24}\ s$ . Hence, this is the required solution.

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Which conclusion can be made based on the information in the table?

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Answer:

The correct option is (b).

Explanation:

The relation between the wavelength and frequency is given by :

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Where

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Diagnostic ultrasound of frequency 4.50 MHz is used to examine tumors in soft tissue. (a) What is the wavelength in air of such

mafiozo [28]

(a) $7.62 \times 10^{-5} m$ is the wavelength in air of such a sound wave.

(b) $3.33 \times 10^{-4}\ m$ is the wavelength of this wave in tissue.

<u>Explanation:</u>

Frequency and wavelength can be related by the equation,

Velocity = Wavelength x Frequency

$v=\lambda \times f$

where,

v - velocity of light for all EM (electromagnetic) waves in vacuum

Given:

f - 4.50 MHz = $4.50 \times 10^{6} \mathrm{Hz}$

a) To find the wavelength in air

We know,

Speed of sound in air = 343 m/s

Apply given frequency and speed of sound in air, we get

$\lambda=\frac{v}{f}=\frac{343}{4.5 \times 10^{6}}=76.2 \times 10^{-6}=7.62 \times 10^{-5}\ \mathrm{m}$

b) If the speed of sound in tissue is 1500 m/s, find the wavelength of this wave in tissue

Speed of sound in tissue, v = 1500 m/s

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A charge of -8.00 nC is spread uniformly over the surface of one face of a nonconducting disk of radius 1.05 cm.

gavmur [86]

Answer:

(a) $E = -1.02 \times 10^5~N/C$

(b) $E = -9.7 \times 10^4~N/C$

Explanation:

(a) The electric field for a point charge is given by the following formula:

$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\^r$

Since this formula is valid for point charges, we have to choose an infinitesimal area, da, from the disk. Then we will calculate the E-field (dE) created by this small area using the above formula, then we will integrate over the entire disk to find the E-field created by the disk.

$dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{(\sqrt{z^2 + r^2})^2}$

Here, z = 0.025 m. And r is the distance of the infinitesimal area from the axis. dQ is the charge of the small area, and should be written in terms of the given variables.

In cylindrical coordinates, da = r dr dθ. So,

$\frac{Q}{\pi R^2} = \frac{dQ}{da}\\\frac{Q}{\pi R^2} = \frac{dQ}{rdrd\theta}\\dQ = \frac{Qrdrd\theta}{\pi R^2}$

Hence, dE is now:

$dE = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi R^2}\frac{rdrd\theta}{z^2 + r^2}$

The surface integral over the disk can now be taken, but there is one more thing to be considered. This dE is a vector quantity, and it needs to be separated its components.

It has two components, one in the vertical direction and another in the horizontal direction. By symmetry, the horizontal components cancel out each other in the end (since it is a disk, each horizontal vector has an equal but opposite counterpart), so only the vertical component should be considered.

Let us denote the angle between dE and the horizontal axis as α. This angle can be found by the geometry of the triangle formed by dE, vertical axis of the disk, and horizontal plane. So,

$\sin(\alpha) = \frac{z}{\sqrt{z^2 + r^2}}$

Therefore, vertical component of dE now becomes

$dE_z = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi R^2}\frac{rdrd\theta}{z^2 + r^2}\frac{z}{\sqrt{z^2+r^2}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\frac{rdrd\theta}{(z^2+r^2)^{3/2}}\\E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\int\limits^{2\pi}_0 \int\limits^R_0 {\frac{rdrd\theta}{(z^2+r^2)^{3/2}}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2} 2\pi(\frac{1}{z} - \frac{1}{\sqrt{z^2+R^2}})$

Substituting the parameters, z = 0.025 m, Q = - 8 x 10^(-9) C, and R = 0.0105 m, yields the final result:

$E_z = \frac{1}{2\epsilon_0}\frac{Qz}{\pi R^2}(\frac{1}{z} - \frac{1}{\sqrt{z^2+R^2}}) = -1.02 \times 10^5~N/C$

(b) We will have a similar approach, but a simpler integral.

$dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{z^2 + R^2}\\\frac{Q}{2\pi R} = \frac{dQ}{Rd\theta}\\dQ = \frac{Qd\theta}{2\pi}\\dE = \frac{1}{4\pi\epsilon_0}\frac{Qd\theta}{2\pi(z^2 + R^2)}\\dE_z = \frac{1}{4\pi\epsilon_0}\frac{Qd\theta}{2\pi(z^2 + R^2)}\frac{z}{\sqrt{z^2+R^2}} = \frac{1}{4\pi\epsilon_0}\frac{Qzd\theta}{2\pi(z^2 + R^2)^{3/2}}\\E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{2\pi(z^2 + R^2)^{3/2}}\int\limits^{2\pi}_0 {} \, d\theta = \frac{1}{4\pi\epsilon_0}\frac{Qz}{2\pi(z^2 + R^2)^{3/2}}2\pi$

$E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{(z^2 + R^2)^{3/2}} = -9.07\times 10^4~N/C$

Note that, in this case the source object is a one dimensional hoop rather than a two dimensional disk.

3 0

3 years ago