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Anna71 [15]
3 years ago
5

A helium–neon laser emits a beam of circular cross section with a radius r and a power P.?. (a) Find the maximum electric field

in the beam. (Use any variable or symbol stated above along with the following as necessary: μ0 and c.. (b) What total energy is contained in a length of the beam? (Use the following as necessary: P,l , and c.)(c) Find the momentum carried by a length of the beam. (Use the following as necessary: P, l, and c.)
Physics
1 answer:
olchik [2.2K]3 years ago
5 0

there is a relation between intensity of light beam and the magnitude of electric field.<span>I=(1/2)c<span>ϵo</span>n<span>E2</span>=P/π<span>r2</span></span> <span><span>E2</span>=2P/c<span>ϵo</span>nπ<span>r2</span></span> E= magnitude of electric field n= refractive index of medium <span><span>μo</span><span>ϵ0</span>=1/<span>c2
</span></span>energy= power*time = P*(1m/speed of light)<span><span>energy=(P∗1m)/c</span></span>
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Explanation: I apologize if I am incorrect.

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In a fast-pitch softball game the pitcher is impressive to watch, as she delivers a pitch by rapidly whirling her arm around so
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Answer:

(a) 181.05 m/s²

(b) 13.2°

Explanation:

Given:

Radius of the circle (R) = 0.610 m

Angular acceleration (α) = 67.6 rad/s²

Angular speed (ω) = 17.0 rad/s

(a)

Radial acceleration of the ball is given as:

a_r=\omega^2R

Plug in the given values and solve for a_r. This gives,

a_r=(17.0\ rad/s)^2\times (0.610\ m)\\\\a_r=289\times 0.610\ m/s^2\\\\a_r=176.29\ m/s^2

Now, tangential acceleration is given by the formula:

a_t=R\alpha

Plug in the given values and solve for a_t. This gives,

a_t=(0.610\ m)(67.6\ rad/s^2)\\\\a_t=41.236\ m/s^2

Now, the magnitude of total acceleration is given as the square root of the sum of the squares of tangential and centripetal accelerations. Therefore,

a_{Total}=\sqrt{(a_r)^2+(a_t)^2}

Plug in the given values and solve for total acceleration, a_{Total}. This gives,

a_{Total}=\sqrt{(176.29)^2+(41.236)^2}\\\\a_{Total}=181.05\ m/s^2

Therefore, the magnitude of total acceleration is 181.05 m/s².

(b)

Angle of total acceleration relative to radial direction is given by the formula:

\theta=\tan^{-1}(\frac{a_t}{a_r})\\\\\theta=\tan^{-1}(\frac{41.236}{176.29})\\\\\theta=13.2\°

Therefore, the total acceleration makes an angle of 13.2° relative to radial direction.

4 0
3 years ago
Martije has made a slight error in naming a compound monocarbon tetrabromide. What compound is she most likely naming, and what
Alik [6]

Answer: CBr_4 : carbon tetrabromide

Explanation:

CBr_4 is a covalent compound because in this compound the sharing of electrons takes place between carbon and bromine. Both the elements are non-metals. Hence, it will form covalent bond.

The naming of covalent compound is given by:

1. The less electronegative element is written first.

2. The more electronegative element is written second. Then a suffix is added with it. The suffix added is '-ide'.

3. If atoms of an element is greater than 1, then prefixes are added which are 'mono' for 1 atom, 'di' for 2 atoms, 'tri' for 3 atoms and so on.

Hence, the correct name for CBr_4 is carbon tetrabromide.

7 0
3 years ago
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