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3 years ago

What would be the result of an alpha particle coming into a magnetic field?

2 answers:

8 0

The right answer for the question that is being asked and shown above is that: "C) The alpha particle will be deflected in a curve path. " the result of an alpha particle coming into a magnetic field is that <span>C) The alpha particle will be deflected in a curve path. </span>

Anarel [89]3 years ago

7 0

Answer: C) The alpha particle will be deflected in a curve path.

Explanation:

Alpha particle carries +2 charge. It is a Helium ion: He⁺²

When any charged particle enters a magnetic field, it experiences a magnetic force perpendicular to direction of the velocity of the particle and the direction of the magnetic field. Thus, the alpha particle moves in a circular path.

Hence, the correct answer is: The result of an alpha particle coming into a magnetic field is C) The alpha particle will be deflected in a curve path.

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Answer:

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Explanation:

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Molecules are made up of: A. elements B. compounds C. protons

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What is the speed of a wave frequency of 300hz and a wavelength of 25M

tia_tia [17]

Answer:

7500 m/s

Explanation:

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3 years ago

Draw sodium formate by placing atoms on the grid and connecting them with bonds. Include all lone-pair electrons.

nikklg [1K]

Answer:

Sodium formate is the sodium salt of formic acid which is given as HCOONa.

Explanation:

The basic structure of Sodium formate consists of following bonds:

The main Ionic bond between the $HCOO^-$ radical and $Na^+$ .
The sigma covalent bonds between atom of H, atom of C and both atoms of O.
The pi bond between atom of C and one atom of O.

The structure along with lone pairs is given as attached

7 0

3 years ago

A typical helium-neon laser found in supermarket checkout scanners emits 633-nm-wavelength light in a 1.0-mm-diameter beam with

Fofino [41]

Answer:

Eo = 9.796 x 10^2 N/C

Bo = 3.266 x 10^-6 T

Explanation:

Given

Wavelength λ = 633 nm

Diameter of the beam D = 1.0 mm

Power P = 1.0 mW

Solution

Radius of the beam r = D/2 = 0.5 mm = 0.0005 m

Area of cross section

$A = \pi r^{2} \\A = 3.15 \times 0.0005^{2}\\A = 7.58 \times 10^{-7} m^{2}\\$

Intensity

$I = \frac{P}{A} \\I = \frac{0.001}{7.85\times 10^{-7}} \\I = 1273.885 {W}/{m^{2} }$

Amplitude of Electric Field

$E_{o} = \sqrt{\frac{2I}{ \epsilon_{o}c } } \\E_{o} = \sqrt{\frac{2 \times 1273.88}{ 8.85 \times 10^{-12} \times 3 \times 10^{8} } }\\E_{o} = 9.796 \times 10^{2}N/C$

Amplitude of Magnetic Field

$B_{o} = \sqrt{\frac{2 \mu_{o}I}{c } } \\B_{o} = \sqrt{\frac{2 \times 4 \times \pi \times 10^{-7} \times 1273.88}{ 3 \times 10^{8} } }\\B_{o} = 3.266 \times 10^{-6} T$

5 0

4 years ago