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lukranit [14]
3 years ago
7

PLEASE HELP ASAP!!! CORRECT ANSWER ONLY PLEASE!!!

Physics
1 answer:
balandron [24]3 years ago
6 0

Some examples you could look up on-line so I don't have to:

Speed of sound in sea water . . . 1,531 m/s

Speed of sound in iron . . . 5,130 m/s

Speed of sound in air . . . 340 m/s

This gives us

fastest . . . medium . . . slowest

<em>solid . . . . . liquid . . . . . gas</em>

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Answer:

1)

When a charge is in motion in a magnetic field, the charge experiences a force of magnitude

F=qvB sin \theta

where here:

For the proton in this problem:

q=1.602\cdot 10^{-19}C is the charge of the proton

v = 300 m/s is the speed of the proton

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F=(1.602\cdot 10^{-19})(300)(19)(sin 65^{\circ})=8.28\cdot 10^{-16} N

2)

The magnetic field produced by a bar magnet has field lines going from the North pole towards the South Pole.

The density of the field lines at any point tells how strong is the magnetic field at that point.

If we observe the field lines around a magnet, we observe that:

- The density of field lines is higher near the Poles

- The density of field lines is lower far from the Poles

Therefore, this means that the magnetic field of a magnet is stronger near the North and South Pole.

3)

The right hand rule gives the direction of the  force experienced by a charged particle moving in a magnetic field.

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- Direction of index finger = direction of motion of the charge

- Direction of middle finger = direction of magnetic field

- Direction of thumb = direction of the force (for a negative charge, the direction must be reversed)

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4)

The radius of a particle moving in a magnetic field is given by:

r=\frac{mv}{qB}

where here we have:

m=6.64\cdot 10^{-22} kg is the mass of the alpha particle

v=2155 m/s is the speed of the alpha particle

q=2\cdot 1.602\cdot 10^{-19}=3.204\cdot 10^{-19}C is the charge of the alpha particle

B = 12.2 T is the strength of the magnetic field

Substituting, we find:

r=\frac{(6.64\cdot 10^{-22})(2155)}{(3.204\cdot 10^{-19})(12.2)}=0.366 m

5)

The cyclotron frequency of a charged particle in circular motion in a magnetic field is:

f=\frac{qB}{2\pi m}

where here:

q=1.602\cdot 10^{-19}C is the charge of the electron

B = 0.0045 T is the strength of the magnetic field

m=9.31\cdot 10^{-31} kg is the mass of the electron

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f=\frac{(1.602\cdot 10^{-19})(0.0045)}{2\pi (9.31\cdot 10^{-31})}=1.23\cdot 10^8 Hz

6)

When a charged particle moves in a magnetic field, its path has a helical shape, because it is the composition of two motions:

1- A uniform motion in a certain direction

2- A circular motion in the direction perpendicular to the magnetic field

The second motion is due to the presence of the magnetic force. However, we know that the direction of the magnetic force depends on the sign of the charge: when the sign of the charge is changed, the direction of the force is reversed.

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7)

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8)

The flux is calculated as

\Phi = BA sin \theta

where

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Here the loop is rectangular with lenght 15 cm and width 8 cm, so the area is

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So the flux is

\Phi = (5.5)(0.012)(sin 18^{\circ})=0.021 Wb

See the last 7 answers in the attached document.

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