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3 years ago

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An electron enters a magnetic field of 0.43 T with a velocity perpendicular to the direction of the field. At what frequency doe

s the electron traverse a circular path? ( m el = 9.11 × 10-31 kg, e = 1.60 × 10-19 C)

1 answer:

Mrac [35]3 years ago

6 0

Answer:

Frequency, $f=1.2\times 10^{10}\ Hz$

Explanation:

It is given that,

Magnetic field, B = 0.43 T

We need to find the frequency the electron traverse a circular path. It is also known as cyclotron frequency. It is given by :

$f=\dfrac{qB}{2\pi m}$

$f=\dfrac{1.6\times 10^{-19}\ C\times 0.43\ T}{2\pi \times 9.11\times 10^{-31}\ Kg}$

$f=1.2\times 10^{10}\ Hz$

Hence, this is the required solution.

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Answer:

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Explanation:

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A wooden block with mass 1.60 kg is placed against a compressed spring at the bottom of a slope inclined at an angle of 30.0° (p

andreyandreev [35.5K]

Answer:

The amount of potential energy that was initially stored in the spring is 88.8 J.

Explanation:

Given that,

Mass of block = 1.60 kg

Angle = 30.0°

Distance = 6.55 m

Speed = 7.50 m/s

Coefficient of kinetic friction = 0.50

We need to calculate the amount of potential energy

Using formula of conservation of energy between point A and B

$U_{A}+k_{A}+w_{A}=U_{B}+k_{B}$

$U_{A}+0-fd=mgy+\dfrac{1}{2}mv^2$

$U_{A}=\mu mg\cos\theta\times d+mg h\sin\theta+\dfrac{1}{2}mv^2$

Put the value into the formula

$U_{A}=0.50\times1.60\times9.8\cos30\times6.55+1.60\times9.8\times6.55\sin30+\dfrac{1}{2}\times1.60\times(7.50)^2$

$U_{A}=88.8\ J$

Hence, The amount of potential energy that was initially stored in the spring is 88.8 J.

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3 years ago