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matrenka [14]
3 years ago
12

Medical cyclotrons need efficient sources of protons to inject into their center. In one kind of ion source, hydrogen atoms (i.e

., protons with one orbiting electron) are fed into a chamber where there is a strong magnetic field. Electrons in this chamber are trapped in tight orbits, which greatly increases the chance that they will collide with a hydrogen atom and ionizes it. One such source uses a magnetic field of 90 mT, and the electrons' kinetic energy is 2.0 eV.If the electrons travel in a plane perpendicular to the field, what is the radius of their orbits?
Physics
1 answer:
blondinia [14]3 years ago
3 0

Answer:

Explanation:

Electron's kinetic energy = 2 eV

= 2 x 1.6 x 10⁻¹⁹ J

1/2 m v² = 3.2 x 10⁻¹⁹

1/2 x 9.1 x 10⁻³¹ x v² = 3.2 x 10⁻¹⁹

v² = .703 x 10¹²

v = .8385 x 10⁶ m/s

Electrons revolve in a circular orbit when forced to travel in a magnetic field whose radius can be expressed as follows

r = mv / Bq

where m , v and q are mass , velocity and charge of electron .

here given magnetic field B = 90 mT

= 90 x 10⁻³ T

Putting these values in the expression above

r =  mv / Bq

= \frac{9\times10^{-31}\times.8385\times10^6}{90\times10^{-3}\times1.6\times10^{-19}}

= .052 mm.

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Over [174]
The answer is voltage. :)
this is because voltage is the rate that electric charges move through a conductor.
6 0
3 years ago
What is the difference between the velocity and the speed of an object?
timurjin [86]

Answer:

Velocity has a direction associated with it, while speed has no specific direction.

Explanation:

Velocity is a vector, while speed is a scalar.

3 0
3 years ago
An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.220 rev/s . The magnitude
matrenka [14]

Answer:

1) The fan's angular velocity after 0.208 seconds is approximately 2.585 rad/s

2) The number of revolutions the blade has travelled in 0.208 s is approximately 0.066 revolutions

3) The tangential speed of a point on the tip of the blade at time t = 0.208 s is approximately 1.034 m/s

4) The magnitude of the tangential acceleration of a point on the tip of the blade at time t = 0.208 seconds is approximately 2.312 m/s²

Explanation:

The given parameters are;

The initial velocity of the fan, n = 0.220 rev/s

The magnitude of the angular acceleration = 0.920 rev/s²

The direction of the angular acceleration and the angular velocity = Clockwise

The diameter of the circle formed by the electric ceiling fan blades, D = 0.800 m

1) The initial angular velocity of the fan, ω₀ = 2·π × n = 2·π × 0.220 rev/s = 1.38230076758 rad/s

The angular acceleration of the fan, α = 2·π×0.920 rad/s² = 5.78053048261 rad/s²

The fan's angular velocity, 'ω', after a time t = 0.208 seconds has passed is given as follows;

ω = ω₀ + α·t

From which we have;

ω = 1.38230076758 rad/s + 5.78053048261 rad/s × 0.208 s = 2.58465110796 rad/s

The fan's angular velocity after 0.208 seconds is ω ≈ 2.585 rad/s

2) The number of revolutions the blade has travelled in the given time interval is given from the angle turned, 'θ', in the given time as follows;

θ = ω₀·t + 1/2·α·t²

θ = 1.38230076758 × 0.208 + 1/2 × 5.78053048261 × 0.208² = 0.41256299505 radians

2·π radians = 1 revolution

∴ 0.41256299505 radians = 0.41256299505 radian× 1 revolution/(2·π radian) = 0.06566144 revolution

The number of revolutions the blade has travelled in 0.208 s ≈ 0.066 revolutions

3) The tangential speed of a point on the tip of the blade at time t = 0.208 s is given as follows;

The tangential speed, v_t = ω × r = ω × D/2

At t = 0.208 s, ω = 2.58465110796 rad/s, therefore, we have;

v_t = ω × D/2 = 2.58465110796 × 0.800/2 = 1.0338604413

The tangential speed, v_t = 1.0338604413 m/s

The tangential speed ≈ 1.034 m/s

4)  The magnitude of the tangential acceleration of a point on the tip of the blade at time t = 0.208 seconds, 'a' is given as follows;

a = α × r = α × D/2

a = 5.78053048261 × 0.800/2 = 2.31221219304

The tangential acceleration, a ≈ 2.312 m/s²

4 0
3 years ago
a ball is thrown horizontally from a 20 m high building with a speed of 5.0 m/s. How far from the base of the building does the
kipiarov [429]
Given:
Dy= 20 m
Vi = 5.0 m/s horizontally
A=9.81 m/s^2

Find:
Horizontal displacement

Solution:
D=ViT+(1/2)AT^2
Dy=(1/2)AT^2
T^2=Dy/(1/2)A
T=sqrt(Dy/(1/2)A)
T=sqrt(20/4.905)
T=2.0s

Dx=ViT
Dx=(5.0)(2.0)
Dx=10. meters
7 0
3 years ago
Which characteristic should a good scientific question have?
Brilliant_brown [7]
The best answer to this question is D. a god scientific question should be based on observations instead of just imagination.

Hope this helps.
7 0
3 years ago
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