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aniked [119]
3 years ago
12

A mass spectrometer applies a voltage of 2.00 kV to accelerate a singly charged positive ion. A magnetic field of B = 0.400 T th

en bends the ion into a circular path of radius 0.305 m. What is the mass of the ion?
Physics
1 answer:
N76 [4]3 years ago
3 0

The mass of the ion is 5.96 X 10⁻²⁵ kg

<u>Explanation:</u>

The electrical energy given to the ion Vq will be changed into kinetic energy \frac{1}{2}mv^2

As the ion moves with velocity v in a magnetic field B then the magnetic Lorentz force Bqv will be balanced by centrifugal force \frac{mv^2}{r}.

So,

Vq = \frac{1}{2}mv^2

and

Bqv = \frac{mv^2}{r}

Right from these eliminating v, we can derive

m = \frac{B^2r^2q}{2V}

On substituting the value, we get:

m = \frac{(0.4)^2X (0.305)^2 X1.602X 10^-^1^9}{2X 2000}\\\\

m = 5.96 X 10⁻²⁵ kg.

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1) Focal length

We can find the focal length of the mirror by using the mirror equation:
\frac{1}{f}= \frac{1}{d_o}+ \frac{1}{d_i} (1)
where 
f is the focal length
d_o is the distance of the object from the mirror
d_i is the distance of the image from the mirror

In this case, d_o = 18 cm, while d_i=-4 cm (the distance of the image should be taken as negative, because the image is to the right (behind) of the mirror, so it is virtual). If we use these data inside (1), we find the focal length of the mirror:
\frac{1}{f}= \frac{1}{18 cm}- \frac{1}{4 cm}=- \frac{7}{36 cm}
from which we find
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2) The mirror is convex: in fact, for the sign convention, a concave mirror has positive focal length while a convex mirror has negative focal length. In this case, the focal length is negative, so the mirror is convex.

3) The image is virtual, because it is behind the mirror and in fact we have taken its distance from the mirror as negative.

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r=2f=2 \cdot 5.1 cm=10.2 cm
3 0
3 years ago
The distance between two particles is 2 centimeters. If the distance is increased to 4 centimeters, the force will be ?
postnew [5]

Answer:

The new force is 1/4 of the previous force.

Explanation:

Given

Initial\ Distance = 2cm ---- r_1

New\ Distance = 4cm --- r_2

Required

Determine the new force

Let the two particles be q1 and q2.

The initial force F1 is:

F_1 = \frac{kq_1q_2}{r_1^2} --- Coulomb's law

Substitute 2 for r1

F_1 = \frac{kq_1q_2}{2^2}

F_1 = \frac{kq_1q_2}{4}

The new force (F2) is

F_2 = \frac{kq_1q_2}{r_2^2}

Substitute 4 for r2

F_2 = \frac{kq_1q_2}{4^2}

F_2 = \frac{kq_1q_2}{4*4}

F_2 = \frac{1}{4}*\frac{kq_1q_2}{4}

Substitute F_1 = \frac{kq_1q_2}{4}

F_2 = \frac{1}{4}*F_1

F_2 = \frac{F_1}{4}

The new force is 1/4 of the previous force.

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