All categories

3 years ago

An ion in a mass spectrometer follows a semicircular path of radius 14.8. What is the distance it travels?

2 answers:

7 0

The circumference of a circle is (2π · the circle's radius).

The length of a semi-circle is (1π · the circle's radius) =

(π · 14.8) = 46.5 (rounded)

(The unit is the same as whatever the unit of the 14.8 is.)

Y_Kistochka [10]3 years ago

4 0

Is there a time frame in which it travels?

You might be interested in

An electron moves at 0.130 c as shown in the figure (Figure 1). There are points: A, B, C, and D 2.10 μm from the electron.

Olegator [25]

Hi there!

We can use Biot-Savart's Law for a moving particle:
$B= \frac{\mu_0 }{4\pi}\frac{q\vec{v}\times \vec{r}}{r^2 }$

B = Magnetic field strength (T)
v = velocity of electron (0.130c = 3.9 × 10⁷ m/s)

q = charge of particle (1.6 × 10⁻¹⁹ C)

μ₀ = Permeability of free space (4π × 10⁻⁷ Tm/A)

r = distance from particle (2.10 μm)

There is a cross product between the velocity vector and the radius vector (not a quantity, but specifies a direction). We can write this as:

$B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }$

Where 'θ' is the angle between the velocity and radius vectors.

a)
To find the angle between the velocity and radius vector, we find the complementary angle:

θ = 90° - 60° = 30°

Plugging 'θ' into the equation along with our other values:

$B= \frac{\mu_0 }{4\pi}\frac{q\vec{v} \vec{r}sin\theta}{r^2 }\\\\B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(30)}{(2.1*10^{-5})^2 }$

$B = \boxed{7.07 *10^{-10} T}$

b)
Repeat the same process. The angle between the velocity and radius vector is 150°, and its sine value is the same as that of sin(30°). So, the particle's produced field will be the same as that of part A.

In this instance, the radius vector and the velocity vector are perpendicular so

'θ' = 90°.

$B= \frac{(4\pi *10^{-7})}{4\pi}\frac{(1.6*10^{-19})(3.9*10^{7}) \vec{r}sin(90)}{(2.1*10^{-5})^2 } = \boxed{1.415 * 10^{-9}T}$

d)
This point is ALONG the velocity vector, so there is no magnetic field produced at this point.

Aka, the radius and velocity vectors are parallel, and since sin(0) = 0, there is no magnetic field at this point.

$\boxed{B = 0 T}$

3 0

2 years ago

Where is the near point of an normal eye when accidentally wear a contact lens with a power of +2.0 diopters?

Lerok [7]

Answer:

The near point of an eye with power of +2 dopters, u' = - 50 cm

Given:

Power of a contact lens, P = +2.0 diopters

Solution:

To calculate the near point, we need to find the focal length of the lens which is given by:

Power, P = $\frac{1}{f}$

where

f = focal length

Thus

f = $\frac{1}{P}$

f = $\frac{1}{2}$ = + 0.5 m

The near point of the eye is the point distant such that the image formed at this point can be seen clearly by the eye.

Now, by using lens maker formula:

$\frac{1}{f} = \frac{1}{u} + \frac{1}{u'}$

where

u = object distance = 25 cm = 0.25 m = near point of a normal eye

u' = image distance

Now,

$\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}$

$\frac{1}{u'} = \frac{1}{0.5} - \frac{1}{0.25}$

$\frac{1}{u'} = \frac{1}{f} - \frac{1}{u}$

Solving the above eqn, we get:

u' = - 0.5 m = - 50 cm

7 0

3 years ago

A(n) 131 g ball is dropped from a height

larisa [96]

Answer:

26.59 N/m

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 131 g

Extention (e) = 4.82755 cm

Acceleration due to gravity (g) = 9.8 m/s²

Spring constant (K) =?

Next, we shall convert 131 g to Kg. This can be obtained as follow:

1000 g = 1 Kg

Therefore,

131 g = 131 g × 1 Kg / 1000 g

131 g = 0.131 Kg

Thus, 131 g is equivalent to 0.131 Kg.

Next, we shall the force exerted by the ball on the spring. This can be obtained as follow:

Mass (m) = 0.131 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Force (F) =?

F = ma

F = 0.131 × 9.8

F = 1.2838 N

Next, we shall convert 4.82755 cm to metre (m)

This can be obtained as follow:

100 cm = 1 m

Therefore,

4.82755 cm = 4.82755 cm × 1 m / 100 cm

4.82755 cm = 0.0482755 m

Thus, 4.82755 cm is equivalent to 0.0482755 m

Finally, we shall determine the spring constant as follow:

Force (F) = 1.2838 N

Extention (e) = 0.0482755 m

Spring constant (K) =?

F = Ke

1.2838 = K × 0.0482755

Divide both side by 0.0482755

K = 1.2838 / 0.0482755

K = 26.59 N/m

Thus the spring constant is 26.59 N/m

7 0

3 years ago

According to Kepler's Third Law, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about

NARA [144]

Answer:

Orbital period, T = 1.00074 years

Explanation:

It is given that,

Orbital radius of a solar system planet, $r=4\ AU=1.496\times 10^{11}\ m$

The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :

$T^2=\dfrac{4\pi^2}{GM}r^3$

M is the mass of the sun

$T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 1.989\times 10^{30}}\times (1.496\times 10^{11})^3$

$T^2=\sqrt{9.96\times 10^{14}}\ s$

T = 31559467.6761 s

T = 1.00074 years

So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.

6 0

3 years ago

Calculate the current through a 3.0 Ď‰ resistor with a voltage of 9.0 v across it.

gulaghasi [49]

Answer: 3 A

Explanation:

According to<u> Ohm's law</u>:

$V=R.I$

Where:

$V=9 V$ is the voltage

$R=3\Omega$ is the resistance of the resistor

$I$ is the electric current (the value we want to find)

Isolating $I$ :

$I=\frac{V}{R}$

$I=\frac{9 V}{3\Omega}$

Finally:

$I=3 A$

7 0

3 years ago