All categories

2 years ago

A singly charged ion (q=−1.6×10−19) makes 7.0 rev in a 45 mT magnetic field in 1.29 ms. The mass of the ion in kg is

Physics

1 answer:

insens350 [35]2 years ago

4 0

Answer:

$m=1.47\times 10^{-24}\ Kg$

Explanation:

Given that,

Charge, $q=1.6\times 10^{-19}\ C$

Revolution = 7 rev

magnetic field, B = 45 mT

Time, t = 1.29 ms

We need to find the mass of the ion. Let m be the mass. The formula for the mass in terms of time period is given by :

$m=\dfrac{qBT}{2\pi}\\\\m=\dfrac{1.6\times 10^{-19}\times 45\times 10^{-3}\times 1.29\times 10^{-3}}{2\pi}\\\\m=1.47\times 10^{-24}\ Kg$

So, the mass of the ion is equal to $1.47\times 10^{-24}\ Kg$ .

You might be interested in

Marc and Linh stretch out a long spring on the classroom floor. Marc holds one end of the spring still. Linh creates waves in th

Soloha48 [4]

The speed of the wave created by Linh in the spring by moving the other end right and left with a frequency of 2 Hz is 1m/s.

<h3>How to calculate speed of a wave?</h3>

The speed of a wave can be calculated by using the following formula:

Speed = Wavelength x Frequency

According to this question, Linh creates waves in the spring by moving the other end right and left with a frequency of 2 Hz. If wave crests are 0.5 m apart, the speed can be calculated as follows:

speed = 2Hz × 0.5m

speed = 1m/s

Therefore, the speed of the wave created by Linh in the spring by moving the other end right and left with a frequency of 2 Hz is 1m/s.

Learn more about speed at: brainly.com/question/10715783

#SPJ1

7 0

2 years ago

Use the graph below to answer the following question: if average acceleration is calculated using the equation, “ change in velo

sergiy2304 [10]

Answer:

$a=9\ cm/s^2$

Explanation:

<u>Average Acceleration </u>

Acceleration is a physical magnitude defined as the change of velocity over time. When we have experimental data, we can compute it by calculating the slope of the line in velocity vs time graph.

Note: <em>We cannot see if the time axis is numbered in increments of 1 second, and we'll assume that. </em>

When $t_2=4\ sec$ , the graph shows a value of $v_2=36\ cm/s$

When $t_1=0\ sec$ , the object is at rest, $v_1=0$

We compute the average acceleration as

$\displaystyle a=\frac{v_2-v_1}{t_2-t_1}$

$\displaystyle a=\frac{36\ cm/s-0\ cm/s}{4\ sec-0\ sec}$

$\displaystyle a=\frac{36\ cm/s}{4\ s}$

$\boxed{a=9\ cm/s^2}$

6 0

2 years ago

Peg P is driven by the forked link OA along the path described by r = eu, where r is in meters. When u = p4 rad, the link has an

8_murik_8 [283]

Answer:

The transverse component of acceleration is 26.32 $m/s^2$ where as radial the component of acceleration is 8.77 $m/s^2$

Explanation:

As per the given data

u=π/4 rad

ω=u'=2 rad/s

α=u''=4 rad/s

$r=e^u$

So the transverse component of acceleration are given as

$a_{\theta}=(ru''+2r'u')\\$

Here

$r=e^u\\r=e^{\pi/4}\\r=2.1932 m$

$r'=e^u.u'\\r'=2.1932 \times 2\\r'=4.3864 m$

$a_{\theta}=(ru''+2r'u')\\a_{\theta}=(2.1932\times 4+2\times 4.3864 \times 2)\\a_{\theta}=26.32 m/s\\$

The transverse component of acceleration is 26.32 $m/s^2$

The radial component is given as

$a_r=r''-r\theta'^2$

Here

$r''=e^u.u'^2+e^u u''\\r''=2.1932 \times (2)^2+2.1932\times 4\\r''=17.5456 m$

$a_r=r''-ru'^2\\a_r=17.5456-2.1932\times (2)^2\\a_r=8.7728 m/s^2$

The radial component of acceleration is 8.77 $m/s^2$

6 0

3 years ago

a seismic wave has an amplitude of 0.012 Meters.If the amplitude of this wave reduces to 0.006 meters, what happens to the energ

irina1246 [14]

Answer:The energy of the wave by a factor of 4

Explanation:

5 0

3 years ago

A sinusoidal wave of angular frequency 1,203 rad/s and amplitude 3.1 mm is sent along a cord with linear density 3.9 g/m and ten

kobusy [5.1K]

Answer:

18.7842493212 W

Explanation:

T = Tension = 1871 N

$\mu$ = Linear density = 3.9 g/m

y = Amplitude = 3.1 mm

$\omega$ = Angular frequency = 1203 rad/s

Average rate of energy transfer is given by

$P=\dfrac{1}{2}\sqrt{T\mu}\omega^2y^2\\\Rightarrow P=\dfrac{1}{2}\sqrt{1871\times 3.9\times 10^{-3}}\times 1203^2\times (3.1\times 10^{-3})^2\\\Rightarrow P=18.7842493212\ W$

The average rate at which energy is transported by the wave to the opposite end of the cord is 18.7842493212 W

7 0

3 years ago