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Marina86 [1]
3 years ago
11

A mass spectrometer is to be used to separate protons from deuterium nuclei. A deuterium nucleus has the same charge and twice t

he mass as a proton, since it contains an extra neutron. Both the deuterium and the proton nuclei are accelerated by the same voltage. Which of the following statements is true?
Answers:
The deuterium will not be deflected in the mass spectrometer, since it contains a neutron.
Both the proton and the deuterium will move along the same circular path.
The deuterium will have a larger radius of curvature, since it is more massive.
Mass spectrometers cannot be used to analyze nuclei, only molecular ions.
The deuterium will have a smaller radius of curvature, since it is more massive.
Physics
1 answer:
Lina20 [59]3 years ago
3 0

Answer:

The deuterium will have a larger radius of curvature, since it is more massive.

Explanation:

Suppose that the initial velocity of deuterium and protons is the same. According to the relation for centripetal acceleration:

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

The force is the same due to same electric charge between proton and deuterium. Solving for r:

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

It means that, the more the mass, the larger the radius due to the directly proportional relation.

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5 0
3 years ago
The cyclotron frequency is the number of revolutions the particle makes in one second. calculate the cyclotron frequency for thi
Roman55 [17]

Answer:

f=qB/2\pim

Explanation:

The cyclotron frequency is the number of revolutions the particle makes in one second. calculate the cyclotron frequency for this particle

solution

The Lorentz force  F_{lorentz} =F_{centripetal}

is the centripetal force  

and makes the particles path to revolve in a circle:

qvB=mv^2/r

radius=r

m=mas of the particle

B=magnetic flux

q=quantity of charge

v=velocity of the particle

v=qBr/m

where v is the velocity of the particle

recall the velocity v=rω

v=2*Pi*f*r

2\pi *f*r=qBr/m

f=qB/2\pim

the cyclotron frequency is therefore f=qB/2\pim

the period(the time it take to make a complete oscillation) will be the inverse of frequency

T=2\pim/qB

4 0
3 years ago
A small, 2.00-mm-diameter circular loop with R = 1.00×10^−2 Ω is at the center of a large 100-mm-diameter circular loop. Both lo
Assoli18 [71]

Answer:

i=2.1\times 10^{-8}}\ A

Explanation:

Given that

Diameter of small loop d= 2 mm  or r=1 mm

Diameter of large loop D= 100 mm  or R=50 mm

We know that induce emf given as

\varepsilon =-\dfrac{d\phi }{dt}

\varepsilon =-\dfrac{BA }{dt}

B=\dfrac{\mu _oI}{2\pi R}

\varepsilon =-\pi \times r^2\times \dfrac{4\pi \times 10^{-7}(I_2-I_1)}{2\pi Rdt}

\varepsilon =-\pi \times 0.001^2\times \dfrac{4\pi \times 10^{-7}(-1-1)}{2\pi \times 0.05\times 0.1}

\varepsilon =2.1\times 10^{-10}\ V

So induce current

i=emf/R

i=\dfrac{2.1\times 10^{-10}}{10^{-2}}\ A

i=2.1\times 10^{-8}}\ A

5 0
3 years ago
If the pendulum took longer to complete one oscillation, how would the graph change?
Sindrei [870]

Answer:

took longer to complete one oscillation, that means its PERIOD increased, and the distance between the peaks of the graph would be longer.

line would be less. the period of oscillation would have any effect on the graph

7 0
3 years ago
Read 2 more answers
An LP turntable must spin at 3.500 rad/s to play a record. How much torque must the motor deliver if the turntable is to reach i
zmey [24]

Answer:

0.00321 Nm

Explanation:

\omega_f = Final angular velocity = 3.5 rad/s

\omega_i = Initial angular velocity = 0

\alpha = Angular acceleration

\theta = Angle of rotation = 2 rev

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\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}\\\Rightarrow \alpha=\frac{3.5^2-0^2}{2\times 2\pi \times 1.8}\\\Rightarrow \alpha=0.54156\ rad/s^2

Moment of inertia is given by

I=mr^2\\\Rightarrow I=0.25\times 0.154^2\\\Rightarrow I=0.005929\ kgm^2

Torque is given by

\tau=I\alpha\\\Rightarrow \tau=0.005929\times 0.54156\\\Rightarrow \tau=0.00321\ Nm

The torque required is 0.00321 Nm

8 0
4 years ago
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