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3 years ago

6

After being accelerated through a potential difference of 5.33 kv, a singly charged carbon ion (12c) moves in a circle of radius

22.5 cm in the magnetic field of a mass spectrometer. what is the magnitude of the field?

1 answer:

LekaFEV [45]3 years ago

4 0

I'm stuck on the same question, as well :(

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A ____________ is a condition or behavior that may threaten an individual's well-being

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It would be risk factor

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2 years ago

Read 2 more answers

Help wit these questions someone.

romanna [79]

In series circuit, Req = R₁ + R₂ + R₃ + ···

In parallel circuit, $\frac{1}{Req} = \frac{1}{R1} + \frac{1}{R2} + \frac{1}{R3} +...$

<h3>Q7.</h3>

total resistance in the upper branch = R₂ + R₃ = R₂ + 2

$\frac{1}{Req} = \frac{1}{R2+R3} + \frac{1}{R1}$

$\frac{1}{4} = \frac{1}{R2+2} +\frac{1}{6}$

R₂ + 2 = 12

R₂ = 10Ω

<h3>Q8.</h3>

$\frac{1}{Req} = \frac{1}{R2+R3} + \frac{1}{R1}$

$\frac{1}{Req} = \frac{1}{2+1} + \frac{1}{4}$

Req = 1.7Ω

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3 years ago

Which property describes the ability of one substance to dissolve in another substance?

oee [108]

The answer would be solubility

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3 years ago

Sally takes two bar magnets and randomly tapes one end of each bar magnet. she labels the magnets A and B. She brings the taped

Artyom0805 [142]

I already answered this quesiton. The fact is that there are only two kind of poles and since the two taped poles of the magnets labeled A and B attracts one to each other, we know that the two taped poles of the first two magnets are oppsosite.

Then, the taped pole of the third magnet has to be equal to one of the first two taped poles and opposite to the other of the first two taped poles.

That drives you to conclude (predict) that when she brings the taped end of the third magnet (magnet C) near each of the first two magntes, in one case they will attract each other and in the other case they will repele mutually.

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3 years ago

A 0.900-V potential difference is maintained across a 1.5m length of 2

Daniel [21]

Answer:

I = 6.42 A

Explanation:

Given that,

Potential difference, V = 0.9 V

Length of the wire, l = 1.5 m

Area of cross section, $A=0.6\ mm^2=6\times 10^{-7}\ m^2$

We need to find the current in the wire. Let I is current. We can find it using Ohm's law as follows :

V = IR

Where R is the resistance of the wire

$I=\dfrac{V}{R}\\\\I=\dfrac{V}{\rho \dfrac{l}{A}}\\\\I=\dfrac{0.9}{5.6\times 10^{-8}\times \dfrac{1.5}{6\times 10^{-7}}}\\\\I=6.42\ A$

So, the current in the wire is 6.42 A.

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3 years ago