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ehidna [41]
3 years ago
6

The electric field inside a cell membrane is 8.0 MN/C. Part A What's the force on a singly charged ion in this field?

Physics
1 answer:
dlinn [17]3 years ago
7 0

Answer:

Force on the singly charged ion will be 12.6\times 10^{-13}N

Explanation:

We have given electric field E = 8 MN/C

So electric field in N/C will be E=8MN/C=8\times 10^6N/C

It is given that ion so charge on ion will be equal to e=1.6\times 10^{-19}C

We have to find the electric force on the ion

Electric force is equal to F=qE, here q is charge and E is electric field

So force on the charge will be equal to F=8\times 10^6\times 1.6\times 10^{-19}=12.6\times 10^{-13}N

So force on the singly charged ion will be 12.6\times 10^{-13}N

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While validity of an instrument is the idea that the instrument measures what it intends to measure, reliability is all about consistency and whether or not results could be replicated.

While your friend's answer may be valid, it can't reliable because of the differences in her response within a short time.

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4 years ago
A 45 kg merry go round worker stands on the rides platform 6.3 m from the center. If her speed as she goes around the circle is
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I’ve got 120 n, hope it helps!

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3 years ago
Describe how Rutherford's experiments changed the accepted scientific model of the atom.
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3 years ago
What is the freezing point of radiator fluid that is 50% antifreeze by mass? k f for water is 1.86 ∘ c/m?
densk [106]
Ethylene glycol is termed as the primary ingredients in antifreeze.
The ethylene glycol molecular formula is C₂H₆O₂.
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Now that antifreeze by mass is 50%, then there is 1kg of ethylene glycol which is present in 1kg of water.
ΔTf = Kf×m
ΔTf = depression in the freezing point.
= freezing point of water freezing point of the solution
= O°c - Tf
= -Tf
Kf = depression in freezing constant of water = 1.86°C/m
M is the molarity of the solution.
=(mass/molar mass) mass of solvent in kg
=1000g/62 (g/mol) /1kg
=16.13m
If we plug the value we get 
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7 0
3 years ago
Read 2 more answers
How to do this, i'm completely lost
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There are two torques t1 and t2 on the beam due to the weights, one torque t3 due to the weight of the beam, and one torque t4 due to the string.

You need to figure out t4 to know the tension in the string.

Since the whole thing is not moving t1 + t2 + t3 = t4.

torque t = r * F * sinФ = distance from axis of rotation * force * sin (∡ between r and F)

t1 =3.2 * 44g 
t2 = 7 * 49g 
t3 = 3.5 * 24g 

t4 = t1 + t2 + t3 = 5570,118

The t4 also is given by:

t4 = r * T * sin Ф

r = 7
Ф = 32°
T: tension in the string

T = t4 / (r * sinФ)

T = t4 / (7 * sin(32°)) 

T = 1501,6 N

8 0
3 years ago
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