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SOVA2
3 months ago
5

The magnitude of the electrical force acting between a +2.4 × 10–8 C charge and a +1.8 × 10–6 C charge that are separated by 0.0

08 m is N, rounded to the tenths place.
Physics
2 answers:
Keith_Richards [3.2K]3 months ago
8 0

On Ed, it's 6.1 N. I just completed the test and got it right!

kicyunya [3.2K]3 months ago
8 0
To solve this problem, Coulomb's law will be applied as follows:
F = k*q1*q2 / r^2 where:
F indicates the force magnitude between the charges
k is a constant = 9.00 * 10^9 N.m^2/C^2
q1 = <span>+2.4 × 10–8 C
q2 = </span><span>+1.8 × 10–6 C
r represents the distance separating the charges = </span><span>0.008 m

By substituting these values, we derive:
F = (9*10^9)(2.4*10^-8)(1.8*10^-6) / (0.008)^2 = 6.075, which rounds to 6.1 Newtons

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The Hall effect can be used to calculate the charge-carrier number density in a conductor. If a conductor carrying a current of
kicyunya [3294]

Answer:

6.6*10^{27}e/m^3

Explanation:

When calculating Hall voltage, it is crucial to have the current, magnetic field strength, length, area, and number of charge carriers available. The Hall voltage can be expressed using the equation:

V_h = \frac{iB}{neL}

Where:

i= the current

B= the magnetic field strength

L = the length

n = the number of charge carriers

e= charge of an electron

We need to replace values and solve for n:

n= \frac{iB}{V_h e L}

n= \frac{2*1.2}{4.5*10^{-6}*5^10^{-3}*1.6*10^{-19}}

n= 6.6*10^{27}electron.m^{-3}

As a result, the charge carrier density is 6.6*10^{27}e/m^3

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3 months ago
A compressed spring has 16.2J of elastic potential energy when it is compressed 0.30m . What is the spring constant of the sprin
ValentinkaMS [3465]
I hope this provides the assistance you need.

3 0
3 months ago
Radiation emitted from human skin reaches its peak at λ = 940 µm. (a) What is the frequency of this radiation? (b) What type of
inna [3103]

Answer:

a) The frequency is 3.191x10^11.

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3 months ago
A 0.050-kg lump of clay moving horizontally at 12 m/s strikes and sticks to a stationary 0.15-kg cart that can move on a frictio
Ostrovityanka [3204]

Answer:

Explanation:

According to the parameters provided,

mass of the clay lump, m₁ = 0.05 kg

initial velocity of the lump, u₁ = 12 m/s

mass of the cart, m₂ = 0.15 kg

initial speed of the cart, u₂ = 0

As the clay adheres to the cart, we have an inelastic collision scenario. Let v represent the combined speed of both the cart and lump post-collision. Given that momentum is conserved, we have:

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The resultant speed is v = 3 m/s.

Thus, the final speed of both cart and lump following the collision is 3 m/s. This concludes the solution.

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3 months ago
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