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2 years ago

what is the electrical potential at the surface of gold nucleus? The radius of a gold atom is 6.6*10

Physics

1 answer:

MaRussiya [10]2 years ago

3 0

Complete question is;

What is the electrical potential at the surface of gold nucleus? The radius of a gold atom is 6.6 × 10^(-5) m and atomic number z = 79.

Answer:

172.36 × 10^(-5) V

Explanation:

We are given;

Radius; r = 6.6 × 10^(-5) m

Atomic number; Z = 79

Formula for Electric potential here is;

V = kZe/r

Where;

e is charge on proton = 1.6 × 10^(-19) C

k has a constant value of 9 × 10^(9) N⋅m²/C²,

Thus;

V = (79 × 1.6 × 10^(-19) × 9 × 10^(9))/(6.6 × 10^(-5))

V = 172.36 × 10^(-5) V

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3 years ago

abody starts from rest and accelerate uniformly at 8m/s/s.calculate the distance travelled by the body in 10s?

vagabundo [1.1K]

Explanation:

using the formula: S=ut+½gt², where u=0, S=?, g=8m/s², t=10seconds.

S=ut+½gt² ("ut" term will cancel because u=0).

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=>S = ½×8×10²

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Therefore, the distance traveled by the body in 10s is 400m.

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6 0

2 years ago

If a point has 40 J of energy and the electric potential is 8 V, what must be the charge?

Alekssandra [29.7K]

If a point has 40 J of energy and the electric potential is 8 V, the charge must be: A. 5 C

Given the following the details;

Energy = 40 Joules

Electric potential = 8 Volts

To find the quantity of charge;

Mathematically, the quantity of charge with respect to electric potential is given by the formula;

$Quantity \; of \; charge = \frac{Energy}{Electric \; potential}$

Substituting the values into the formula, we have;

$Quantity \; of \; charge = \frac{40}{8}$

Quantity of charge = 5 Coulombs

Therefore, the quantity of charge must be 5 Coulombs.

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3 years ago

Read 2 more answers

A heavy anvil is suspended by a 0.75 m long steel wire that has a mass of 12 g. When the wire is plucked, it hums at its fundame

Dima020 [189]

Explanation:

It is given that,

length of steel wire, l = 0.75 m

Mass of the wire, m = 12 g = 0.012 kg

Fundamental frequency, f = 120 Hz

We need to find the mass of the anvil (m'). The fundamental frequency is given by :

$f=\dfrac{v}{2l}$

v is the speed of the mass

Speed is given by :

$v=\sqrt{\dfrac{T}{\mu}}$

$\mu$ is the mass per unit length, $\mu=\dfrac{m}{l}$

$f=\dfrac{1}{2l}\sqrt{\dfrac{T}{\mu}}$

T is the tension in the wire,

$f=\dfrac{1}{2l}\sqrt{\dfrac{Tl}{m}}$

$T=4f^2lm$

$T=4(120)^2\times 0.75\times 0.012$

T = 518.4 N

Tension in the wire, T = m' g

$m'=\dfrac{T}{g}$

$m'=\dfrac{518.4}{9.8}$

m' = 52.89 kg

So, the mass of the anvil is 52.89 kg. Hence, this is the required solution.

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3 years ago

A test piolot flies with an acceleration of 5

Colt1911 [192]

On Earth, 1 g = 9.8 m/s² .

5 g = 5 · (9.8 m/s²)

5 g = 49 m/s²

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3 years ago

what is the electrical potential at the surface of gold nucleus? The radius of a gold atom is 6.6*10​

what is the electrical potential at the surface of gold nucleus? The radius of a gold atom is 6.6*10