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PSYCHO15rus [73]
3 years ago
9

Protons are found only in the atomic nucleus. TRUE FALSE

Physics
2 answers:
VashaNatasha [74]3 years ago
7 0

Answer:

true

Explanation:

Protons and neutrons are found in the nucleus

timama [110]3 years ago
4 0

Answer:True

Explanation:

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A scientist kept 1,000 grams of a radioactive material in a container. After 66 days, he observed that 125 grams of the radioact
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The sample appears to have gone through 3 half-lives
1st half life: 1000 to 500 g
2nd half life: 500 to 250 g
3rd half life: 250 to 125 g
The duration of a half-life, therefore, can be inferred to be 66 ÷ (3) = 22 days. 

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Suppose a straight 1.00-mm-diameter copper wire could just "float" horizontally in air because of the force due to the Earth’s m
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To solve this problem it is necessary to apply the concepts related to the Force since Newton's second law, as well as the concept of Electromagnetic Force. The relationship of the two equations will allow us to find the magnetic field through the geometric relations of density and volume.

F_{mag}= BIL

Where,

B = Magnetic Field

I = Current

L = Length

<em>Note: F_{mag}  is a direct adaptation of the vector relation F=q \times V \times B</em>

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F_g = mg

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m = Mass

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For there to be balance the two forces must be equal therefore

F_{mag} = F_g

BIL = mg

Our values are given as,

Diameter (d) = 1.0mm = 1*10^{-3}m

Radius (r) = \frac{d}{2} = \frac{1*10^{-3}}{2} = 0.5*10^{-3}m

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From the relationship of density another way of expressing mass would be

\rho = \frac{m}{V} \rightarrow m = \rho V

At the same time the volume ratio for a cylinder (the shape of the wire) would be

V = \pi r^2 L \rightarrow L =Length, r= Radius

Replacing this two expression at our first equation we have that:

BIL = mg

BIL = ( \rho V)g

BIL = ( \rho \pi r^2 L)g

Re-arrange to find I

I = \frac{( \rho \pi r^2 L)g}{BL}

I = \frac{( \rho \pi r^2 )g}{B}

We have for definition that the Density of copper is 8.9*10^3 Kg/m^3, gravity acceleration is 9.8m/s^2 and the values of magnetic field (B) and the radius were previously given, then:

I = \frac{( (8.9*10^3 ) \pi (0.5*10^{-3})^2 )(9.8)}{5.0*10^{-5}}

I = 1370.05A

The current is too high to be transported which would make the case not feasible.

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