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balu736 [363]
3 years ago
5

A Carbon-10 nucleus has 6 protons and 4 neutrons. Through radioactive beta decay, it turns into a Boron-10 nucleus, with 5 proto

ns and 5 neutrons, plus another particle. What kind of additional particle, if any, is produced during this decay?A : Positively charged.B : No particle is produced.C : Negatively charged.D : Neutral.
Physics
2 answers:
Studentka2010 [4]3 years ago
6 0
<h2>Answer:</h2>

(a) positively charged

<h2>Explanation:</h2>

Let's take the scenario step by step

(i) Carbon-10 with 6 protons and 4 neutrons. i.e

¹⁰₆C

(ii) Undergoes a radioactive beta decay (β) decay to form Boron-10 with 5 protons and 5 neutrons ¹⁰₅B as follows;

¹⁰₆C  =>  ¹⁰₅B + β   -----------------------(i)

<em>Remember that;</em>

β decay can either be β- or β+

Where;

β -   =   ⁰₋₁ e⁻ (called electron and is negatively charged)

β +  =   ⁰₁ e⁺ (called positron and is positively charged)

Now, to balance the equation (i) above, β+ has to be used as follows;

¹⁰₆C =>  ¹⁰₅B + ⁰₁ e⁺

Therefore, the additional particle produced during this decay is ⁰₁ e⁺ (called positron and is positively charged)

svetoff [14.1K]3 years ago
5 0

Answer:

Positively charged particle

Explanation:

¹⁰₆C → ¹⁰₅B + e⁺ + ve

Since there is a decrease in the atomic number, there has been nuclear transmutation and a proton has be broken into a neutron and positron with the release of neutrino which has zero charge and very small mass. This is a beta plus decay with the release of Positively charged particle

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f a single circular loop of wire carries a current of 45 A and produces a magnetic field at its center with a magnitude of 1.50
Lelu [443]

Answer:

Radius of the loop is 0.18 m or 18 cm

Explanation:

Given :

Current flowing through the wire, I = 45 A

Magnetic field at the center of the wire, B = 1.50 x 10⁻⁴ T

Number of turns in circular wire, N = 1

Consider R be the radius of the circular wire.

The magnetic field at the center of the current carrying circular wire is determine by the relation:

B=\frac{N\mu_{0} I}{2R}

Here μ₀ is vacuum permeability constant and its value is 4π x 10⁻⁷ Tm/A.

Substitute the suitable values in the above equation.

1.50\times10^{-4} =\frac{4\pi \times10^{-7}\times45 }{2R}

R = 0.18 m

4 0
2 years ago
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katovenus [111]

Answer:

I think it’s a

Explanation:

8 0
3 years ago
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Three balls are kicked from the ground level at some angles above horizontal with different initial speeds. All three balls reac
Charra [1.4K]

Answer:

Time of flight  A is greatest

Explanation:

Let u₁ , u₂, u₃ be their initial velocity and θ₁ , θ₂ and θ₃ be their angle of projection. They all achieve a common highest height of H.

So

H = u₁² sin²θ₁ /2g

H = u₂² sin²θ₂ /2g

H = u₃² sin²θ₃ /2g

On the basis of these equation we can write

u₁ sinθ₁ =u₂ sinθ₂=u₃ sinθ₃

For maximum range we can write

D = u₁² sin2θ₁ /g

1.5 D = u₂² sin2θ₂ / g

2 D =u₃² sin2θ₃ / g

1.5 D / D = u₂² sin2θ₂ /u₁² sin2θ₁

1.5 = u₂ cosθ₂ /u₁ cosθ₁      ( since , u₁ sinθ₁ =u₂ sinθ₂ )

u₂ cosθ₂ >u₁ cosθ₁

u₂ sinθ₂ < u₁ sinθ₁

2u₂ sinθ₂ / g < 2u₁ sinθ₁ /g

Time of flight B < Time of flight  A

Similarly we can prove

Time of flight C < Time of flight B

Hence Time of flight  A is greatest .

8 0
3 years ago
Bird bones have air pockets in them to reduce their weight—this also gives them an average density significantly less thanthat o
snow_tiger [21]

Answer:

41.4 g

41.4 cm³

1.08695 g/cm³

Explanation:

\rho = Density of water = 1 g/cm³

Mass of water displaced will be the difference of the

m=45-3.6\\\Rightarrow m=41.4\ g

Mass of water displaced is 41.4 g

Density is given by

\rho=\dfrac{m}{v}\\\Rightarrow v=\dfrac{m}{\rho}\\\Rightarrow v=\dfrac{41.4}{1}\\\Rightarrow v=41.4\ cm^3

So, volume of bone is 41.4 cm³

Average density of the bird is given by

\rho=\dfrac{45}{41.4}\\\Rightarrow \rho=1.08695\ g/cm^3

The average density is 1.08695 g/cm³

3 0
3 years ago
wo lacrosse players collide in midair. Jeremy has a mass of 120 kg and is moving at a speed of 3 m/s. Hans has a mass of 140 kg
Julli [10]

2.71 m/s fast Hans is moving after the collision.

<u>Explanation</u>:

Given that,

Mass of Jeremy is 120 kg (M_J)

Speed of Jeremy is 3 m/s (V_J)

Speed of Jeremy after collision is (V_{JA}) -2.5 m/s

Mass of Hans is 140 kg (M_H)

Speed of Hans is -2 m/s (V_H)

Speed of Hans after collision is (V_{HA})

Linear momentum is defined as “mass time’s speed of the vehicle”. Linear momentum before the collision of Jeremy and Hans is  

= =\mathrm{M}_{1} \times \mathrm{V}_{\mathrm{J}}+\mathrm{M}_{\mathrm{H}} \times \mathrm{V}_{\mathrm{H}}

Substitute the given values,

= 120 × 3 + 140 × (-2)

= 360 + (-280)

= 80 kg m/s

Linear momentum after the collision of Jeremy and Hans is  

= =\mathrm{M}_{\mathrm{J}} \times \mathrm{V}_{\mathrm{JA}}+\mathrm{M}_{\mathrm{H}} \times \mathrm{V}_{\mathrm{HA}}

= 120 × (-2.5) + 140 × V_{HA}

= -300 + 140 × V_{HA}

We know that conservation of liner momentum,

Linear momentum before the collision = Linear momentum after the collision

80 = -300 + 140 × V_{HA}

80 + 300 = 140 × V_{HA}

380 = 140 × V_{HA}

380/140= V_{HA}

V_{HA} = 2.71 m/s

2.71 m/s fast Hans is moving after the collision.

4 0
2 years ago
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