Answer:
No
Explanation:
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Alpha emission is the process results in a change in mass number. Option B is correct.
<h3>What is mass number?</h3>
The total number of protons and neutrons in an atomic nucleus is known as the mass number, often known as the atomic mass number or nucleon number.
It's about the same as the atom's atomic mass, expressed in atomic mass units.
The alpha particle is a helium nucleus with two protons and two neutrons in an alpha decay or alpha emission. The number of protons and neutrons is reduced by two as a result of this action.
The quantity of protons and neutrons is affected by gamma emission descent. Also, while electron capture has no effect on the number of neutrons, it does raise the 1 also number of protons by one.
Alpha emission is the process results in a change in mass number.
Hence option B is correct.
To learn more about the mass number, refer:
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Answer:
The x-component of the electric field at the origin = -11.74 N/C.
The y-component of the electric field at the origin = 97.41 N/C.
Explanation:
<u>Given:</u>
- Charge on first charged particle,
- Charge on the second charged particle,
- Position of the first charge =
- Position of the second charge =
The electric field at a point due to a charge 
at a point 
distance away is given by
where,
= Coulomb's constant, having value 
= position vector of the point where the electric field is to be found with respect to the position of the charge .
= unit vector along .
The electric field at the origin due to first charge is given by
is the position vector of the origin with respect to the position of the first charge.
Assuming, 
are the units vectors along x and y axes respectively.
Using these values,
The electric field at the origin due to the second charge is given by
is the position vector of the origin with respect to the position of the second charge.
Using these values,
The net electric field at the origin due to both the charges is given by
Thus,
x-component of the electric field at the origin = -11.74 N/C.
y-component of the electric field at the origin = 97.41 N/C.
Answer:
51 Ω.
Explanation:
We'll begin by calculating the equivalent resistance of R₁ and R₃. This can be obtained as follow:
Resistor 1 (R₁) = 40 Ω
Resistor 3 (R₃) = 70.8 Ω
Equivalent Resistance of R₁ and R₃ (R₁ₙ₃) =?
Since the two resistors are in parallel connection, their equivalent can be obtained as follow:
R₁ₙ₃ = R₁ × R₃ / R₁ + R₃
R₁ₙ₃ = 40 × 70.8 / 40 + 70.8
R₁ₙ₃ = 2832 / 110.8
R₁ₙ₃ = 25.6 Ω
Finally, we shall determine the equivalent resistance of the group. This can be obtained as follow:
Equivalent Resistance of R₁ and R₃ (R₁ₙ₃) = 25.6 Ω
Resistor 2 (R₂) = 25.4 Ω
Equivalent Resistance (Rₑq) =?
Rₑq = R₁ₙ₃ + R₂ (series connection)
Rₑq = 25.6 + 25.4
Rₑq = 51 Ω
Therefore, the equivalent resistance of the group is 51 Ω.
