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.
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Since the slowest instruction in the SCA executes in 12.5 ns, the maximum system clock frequency is 80 MHz
To answer the question, we need to know what frequency is.
<h3>What is frequency?</h3>
Frequency is the number of oscillations per second of a wave.
It is given by f = 1/T where T = period of wave
Now, given that the slowest instruction in the SCA executes in t = 12.5 ns, we need to calculate maximum system clock frequency, f.
<h3>What is the maximum system clock frequency?</h3>
So, f = 1/t
= 1/12.5 ns
= 1/(12.5 × 10⁻⁹ s)
= 1/12.5 × 10⁹ Hz
= 0.08 × 10⁹ Hz
= 80 × 10⁻³ × 10⁹ Hz
= 80 × 10⁶ Hz
= 80 MHz
So, the maximum system clock frequency is 80 MHz
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Answer:
1.The main advantage of an electromagnet over a permanent magnet is that the magnetic field can be quickly changed by controlling the amount of electric current in the winding. However, unlike a permanent magnet that needs no power, an electromagnet requires a continuous supply of current to maintain the magnetic field.
P=mv
0.25v=0.05*500
v=100 m/s
Pretty fast...
Answer:
a. the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N
b. this force compare to the weight of the muon; the force is 1.38 × 10⁸ greater than muon
Explanation:
F= ma
v²=u² -2aS
(1.56 ✕ 10⁶)²=(2.40 ✕ 10⁶)²-2a(1220)
a=1.36×10⁹m/s²
recall
F=ma
F = 1.88 ✕ 10⁻²⁸ kg × 1.36×10⁹m/s²
F= 2.55 × 10⁻¹⁹N
the magnitude of the force experienced by the muon is 2.55 × 10⁻¹⁹N
b. this force compare to the weight of the muon
F/mg= 2.55 × 10⁻¹⁹/ (1.88 ✕ 10⁻²⁸ × 9.8)
= 1.38 × 10⁸
