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An electron is emitted from a nucleus- State the effect this has on the charge of the nucleus
Answer:
an electron orbiting around the nucleus combines with a nuclear proton to produce a neutron, which remains in the nucleus, and a neutrino, which is emitted. As in positron emission, the nuclear positive charge and hence the atomic number decreases by one unit, and the mass number remains the same.
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Answer:
The kinetic energy of the more massive ball is greater by a factor of 2.
Explanation:
By conservation of energy, we know that the initial energy = final energy. At first, the balls are dropped from a height with no initial velocity so their initial energy is all potential energy. When they reach the bottom, all their energy is kinetic energy. So all of their energy is changed from potential to kinetic energy. This means that the ball with greater potential energy will have a greater kinetic energy.
Potential energy = mgh. Since g = gravity is a constant and h = height is the same, the only difference is mass. Since mass is directly proportional to potential energy, the greater the mass, the greater the potential energy, so the more massive ball has a greater initial potential energy and will have a greater kinetic energy at the bottom.
Additionally, let B1 = lighter ball with mass m and let B2 = heavier ball with mass m2. Since we know that intial potential energy = final kinetic energy. We can rewrite it as potential energy = kinetic energy = mass * gravity constant * height. For B1, it is mgh and for B2 it is 2mgh, so B2's kinetic energy is twice that of B1.
Answer:
The frequency heard by the motorist is 4313.2 Hz.
Explanation:
let f1 be the frequency emited by the police car and f2 be the frequency heard by the motorist, let v1 be the speed of the police car and v2 be the speed of the motorist and v = 343 m/s be the speed of sound.
because the police car is moving towards the motorist at a higher speed, then the motorist will hear a increasing frequency and according to Dopper effect, that frequency is given by:
f1 = [(v + v2/(v - v1))]×(f2)
= [( 343 + 30)/(343 - 36)]×(3550)
= 4313.2 Hz
Therefore, the frequency heard by the motorist is 4313.2 Hz.
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