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Answer:
25.71 kgm/s
Explanation:
Let K₁ and K₂ be the initial and final kinetic energies of object A and v₁ and v₂ its initial and final speeds.
Given that K₂ = 0.7K₁
1/2mv₂² = 0.7(1/2mv₁²)
v₂ = √0.7v₁ = √0.7 × 20 m/s = ±16.73 m/s
Since A rebounds, its velocity = -16.73 m/s and its momentum change, p₂ = mΔv = m(v₂ - v₁) = 0.7 kg (-16.73 - 20) m/s = 0.7( -36.73) = -25.71 kgm/s.
Th magnitude of object A's momentum change is thus 25.71 kgm/s
Answer and Explanation:
the electronic devices always have some noises present in the signal
there are some important considerations in optical fiber communications these are.
- the noise which is contributed by transmitter are electronic random noise, low frequency noise
- noise which is contributed by laser are relative intensity noise, mode partition noise, conversion of phase noise to amplitude noise.
- noise contributed by photo detector are quantum shot noise, shot noise from dark current, avalanche multiplication noise.
PRINCIPLE OF POPULATION INVERSION :
The principle of population inversion is defined as for production of high percentage of simulated emission for a laser beam the number of atoms in higher state should be greater than lower energy state
The correct answer is A. 32.5
Mechanical advantage is the ratio of force that is input into a machine to the force output.
Mechanical advantage of a wheel and axle is calculated by dividing the radius of the wheel by that of the axle.
MA=R/r where R is the radius of the wheel and r is the radius of the axle.
Substituting for the values in the question gives:
MA=26cm/0.8cm
=32.5
As the distance from a charged particle, "q", increases, the electric potential decreases.
<h3>
Electric potential between particles</h3>
The electric potential between particles is the work done in moving a unit charge from infinity to a certain point against the electrical resistance of the field.
V = Kq/r
where;
- K is Coulomb's constant
- q is the magnitude of the charge
- r is the distance between the charges
Thus, from the formula above, as the distance from a charged particle, "q", increases, the electric potential decreases.
Learn more about electric potential here: brainly.com/question/14306881
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