The final velocity is 2.7 m/s
Explanation:
We can solve this problem by using the principle of conservation of momentum: in fact, in absence of external forces, the total momentum of the system must be conserved before and after the collision.
Therefore we can write:
where:
is the mass of the putty
is the initial velocity of the putty (we take its direction as positive direction)
is the mass of the ball
is the initial velocity of the ball (at rest)
is the final combined velocity of the two putty+ball
Re-arranging the equation and substituting the values, we find the final combined velocity:
And the positive sign indicates their final direction is the same as the initial direction of the putty.
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The magnetic field strength of a very long current-carrying wire is proportional to the inverse of the distance from the wire. The farther you go from the wire, the weaker the magnetic field becomes.
B ∝ 1/d
B = magnetic field strength, d = distance from wire
Calculate the scaling factor for d required to change B from 25μT to 2.8μT:
2.8μT/25μT = 1/k
k = 8.9
You must go to a distance of 8.9d to observe a magnetic field strength of 2.8μT
Answer:
The frequency is 302.05 Hz.
Explanation:
Given that,
Speed = 18.0 m/s
Suppose a train is traveling at 30.0 m/s relative to the ground in still air. The frequency of the note emitted by the train whistle is 262 Hz .
We need to calculate the frequency
Using formula of frequency

Where, f = frequency
v = speed of sound
= speed of passenger
= speed of source
Put the value into the formula


Hence, The frequency is 302.05 Hz.
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Answer:
- The energy that must be added to the electron to move it to the third excited state is -1.153 eV
- The energy that must be added to the electron to move it to the fourth excited state is -1.181 eV
Explanation:
Given;
Energy of electron in ground state (n = 1 ) = 1.23 eV
E₁ = 1.23 eV
Eₙ = E₁ /n²
where;
E₁ is the energy of the electron in ground state
n is the energy level,
For third excited state, n = 4
E₄ = E₁ /4²
E₄ = (1.23 eV) / 16
E₄ = 0.077 eV
Change in energy level, = E₄ - E₁ = 0.077 eV - 1.23 eV = -1.153 eV
The energy that must be added to the electron to move it to the third excited state is -1.153 eV
For fourth excited state, n = 5
E₅ = E₁ /5²
E₄ = (1.23 eV) / 25
E₄ = 0.049 eV
Change in energy level, = E₅ - E₁ = 0.049 eV - 1.23 eV = -1.181 eV
The energy that must be added to the electron to move it to the fourth excited state is -1.181 eV