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Artyom0805 [142]

2 years ago

5

A small block on a frictionless, horizontal surface has a mass of 0.0250 kg. It is attached to a massless cord passing through a

hole in the surface (Fig. E10.42). The block is originally revolving at a distance of 0.300 m from the hole with an angular speed of 2.85 rad>s. The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.150 m. Model the block as a particle.
a. Is angular momentum conserved?
b. Find the change in kinetic energy of the block, in J.
c. How much work was done in pulling the cord? in J.

1 answer:

sweet [91]2 years ago

8 0

Answer:

W= $K_2-K_1==9.12\times10^{-3} J$

Explanation:

a) Yes, In the absence of external torques acting on the system, the angular momentum is conserved.

b) By the law of conservation of energy angular momentum

$L_1=L_2$

$I_1\omega_1=I_2\omega_2$

$mr_1^2\omega_1=mr_2^2\omega_2\\\omega_2=(\frac{r_1}{r_2} )^2\omega_1$

$\omega_2=(\frac{0.3}{0.15})^2\times2.85$

$\omega_2=5.7\text{ rad/sec}$

c) work done in pulling the chord W= Final kinetic energy(K_2)-Initial Kinetic energy(K_1)

$K_1=\frac{1}{2} mr_1^2\omega_1^2$

$K_1=\frac{1}{2} \times0.025\times0.3^2\times2.85^2$

$=9.12\times10^{-3} J$

Now,

$K_2=\frac{1}{2} mr_2^2\omega_2^2$

$K_2=\frac{1}{2} \times0.025\times0.15^2\times5.7^2$

$K_2=18.24\times10^{-3}$ J

Therefore, Work done W= $K_2-K_1==9.12\times10^{-3} J$

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yarga [219]

Answer

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Explanation

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$\frac{d \phi }{dt} = \int\limits^{} {E \cdot dl}$

Where $dl$ is change of a unit length

$\frac{d \phi}{dt} = A * \frac{dB}{dt}$

Where A is the area which is mathematically represented as

$A = \pi r^2$

So

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where L is the circumference of the circle which is mathematically represented as

$L = 2 \pi r$

So

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$[\frac{dB}{dt} ] = \frac{E}{ \frac{r}{2} }$

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2 years ago

Which one of the following represents an acceptable set of quantum numbers for an electron in an atom? (arranged as n, l, m l ,

Vitek1552 [10]

Answer:

The correct option that represents an acceptable set of quantum numbers for an electron in an atom is;

(b) 4, 3, -3, 1/2.

Explanation:

To solve the question, we note that the available options where the set of quantum numbers for an electron in an atom are arranged as n, l, m l , and ms are;

4, 4, 4, 1/2

4, 3, -3, 1/2

4, 3, 0, 0

4, 5, 7, -1/2

4, 4, -5, 1/2

Let us label them as a to as follows

(a) 4, 4, 4, 1/2

(b) 4, 3, -3, 1/2

(c) 4, 3, 0, 0

(d) 4, 5, 7, -1/2

(e) 4, 4, -5, 1/2

Next we note the rules for the assignment and arrangement of quantum numbers are as follows

Number Symbol Possible values

Principal Quantum Number .......n........................1, 2, 3, ......n

Angular momentum quantum

number...............................................l.........................0, 1, 2, .......(n - 1)

Magnetic Quantum Number........m₁......................-l, ..., -1, 0, 1,.....,l

Spin Quantum Number.................m $_s$ .....................+1/2, -1/2

We are meant to analyze each of the arrangement for acceptability.

Therefore for (a),

we note that the angular momentum quantum number, l =4 , is equal to the principal quantum number n =4 which violates the rule as the maximum value of the angular momentum quantum number is (n-1) where the maximum value of the principal quantum number is n.

Therefore (a) is not acceptable.

(b) Here we note that

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 3 ∈ (0, 1, 2, .......(n - 1)) → acceptable

The magnetic quantum number m₁ = -3 ∈ (-l, ..., -1, 0, 1,.....,l) → acceptable

The spin quantum number m $_s$ = 1/2 ∈ (+1/2, -1/2) → acceptable

Therefore (b) 4, 3, -3, 1/2 represents an acceptable set of quantum numbers for an electron in an atom.

(c) Here we have

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 3 ∈ (0, 1, 2, .......(n - 1)) → acceptable

The magnetic quantum number m₁ = 0 ∈ (-l, ..., -1, 0, 1,.....,l) → acceptable

The spin quantum number m $_s$ = 0 ∉ (+1/2, -1/2) → not acceptable

Therefore (c) 4, 3, 0, 0 does not represents an acceptable set of quantum numbers for an electron in an atom.

(d) Here we have;

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 5 ∉ (0, 1, 2, .......(n - 1)) → not acceptable

The magnetic quantum number m₁ = 7 ∉ (-l, ..., -1, 0, 1,.....,l) → acceptable

The spin quantum number m $_s$ = -1/2 ∈ (+1/2, -1/2) → acceptable

Therefore (d) 4, 5, 7, -1/2 does not represents an acceptable set of quantum numbers for an electron in an atom.

(e) Here we have;

The principal quantum number n = 4 ∈ (1, 2, 3, ......n) → acceptable

The angular momentum quantum number l = 4 ∉ (0, 1, 2, .......(n - 1)) → not acceptable

The magnetic quantum number m₁ = -5 ∉ (-l, ..., -1, 0, 1,.....,l) → acceptable

The spin quantum number m $_s$ = 1/2 ∈ (+1/2, -1/2) → acceptable

Therefore (e) 4, 4, -5, 1/2 does not represents an acceptable set of quantum numbers for an electron in an atom.

3 0

2 years ago