A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end. The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its center and becomes embedded in it. The mass of the bullet is one-fourth the mass of the rod. What is thefinal angular speed of the rod?

Answers

Answer 1

Answer:

The value of final angular speed of the uniform rod which rests on the frictionless horizontal surface is,

$\omega=(6v)/(19L)$

What is angular speed of a body?

The angular speed of a body is the rate by which the body changed its angle with respect to the time. It can be given as,

$\omega= (\Delta \theta)/(\Delta t)$

A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end.

The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its center and becomes embedded in it.

The mass of the bullet is one-fourth the mass of the rod. The diagram for the above condition is attached below.

In the attached image the angular momentum about the point A is constant just before and after the collision. Thus,

$L_i=L_f\nL_i=(m)/(4)*(vL)/(2)=I\omega\n(m)/(4)*(vL)/(2)=I\omega$

Put the value of inertia as,

$(m)/(4)*(vL)/(2)=\left((mL^2)/(3)+(mL^2)/(4*4)\right)\n(mvL)/(8)*(vL)/(2)=\left((19mL^2)/(48)\right)$

Solving it further we get,

$(v)/(8)=(19)/(48)L\omega\n\omega=(6v)/(19L)$

Hence, the value of final angular speed of the uniform rodwhich rests on the frictionless horizontal surface is,

$\omega=(6v)/(19L)$

Learn more about the angular speed here;

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Answer 2

Answer:

Answer: a) 0.315 (V/L)

Explanation:

From Conservation of angular momentum, we know that

L1 = L2 ,

Therefore MV L/2 = ( Irod + Ib) x W

M/4 x V x L/2 = (M (L/2)^2 + 1/3xMxL^2) x W

M/8 X VL = (ML^2/16 + ML^2 /3 )

After elimination we have,

V/8 = 19/48 x L x W

W = 48/8 x V/19L = 6/19 x V/L

Therefore W = (0.136)X V/L

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Ted Clubber Lang. A hook in boxing primarily involves horizontal flexion of the shoulder while maintaining a constant angle at the elbow. During this punch, the horizontal flexor muscles of the shoulder contract and shorten at an average speed of 75 cm/s. They move through an arc length of 5 cm during the hook, while the first moves through an arc length of 100 cm. What is the average speed of the first during the hook?

Answers

Answer:

15 m/s or 1500 cm/s

Explanation:

Given that

Speed of the shoulder, v(h) = 75 cm/s = 0.75 m/s

Distance moved during the hook, d(h) = 5 cm = 0.05 m

Distance moved by the fist, d(f) = 100 cm = 1 m

Average speed of the fist during the hook, v(f) = ? cm/s = m/s

This can be solved by a very simple relation.

d(f) / d(h) = v(f) / v(h)

v(f) = [d(f) * v(h)] / d(h)

v(f) = (1 * 0.75) / 0.05

v(f) = 0.75 / 0.05

v(f) = 15 m/s

Therefore, the average speed of the fist during the hook is 15 m/s or 1500 cm/s

What is the magnitude of the line charge density on the power line? Express your answer using two significant figures.

Answers

Answer:

λ = -47 nC / m

Explanation:

The missing question is as follows:

" The potential difference between the surface of a 2.2 cm -diameter power line and a point 1.9 m distant is 3.8 kV. What is the magnitude of the line charge density on the power line? Express your answer using two significant figures. "

Given:

- The Diameter of the power line D = 2.2 cm

- The distance between two ends of power line L = 1.9m

- The potential difference across two ends V = 3.8 KV

Find:

What is the magnitude of the line charge density on the power line?

Solution:

- The derivation of the line of charges for a length L oriented along any axis centered at origin and the potential difference between two ends is as follows:

V = 2*k*λ*Ln( D / L )

Where,

k : Coulomb's Constant = 8.99*10^9

λ : The line charge density

- Re-arrange and solve for λ:

λ = V / 2*k*Ln( D / L )

Plug in the values:

λ = 3800 / 2*8.99*10^9*Ln( 2.2 / 190 )

λ = -4.74022*10^-8 C / m

λ = -47 nC / m

Final answer:

Line charge density is the total charge distributed along the length of a wire, expressed in coulombs per meter. To calculate it, divide the total charge by the total length of the wire. Without specific numbers for charge and length, a numerical value can't be given.

Explanation:

To calculate the magnitude of the line charge density of a power line, you need to know the total charge (Q) distributed along the total length (L) of the wire. The line charge density (λ) is then defined as λ = Q/L. Unfortunately, without any specific numbers provided for these parameters, I can't provide a numerical answer.

Line charge density is a significant concept in electromagnetism and is measured in coulombs per meter (C/m).

Remember that the charge can be uniform or non-uniform along the length of the line.

For example, if a power line has a total charge of 0.02 C spread along its length of 50 m, it would have a line charge density of λ = Q/L = 0.02 C / 50 m = 0.0004 C/m

Learn more about Line Charge Density here:

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Why do electrical devices have resistance

Answers

As electrons move through the conductor, some collide with atoms, other electrons, or impurities in the metal.

What is the wavelength of the photons emitted by hydrogen atoms when they undergo n =5 to n =3 transitions?

Answers

Answer:

$\lambda=1282nm$

Explanation:

The wavelength of the photons emitted due to an atomic electron transition in a hydrogen atom, is given by the Rydberg formula:

$(1)/(\lambda)=R_H((1)/(n_1^2)-(1)/(n_2^2)})$

Here $R_H$ is the Rydberg constant for hydrogen and $n_1,n_2$ are the lower and higher quantum number for the energy levels of the atomic electron transition, respectively. Replacing the given values and solving for $\lambda$

$(1)/(\lambda)=1.097*10^7m^(-1)((1)/(3^2)-(1)/(5^2)})\n(1)/(\lambda)=7.81*10^5m^(-1)\n\lambda=(1)/(7.81*10^5m^(-1))\n\lambda=1.282*10^(-6)m\n\lambda=1.282*10^(-6)m*(1nm)/(10^(-9)m)\n\lambda=1282nm$

Which statement best describes how the first quatrain relates to the second quatrain? The first shows the beloved’s actions; the second describes how she imitates them. Both the first and the second show the actions of the speaker and the beloved. The first shows the speaker’s actions; the second shows the beloved’s opposition to them. The first shows the speaker’s sadness; the second shows the beloved’s anger.

Answers

Answer: the first shows the speakers actions; the second shows the beloveds opposition to them

Explanation:

What happens to the pressure in all parts of a confined fluid if the pressure in one part is increased? The pressure in the other parts remains the same.The pressure everywhere increases by different amounts depending on the area of each part.
The pressure everywhere increases by the same amount.
The pressure everywhere decreases to conserve total pressure.

Answers

Answer:

option C

Explanation:

the correct answer is option C

When in a confined fluid the pressure is increased in one part than the pressure will equally distribute in the whole system.

According to Pascal's law when pressure is increased in the confined system then the pressure will equally transfer in the whole system.

This law's application is used in machines like hydraulic jacks.

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