Ask question

Ask question

All categories

2 years ago

15

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?

1 answer:

Svet_ta [14]2 years ago

6 0

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

You might be interested in

A soccer ball of diameter 22.6cm and mass 426g rolls up a hill without slipping, reaching a maximum height of 5m above the base

-Dominant- [34]

Answer with Explanation:

We are given that

Diameter=d=22.6 cm

Mass,m=426 g= $426\times 10^{-3} kg$

1 kg=1000 g

Radius,r= $\frac{d}{2}=\frac{22.6}{2}=11.3 cm=11.3\times 10^{-2} m$

1m=100 cm

Height,h=5m

$I=\frac{2}{2}mr^2$

a.By law of conservation of energy

$\frac{1}{2}I\omega^2+\frac{1}{2}mv^2=mgh$

$\frac{1}{2}\times \frac{2}{3}mr^2\omega^2+\frac{1}{2}mr^2\omega^2=mgh$

$v=\omega r$

$gh=\frac{1}{3}r^2+\frac{1}{2}r^2=\frac{5}{6}r^2\omega^2$

$\omega^2=\frac{6}{5r^2}gh$

$\omega=\sqrt{\frac{6gh}{5r^2}}=\sqrt{\frac{6\times 9.8\times 5}{5(11.3\times 10^{-2})^2}}=67.86 rad/s$

Where $g=9.8m/s^2$

b.Rotational kinetic energy= $\frac{1}{2}I\omega^2=\frac{1}{2}\times \frac{2}{3}mr^2\omega^2=\frac{1}{2}\times \frac{2}{3}(426\times 10^{-3})(11.3\times 10^{-2})^2(67.86)^2=8.35 J$

Rotational kinetic energy=8.35 J

6 0

3 years ago

A boat takes off from a dock at 2.5 m/s and speeds up at 4.2 m/s squared for six seconds how far has the most traveled

GaryK [48]

The boat's position $x$ relative to its starting point $x_0=0$ is determined by

$x=x_0+v_0t+\dfrac12at^2$

where $v_0$ is its initial velocity, $a$ is its acceleration, and $t$ is time. After $t=6\,\mathrm s$ , the boat has traveled

$x=\left(2.5\,\dfrac{\mathrm m}{\mathrm s}\right)(6\,\mathrm s)+\dfrac12\left(4.2\,\dfrac{\mathrm m}{\mathrm s^2}\right)(6\,\mathrm s)^2$

$\implies x=91\,\mathrm m$

3 0

2 years ago

How many lines per mm are there in the diffraction grating if the second order principal maximum for a light of wavelength 536 n

grandymaker [24]

To solve this problem it is necessary to apply the concepts related to the principle of superposition and the equations of destructive and constructive interference.

Constructive interference can be defined as

$dSin\theta = m\lambda$

Where

m= Any integer which represent the number of repetition of spectrum

$\lambda$ = Wavelength

d = Distance between the slits.

$\theta$ = Angle between the difraccion paterns and the source of light

Re-arrange to find the distance between the slits we have,

$d = \frac{m\lambda}{sin\theta }$

$d = \frac{2*536*10^{-9}}{sin(24)}$

$d = 2.63*10^{-6}m$

Therefore the number of lines per millimeter would be given as

$\frac{1}{d} = \frac{1}{2.63*10^{-6} }$

$\frac{1}{d} = 379418.5\frac{lines}{m}(\frac{10^{-3}m}{1 mm})$

$\frac{1}{d} = 379.4 lines/mm$

Therefore the number of the lines from the grating to the center of the diffraction pattern are 380lines per mm

6 0

2 years ago

How does the government involve in a child or young person’s upbringing in your<br> country?

dolphi86 [110]

The government are involved in a child or young person’s upbringing in my country by providing the right amenities and ensure the justice system is fair.

<h3>What is Upbringing?</h3>

This is defined as the way an individual treated at a young age by older adults such as parents.

Providing amenities will ensure the upbringing of the child is easier and a fair justice system will help to control crime.

Read more about Upbringing here brainly.com/question/12397063

#SPJ1

8 0

1 year ago

A depiction of a famous scientific experiment is given. Consider how the beam changes when the magnet is off compared to when th

iogann1982 [59]

Answer:

The beam used is a negatively charged electron beam with a velocity of

v = E / B

Explanation:

After reading this long statement we can extract the data to work on the problem.

* They indicate that when the beam passes through the plates it deviates towards the positive plate, so the beam must be negative electrons.

* Now indicates that the electric field and the magnetic field are contracted and that the beam passes without deviating, so the electric and magnetic forces must be balanced

$F_{e} = F_{m}$

q E = qv B

v = E / B

this configuration is called speed selector

They ask us what type of beam was used.

The beam used is a negatively charged electron beam with a velocity of v = E / B

6 0

3 years ago