Ask question

Ask question

All categories

2 years ago

14

A uniform disk has a mass of 3.7 kg and a radius of 0.40 m. The disk is mounted on frictionless bearings and is used as a turnta

ble. The turntable is initially rotating at 30 rpm. A thin-walled hollow cylinder has the same mass and radius as the disk. It is released from rest, just above the turntable, and on the same vertical axis. The hollow cylinder slips on the turntable for 0.20 s until it acquires the same final angular velocity as the turntable. What is the final angular momentum of the system

1 answer:

yuradex [85]2 years ago

5 0

Answer:

1.25 kgm²/sec

Explanation:

Disk inertia, Jd =

Jd = 1/2 * 3.7 * 0.40² = 0.2960 kgm²

Disk angular speed =

ωd = 0.1047 * 30 = 3.1416 rad/sec

Hollow cylinder inertia =

Jc = 3.7 * 0.40² = 0.592 kgm²

Initial Kinetic Energy of the disk

Ekd = 1/2 * Jd * ωd²

Ekd = 0.148 * 9.87

Ekd = 1.4607 joule

Ekd = (Jc + 1/2*Jd) * ω²

Final angular speed =

ω² = Ekd/(Jc+1/2*Jd)

ω² = 1.4607/(0.592+0.148)

ω² = 1.4607/0.74

ω² = 1.974

ω = √1.974

ω = 1.405 rad/sec

Final angular momentum =

L = (Jd+Jc) * ω

L = 0.888 * 1.405

L = 1.25 kgm²/sec

You might be interested in

Joe wanted to experiment with different factors that affect the freezing rate of water. He put two cups of water into each of tw

creativ13 [48]

the answer is not D ....... the answer is {B} if you got it right give me a 5 stars and a hard

7 0

2 years ago

Read 2 more answers

A wire with a length of 150 m and a radius of 0.15 mm carries a current with a uniform current density of 2.8 x 10^7A/m^2. The c

Mrac [35]

Answer:

The current is 2.0 A.

(A) is correct option.

Explanation:

Given that,

Length = 150 m

Radius = 0.15 mm

Current density $J=2.8\times10^{7}\ A/m^2$

We need to calculate the current

Using formula of current density

$J = \dfrac{I}{A}$

$I=J\timesA$

Where, J = current density

A = area

I = current

Put the value into the formula

$I=2.8\times10^{7}\times\pi\times(0.15\times10^{-3})^2$

$I=1.97=2.0\ A$

Hence, The current is 2.0 A.

7 0

2 years ago

Assume the radius of an atom, which can be represented as a hard sphere, is r = 1.95 Å. The atom is placed in a (a) simple cubic

Nuetrik [128]

Answer:

(a) A = $3.90 \AA$

(b) $A = 4.50 \AA$

(c) $A = 5.51 \AA$

(d) $A = 9.02 \AA$

Solution:

As per the question:

Radius of atom, r = 1.95 $\AA = 1.95\times 10^{- 10} m$

Now,

(a) For a simple cubic lattice, lattice constant A:

A = 2r

A = $2\times 1.95 = 3.90 \AA$

(b) For body centered cubic lattice:

$A = \frac{4}{\sqrt{3}}r$

$A = \frac{4}{\sqrt{3}}\times 1.95 = 4.50 \AA$

(c) For face centered cubic lattice:

$A = 2{\sqrt{2}}r$

$A = 2{\sqrt{2}}\times 1.95 = 5.51 \AA$

(d) For diamond lattice:

$A = 2\times \frac{4}{\sqrt{3}}r$

$A = 2\times \frac{4}{\sqrt{3}}\times 1.95 = 9.02 \AA$

6 0

2 years ago

Medical cyclotrons need efficient sources of protons to inject into their center. In one kind of ion source, hydrogen atoms (i.e

Arlecino [84]

Answer:

The radius is $r = 4.434 *10^{-5} \ m$

Explanation:

From the question we are told that

The magnetic field is $B = 90 mT = 90*10^{-3} \ T$

The electron kinetic energy is $KE = 1.4 eV = 1.4 * (1.60*10^{-19}) =2.24*10^{-19} \ J$

Generally for the collision to occur the centripetal force of the electron in it orbit is equal to the magnetic force applied

This is mathematically represented as

$\frac{mv^2}{r} = qvB$

=> $r = \frac{m* v}{q * B}$

Where m is the mass of electron with values $m = 9.1 *10^{-31} \ kg$

v is the escape velocity which is mathematically represented as

$v = \sqrt{\frac{2 * KE}{m} }$

So

$r = \frac{m}{qB} * \sqrt{\frac{2 * KE}{m} }$

apply indices

$r = \frac{\sqrt{2 * KE * m} }{qB}$

substituting values

$r = \frac{\sqrt{2 * 2.24*10^{-19}* 9.1 *10^{-31}} }{ 1.60 *10^{-19}* 90*10^{-3}}$

$r = 4.434 *10^{-5} \ m$

6 0

2 years ago

How much energy does a 50 kg rock have if it is sitting on the edge of a 15 m cliff?

noname [10]

Answer:

7350 J

Explanation:

The gravitational potential energy of the rock sitting on the edge of the cliff is given by:

$U=mgh$

where

m is the mass of the rock

g is the gravitational acceleration

h is the height of the cliff

In this problem, we have

m = 50 kg

g = 9.8 m/s^2

h = 15 m

Substituting numbers into the formula, we find:

$U=(50 kg)(9.8 m/s^2)(15 m)=7350 J$

3 0

2 years ago