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Elodia [21]
4 years ago
5

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
Physics
1 answer:
nevsk [136]4 years ago
4 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

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Svetradugi [14.3K]

Answer:

No. 67

Peter Street

12th Road

Chennai

24th June 201_

Dear Amrish

I have come to know that since your school has closed for the Autumn Break you have plenty of free time at your disposal at the moment. I would like to tell you that even I am having holidays now.

It has been a long time since we have spent some time together. If you are free, I would welcome to have your company this weekend. Why don’t you come over to my house and spend a day or so with me?

I am anxiously waiting for your reply.

Yours affectionately

your name

4 0
3 years ago
Read 2 more answers
What is the density of the block if it has a mass of 45.5 grams and a volume of 2.4 cm³?
rjkz [21]

Explanation:

density= mass/volume

therefore;45.5/2.4

=19.0 g/cm³

5 0
3 years ago
CISTU U
marusya05 [52]

solution:

radius of steel ball(r)=5cm=0.05m

density of ball =8000kgm

terminal velocity(v)=25m/s^2

density of air( d) =1.29 kgm

now

volume of ball(V)=4/3pir^3=1.33×3.14×0.05^3=0.00052 m^3

density of ball= mass of ball/Volume of ball

or, 8000=m/0.00052

or, m=4.16 kg

weight of the ball (W)= mg=4.16×10=41.6 N

viscous force(F)=6 × pi × eta × r × v

=6×3.14×eta×0.05×25

=23.55×eta

To attain the terminal velocity,

Fiscous force=Weight

or, 23.55× eta = 41.6

or, eta = 1.76

whete eta is the coefficient of viscosity.

5 0
3 years ago
If a satellite weighs 321 lb. on the earth's surface (R = 4,000 miles), how much does it weigh 12,000 miles above the surface? (
Sever21 [200]

The gravitational force between the Earth and the satellite (its "weight") is inversely proportional to the distance between the centers of both objects.

On the surface, their centers are separated by 1 Earth radius.

12,000 miles above the surface, they're separated by 4 Earth radiii.

(4/1) = 4

So after the move, the satellite's weight is (1/4²) = 1/16 of its surface weight.

(321 lb) / (16) = (20 and a hair) lb

The correct choice from the given list is " <em>>20 lb "</em> .  

3 0
3 years ago
You pull on a spring whose spring constant is 22 N/m, and stretch it from its equilibrium length of 0.3 m to a length of 0.7 m.
Liono4ka [1.6K]

Answer:

W= 4.4 J

Explanation

Elastic potential energy theory

If we have a spring of constant K to which a force F that produces a Δx deformation is applied, we apply Hooke's law:

F=K*x  Formula (1): The force F applied to the spring is proportional to the deformation x of the spring.

As the force is variable to calculate the work we define an average force

F_{a} =\frac{F_{f}+F_{i}  }{2}  Formula (2)

Ff: final force

Fi: initial force

The work done on the spring is :

W = Fa*Δx

Fa : average force

Δx :  displacement

W = F_{a} (x_{f} -x_{i} )   :Formula (3)

x_{f} :  final deformation

x_{i}  :initial deformation

Problem development

We calculate Ff and Fi , applying formula (1) :

F_{f} = K*x_{f} =22\frac{N}{m} *0.7m =15.4N

F_{i} = K*x_{i} =22\frac{N}{m} *0.3m =6.6N

We calculate average force applying formula (2):

F_{a} =\frac{15.4N+6.2N}{2} = 11 N

We calculate the work done on the spring  applying formula (3) :         :

W= 11N*(0.7m-0.3m) = 11N*0.4m=4.4 N*m = 4.4 Joule = 4.4 J

Work done in stages

Work is the change of elastic potential energy (ΔEp)

W=ΔEp

ΔEp= Epf-Epi

Epf= final potential energy

Epi=initial potential energy

E_{pf} =\frac{1}{2} *k*x_{f}^{2}

E_{pi} =\frac{1}{2} *k*x_{i}^{2}

E_{pf} =\frac{1}{2} *22*0.7^{2} = 5.39 J

E_{pf} =\frac{1}{2} *22*0.3^{2} = 0.99 J

W=ΔEp=  5.39 J-0.99 J = 4.4J

:

4 0
3 years ago
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