1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
olganol [36]
3 years ago
12

A string is wrapped around a disk of mass 1.7 kg and radius 0.11 m. Starting from rest, you pull the string with a constant forc

e 7 N along a nearly frictionless surface. At the instant when the center of the disk has moved a distance 0.15 m, your hand has moved a distance of 0.22 m.
a. At this instant, what is the speed of the center of mass of the disk?
b. At this instant, how much rotational kinetic energy does the disk have relative to its center of mass?
Physics
2 answers:
V125BC [204]3 years ago
5 0

Answer:

Explanation:

a) work done in moving the disk by 0.15 m = its kinetic energy = 0.5 mv²

Fd = 0.5 mv²

2 fd = mv²

√( \frac{2fd}{m}) = v

v = 1.11 m/s

b) rotational kinetic energy = F ( 0.22 m - 0.15 m) = 0.49 J

Evgen [1.6K]3 years ago
5 0

Answer:

(a) Vcm = 1.43m/s

(b) KEcm = 2.59J

Explanation:

Given

m = mass of the solid disk = 1.7kg

Radius R = 0.11m

F = 7N

Let the distance moved by the the center of mass be x = 0.15m

And the distance moved by in unwinding the rope be d = 0.22m

The total workdone on the disk is causing it to rotate and also move its center of mass is F×(d+x)

W = 7×(0.15+0.22)

= 7× 0.37 = 2.59J

By work-energy theorem,

W = ΔKE = 1/2mVcm² + 1/2×I×ω²....(1)

I = moment of inertia = 1/2MR² and

ω = angular speed = Vcm/R

So substituting these expressions into the equation above we have

W = 1/2MVcm² + 1/2×1/2MR²×(Vcm/R)²

W = 1/2MVcm² + 1/4MVcm²

W = 3/4MVcm²

Vcm² = 4/3×W/M

Vcm = √(4/3×W/M)

Vcm = √(4/3×2.59/1.7)

Vcm = 1.43m/s

KE = workdone = 2.59J.

You might be interested in
A double-slit experiment uses light of wavelength 650 nm with a slit separation of 0.100 mm and a screen placed 4.0 m away. a) W
dezoksy [38]

Answer:

Explanation:

a ) Slit separation d = .1 x 10⁻³ m

Screen distance D = 4 m

wave length of light  λ = 650 x 10⁻⁹ m

Width of central fringe = λ D / d

= \frac{650\times10^{-9}\times4}{.1\times10^{-3}}

= 26 mm

b ) Distance between 1 st and 2 nd bright fringe will be equal to width of dark fringe which will also be equal to 26 mm

c ) Angular separation between the central maximum and 1 st order maximum will be equal to angular width of fringe which is equal to

λ  / d

= \frac{650\times10^{-9}}{.1\times10^{-3}}

= 6.5 x 10⁻³ radian.

8 0
3 years ago
What are the two requirements of a circuit so that current will flow?
antoniya [11.8K]

Answer:

-The battery-the power source

-Closed conducting loop

Explanation:

-To produce an electric current, the following requirements must be met:

-A battery-This is the energy source than will do work on the charge thus moving from a low energy location to high energy location.

-Closed Conducting Loop-The loop is usually made of copper wires due to their high electric conductivity.

8 0
3 years ago
Find the density of seawater at a depth where the pressure is 500 atm if the density at the surface is 1100 kg/m^3 . Seawater ha
mixer [17]

The density of seawater at a depth where the pressure is 500 atm is 1124kg/m^3

Explanation:

The relationship between bulk modulus and pressure is the following:

B=\rho_0 \frac{\Delta p}{\Delta \rho}

where

B is the bulk modulus

\rho_0 is the density at surface

\Delta p is the variation of pressure

\Delta \rho is the variation of density

In this problem, we have:

B=2.3\cdot 10^9 N/m^2 is the bulk modulus

\rho_0 =1100 kg/m^3

\Delta p = p-p_0 = 500 atm - 1 atm = 499 atm = 5.05\cdot 10^7 Pa is the change in pressure with respect to the surface (the pressure at the surface is 1 atm)

Therefore, we can find the density of the water where the pressure is 500 atm as follows:

\rho = \rho_0 + \Delta \rho = \rho_0+\frac{\rho_0 \Delta p}{B}=\rho_0 (1+\frac{\Delta p}{B})=(1100)(1+\frac{5.05\cdot 10^7}{2.3\cdot 10^9})=1124kg/m^3

Learn more about pressure in a fluid:

brainly.com/question/9805263

#LearnwithBrainly

7 0
3 years ago
What is the meaning of smooth?
damaskus [11]

A surface in which is flat or very soft to the touch and reduces splinters or anything sticking out, having an surface which does not have lumps, or indentations.

5 0
3 years ago
A .5 kg air puck moves to the right at 3 m/s, colliding with a 1.5kg air puck that is moving to the left at 1.5 m/s.
arlik [135]

Answer:

part (a) v = 1.7 m/s towards right direction

part (b) Not an elastic collision

part (c) F = -228.6 N towards left.

Explanation:

Given,

  • Mass of the first puck = m_1\ =\ 5\ kg
  • Mass of the second puck = m_2\ =\ 3\ kg
  • initial velocity of the first puck = u_1\ =\ 3\ m/s.
  • Initial velocity of the second puck = u_2\ =\ -1.5\ m/s.

Part (a)

Pucks are stick together after the collision, therefore the final velocities of the pucks are same as v.

From the conservation of linear momentum,

m_1u_1\ +\ m_2u_2\ =\ (m_1\ +\ m_2)v\\\Rightarrow v\ =\ \dfrac{m_1u_1\ +\ m_2u_2}{m_1\ +\ m_2}\\\Rightarrow v\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5}{5\ +\ 1.5}\\\Rightarrow v\ =\ 1.7\ m/s.

Direction of the velocity is towards right due to positive velocity.

part (b)

Given,

Final velocity of the second puck = v_2\ =\ 2.31\ m/s.

Let v_1 be the final velocity of first puck after the collision.

From the conservation of linear momentum,

m_1u_1\ +\ m_2u_2\ +\ m_1v_1\ +\ m_2v_2\\\Rightarrow v_1\ =\ \dfrac{m_1u_1\ +\ m_2u_2\ -\ m_2v_2}{m_1}\\\Rightarrow v_1\ =\ \dfrac{5\times 3\ -\ 1.5\times 1.5\ -\ 1.5\times 2.31}{5}\\\Rightarrow v_1\ =\ 1.857\ m/s.

For elastic collision, the coefficient of restitution should be 1.

From the equation of the restitution,

v_1\ -\ v_2\ =\ e(u_2\ -\ u_1)\\\Rightarrow e\ =\ \dfrac{v_1\ -\ v_2}{u_2\ -\ u_1}\\\Rightarrow e\ =\ \dfrac{1.857\ -\ 2.31}{-1.5\ -\ 3}\\\Rightarrow e\ =\ 0.1\\

Therefore the collision is not elastic collision.

part (c)

Given,

Time of impact = t = 25\times 10^{-3}\ sec

we know that the impulse on an object due to a force is equal to the change in momentum of the object due to the collision,

\therefore I\ =\ \ m_1v_1\ -\ m_1u_1\\\Rightarrow F\times t\ =\ m_1(v_1\ -\ u_1)\\\Rightarrow F\ =\ \dfrac{m_1(v_1\ -\ u_1)}{t}\\\Rightarrow F\ =\ \dfrac{5\times (1.857\ -\ 3)}{25\times 10^{-3}}\\\Rightarrow F\ =\ -228.6\ N

Negative sign indicates that the force is towards in the left side of the movement of the first puck.

3 0
3 years ago
Other questions:
  • The equation below is for potassium oxide.
    10·2 answers
  • What are 5 metals used as either pure substances or alloys
    12·1 answer
  • WILL GIVE BRAINLY!!!
    5·1 answer
  • How can you inhabit saturn
    10·1 answer
  • Pls help me pls I’ll be glad
    11·1 answer
  • Why us an element considered a pure substance
    11·2 answers
  • plz tell fast oktwo paragraphs describing the advantages and disadvantages of living in the town.plz tell​
    14·1 answer
  • An offshore oil platform houses a drill for drilling oil. The drill descends from the platform and enters the ocean. At the bott
    5·2 answers
  • A car accelerates from rest at 2 m/s. what is the speed after 8 sec?
    14·1 answer
  • If frequency increases, then wavelength _________.
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!