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statuscvo [17]
1 year ago
11

A 2.0kg solid disk rolls without slipping on a horizontal surface so that its center proceeds to the right with a speed of 5.0 m

/s. What is the instantaneous speed of the point of the disk that makes contact with the surface?
Physics
1 answer:
Alexandra [31]1 year ago
4 0

Answer:

Instantaneous speed of contact point will be ZERO

Explanation:

As we know that disc is rolling without slipping on horizontal surface

So here the speed of center of the disc is given as

v = 5 m/s

now at the contact point the tangential speed will be in reverse direction

v_t = R\omega

now we know that net contact speed with respect to its lower surface must be zero

v_{net} = v - v_t = 0

so net velocity of contact point with respect to its lower surface must be ZERO here

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svetoff [14.1K]

Complete Question

The complete question is is shown on the first uploaded

Answer:

The elastic potential energy at point B is  PE_{elastic} = 50J

The kinetic energy at point D is KE = 75J

Explanation:

Looking at the given point we can observe that mechanically energy(i.e potential and kinetic energy ) is conserved and it value is E_ {m} = 100J

     So at point B

           E_{m} = PE_{elastic} +KE

           100 = PE_{elastic} + KE

   KE at point B is  50J

So     PE_{elastic} = 100 - 50 = 50J

     Now at point D

          E_{m} = PE_{elastic} +KE

    PE_{elastic} at point D is 25J

 So  KE = 100 - 25 = 75J

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1 year ago
A farsighted girl has a near point at 2.0 m but has forgotten her glasses at home. The girl borrows eyeglasses that have a power
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Find the pictures in the attachment

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1 year ago
Consider a bicycle wheel to be a ring of radius 30 cm and mass 1.5 kg. Neglect the mass of the axle and sprocket. If a force of
Charra [1.4K]

To solve the problem it is necessary to apply the Torque equations and their respective definitions.

The Torque is defined as,

\tau = I \alpha

Where,

I=Inertial Moment

\alpha = Angular acceleration

Also Torque with linear equation is defined as,

\tau = F*d

Where,

F = Force

d= distance

Our dates are given as,

R = 30 cm = 0.3m

m = 1.5 kg

F = 20 N

r = 4.0 cm = 0.04 m

t = 4.0s

Therefore matching two equation we have that,

d*F = I\alpha

For a wheel the moment inertia is defined as,

I= mR2, replacing we have

d*F= \frac{mR^2a}{R}

d*F= mRa

a = \frac{rF}{ mR}

a = \frac{0.04*20}{1.5*0.3}

a=1.77 m/s^2

Then the velocity of the wheel is

V = a *t \\V=1.77*4 \\V=7.11 m/s

Therefore the correct answer is D.

4 0
1 year ago
The newly formed xenon nucleus is left in an excited state. Thus, when it decays to a state of lower energy a gamma ray is emitt
nevsk [136]

Answer:3.87*10^-4

Explanation:

What is the decrease in mass, delta mass Xe , of the xenon nucleus as a result of this deca

We have been given the wavelength of the gamma ray, find the frequency using c = freq*wavelength.

C=f*lambda

3*10^8=f*3.44*10^-12

F=0.87*10^20 hz

Then with the frequency, find the energy emitted using equation

E=hf E = freq*Plank's constant

E=.87*10^20*6.62*10^-34

E=575.94*10^(-16)

With this energy, convert into MeV from joules.

With the energy in MeV, use E=mc^2 using c^2 = 931.5 MeV/u.

Plugging and computing all necessary numbers gives you

3.87*10^-4 u.

6 0
1 year ago
Magnetic fields within a sunspot can be as strong as 0.4T. (By comparison, the earth's magnetic field is about 1/10,000 as stron
WARRIOR [948]

Answer:

The speed of ejection is 2.06\times 10^{4}\ m/s

Solution:

As per the question:

Magnetic field density, B = 0.4 T

Density of the material in the sunspot, \rho = 3\times 10^{4}\ kg/m^{3}

Now,

To calculate the speed of ejection of the material, v:

The magnetic field energy density is given by:

U_{B} = \frac{B^{2}}{2\mu_{o}}

This energy density equals the kinetic energy supplied by the field.

Thus

KE = U_{B}

\frac{1}{2}mv^{2} = \frac{B^{2}}{2\mu_{o}}

where

m = mass of the sunspot in 1\ m^{3} = 3\times 10^{- 4}\ kg/m^{3}

v = \frac{B}{\sqrt{\mu_{o}m}}

v = \frac{0.4}{\sqrt{4\pi \times 10^{- 7}\times 3\times 10^{- 4}}} = 2.06\times 10^{4}\ m/s

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