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damaskus [11]
2 years ago
5

g A CD is spinning on a CD player. You open the CD player to change out the disk and notice that the CD comes to rest after 15 r

evolutions with a constant deceleration of 120 r a d s 2 . What was the initial angular speed of the CD
Physics
1 answer:
Umnica [9.8K]2 years ago
7 0

Answer:

\omega_1=150rads/sec

Explanation:

From the question we are told that:

Number of Revolution N=15=30\pi

Deceleration d= -120 rads/2

Generally the equation for  initial angular speed \omega_1 is mathematically given by

 \omega_2^2=\omega_1^2 +2(d)(N)

 0=\omega_1^2 +2(-120)(20 \pi)

 \omega_1^2=7200 \pi

 \omega_1=150rads/sec

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Samanthawalks along a horizontal path in the direction shown the curved path is a semi circle with a radius of 2 m while the hor
Anna11 [10]

Answer:

Explanation:

Samantha walks along a horizontal path in the direction shown the curved path is a semi circle with a radius of 2 m while the horizontal part is for me what is the magnitude of displacement

Displacement is given by the straight line distance between P and Q. Displacement will be length of straight line joining P and Q

a semi circle with a radius of 2 m

Length of this straight line=4+diameter

=4+(2*2)

=8 m

7 0
3 years ago
1. A mass is on a level plane, it has a weight of 20N. What is the coefficient of kinetic friction if an applied force
Arturiano [62]

Answer:

0.4

Explanation:

F-Fr=ma where F is applied force, Fr is friction, m is mass and a is acceleration.

Since the mass is moving with a constant velocity, there's no acceleration hence

F=Fr=\mu N where N is the weight of object and \mu is coefficient of kinetic friction.

F=\mu N and making [tex]\mu the subject

\mu=\frac {F}{N}

Substituting F for 8 N and N for 20 N

\mu=\frac {8}{20}=0.4

Therefore, coefficient of kinetic friction is 0.4

4 0
3 years ago
Which situation describes a system with decreasing gravitational potential energy?
lord [1]

Answer: a boy jumping down

6 0
3 years ago
Bagaimana cara untuk melatih kemampuan melempar tangkap yang baik dalam bola basket​
anyanavicka [17]

Answer:

pergi ke pertandingan sepak bola dan amati

7 0
3 years ago
X-rays with an energy of 300 keV undergo Compton scattering from a target. If the scattered rays are detected at 30 relative to
lys-0071 [83]

Answer:

a) \Delta \lambda = \lambda' -\lambda_o = \frac{h}{m_e c} (1-cos \theta)

For this case we know the following values:

h = 6.63 x10^{-34} Js

m_e = 9.109 x10^{-31} kg

c = 3x10^8 m/s

\theta = 37

So then if we replace we got:

\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm

b) \lambda_0 = \frac{hc}{E_0}

With E_0 = 300 k eV= 300000 eV

And replacing we have:

\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm

And then the scattered wavelength is given by:

\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm

And the energy of the scattered photon is given by:

E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV

c) E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV

Explanation

Part a

For this case we can use the Compton shift equation given by:

\Delta \lambda = \lambda' -\lambda_0 = \frac{h}{m_e c} (1-cos \theta)

For this case we know the following values:

h = 6.63 x10^{-34} Js

m_e = 9.109 x10^{-31} kg

c = 3x10^8 m/s

\theta = 37

So then if we replace we got:

\Delta \lambda = \frac{6.63x10^{-34} Js}{9.109 x10^{-31} kg *3x10^8 m/s} (1-cos 37) = 4.885x10^{-13} m * \frac{1m}{1x10^{-15} m}= 488.54 fm

Part b

For this cas we can calculate the wavelength of the phton with this formula:

\lambda_0 = \frac{hc}{E_0}

With E_0 = 300 k eV= 300000 eV

And replacing we have:

\lambda_0 = \frac{1240 x10^{-9} eV m}{300000eV}=4.13 x10^{-12}m = 4.12 pm

And then the scattered wavelength is given by:

\lambda ' = \lambda_0 + \Delta \lambda = 4.13 + 0.489 pm = 4.619 pm

And the energy of the scattered photon is given by:

E' = \frac{hc}{\lambda'}= \frac{1240x10^{-9} eVm}{4.619x10^{-12} m}=268456.37 eV - 268.46 keV

Part c

For this case we know that all the neergy lost by the photon neds to go into the recoiling electron so then we have this:

E_f = E_0 -E' = 300 -268.456 kev = 31.544 keV

3 0
3 years ago
Read 2 more answers
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