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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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An electron is released from rest at a distance of 6.00 cm from a proton. If the proton is held in place, how fast will the elec
lana66690 [7]

Answer:

91.87 m/s

Explanation:

<u>Given:</u>

  • x = initial distance of the electron from the proton = 6 cm = 0.06 m
  • y = initial distance of the electron from the proton = 3 cm = 0.03 m
  • u = initial velocity of the electron = 0 m/s

<u>Assume:</u>

  • m = mass of an electron = 9.1\times 10^{-31}\ kg
  • v = final velocity of the electron
  • e = magnitude of charge on an electron = 1.6\times 10^{-19}\ C
  • p = magnitude of charge on a proton = 1.6\times 10^{-19}\ C

We know that only only electric field due to proton causes to move from a distance of 6 cm from proton to 3 cm distance from it. This means the electric force force does work on the electron to move it from one initial position to the final position which is equal to the change in potential energy of the electron due to proton.

Now, according to the work-energy theorem, the total work done by the electric force on the electron due to proton is equal to the kinetic energy change in it.

\therefore \textrm{Kinetic energy change}= \textrm{Change in potential energy}\\\Rightarrow \dfrac{1}{2}m(v^2-u^2)= \dfrac{kpe}{y}-\dfrac{kpe}{x}\\\Rightarrow \dfrac{1}{2}m(v^2-(0)^2)= \dfrac{kpe}{0.03}-\dfrac{kpe}{0.06}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{3}-\dfrac{100kpe}{6}\\\Rightarrow \dfrac{1}{2}mv^2= \dfrac{100kpe}{6}\\

\Rightarrow v^2= \dfrac{100kpe\times 2}{6m}\\\Rightarrow v^2= \dfrac{100kpe}{3m}\\\Rightarrow v^2= \dfrac{100\times 9\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{3\times 9.1\times 10^{-31}}\\\Rightarrow v^2=8.44\times 10^3\\\Rightarrow v=91.87\ m/s\\

Hence, when the electron is at a distance of c cm from the proton, it moves with a velocity of 91.87 m/s.

8 0
3 years ago
In a magnetized substance, the domains __________________________ .
ahrayia [7]
I think the answer you're looking for is -are randomly oriented. if not sorry... i tried.
5 0
3 years ago
1-A train travels 100 km to reach town A in one hour and 15 min. The train stops at station A for 45 minutes. Then it travels 15
kirill [66]

Answer 1) : 62.5 km/hour is the average velocity of the train.

2) The final velocity of the car at the end of 75 m is 14.69 m/s

Explanation:

1) Displacement of the train = 100 km + 150 km = 250 km

Total time train took =1 hour 15 min+ 45 min + 2 hours = 240 min = 4 hours

Average velocity=\frac{Displacement}{time}=\frac{250 km}{4 hour}=62.5 km/h

62.5 km/hour is the average velocity of the train.

2) The acceleration of the car, a= 1.2 m/s^2

Distance covered by the car,s = 75 m

Initial velocity of the car ,v_i = 6 m/s

Final velocity of thre car ,v_f=?

Using third equation of motion:

v_{f}^2=v_{i}^2+2as=(6 m/s)^2+2\times 1.2 m/s\times 75 m=216 m^2/s^2

v_{f}=14.69 m/s

The final velocity of the car at the end of 75 m is 14.69 m/s

8 0
3 years ago
What effect does the vertical acceleration have on the horizontal velocity of the projectile?
KengaRu [80]
Answer:
None, if air resistance is ignored.

Explanation:
At any instant, the projectile has vertical and horizontal components of velocity.
Vertical acceleration due to gravity affects the vertical velocity by accelerating the object toward the center of the earth, and by decreasing the upward vertical velocity.. 
The horizontal component of velocity makes the object travel horizontally as long as the projectile is airborne.
Thsi discussion assumes that air resistance is ignored.
3 0
3 years ago
A bike accelerates uniformly from rest to a speed of 7.10 m/s over a distance of 35.4m .determine the acceleration of the bike.
sergeinik [125]
Given: Change of x is 35.4m, Velocity Final=7.10 m/s, Velocity Initial=0m/s
Find: Acceleration
Analysis:
Vf²=Vi²+2aΔx (Velocity final squared equals Velocity initial squared plus 2 times acceleration times change of x)
(7.10 m²/s)²=(0 m/s)²+2a(35.4 m)
50.41 m/s²=(70.8 m)a
a=0.712 m/s²
5 0
2 years ago
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