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NISA [10]
2 years ago
7

Linear Velocity What is the linear velocity in cm>min for any edge point of a 12-cm-diameter CD (compact disc) spinning at 10

,350 rev>min?
Physics
1 answer:
daser333 [38]2 years ago
8 0

3,89,988 cm/min is the linear velocity

Given,

Diameter of CD = 12 cm

So, Radius of CD = 6 cm

CD is spinning at 10350 rev/min

Firstly , convert rev/min into rad/min

1 rev = 2π radians

10350 rev/min = 10350 × 2π

                        = 64998 rad/min

Formula used,

v=rw where,

v is the Linear velocity

r is the radius

w is the angular velocity

v = 6 cm × 64998rad/min

  = 3,89,988 cm/min

Thus, linear velocity for any edge point of a 12-cm-diameter CD (compact disc) spinning at 10,350 rev/min is 389988 cm/min.

Learn more about Angular speed here brainly.com/question/540174

#SPJ4

                 

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When an electron falls from a higher to a lower energy level in an atom, the photon released has a wavelength of 121.6 nm. What
yarga [219]

Answer:

\Delta E=1.64*10^{-18}J

Explanation:

The energy difference between the energy levels involved in the transition of the electron is directly proportional to the frequency of the emitted photon:

\Delta E=h\nu(1)

Where h is the Planck constant. The photon's frequency is inversely proportional to its wavelegth:

\nu=\frac{c}{\lambda}(2)

Here c is the speed of light. Replacing (2) in (1):

\Delta E=\frac{hc}{\lambda}\\\Delta E=\frac{(6.63*10^{-34}J\cdot s)(3*10^8\frac{m}{s})}{121.6*10^{-9}m}\\\Delta E=1.64*10^{-18}J

6 0
3 years ago
An electron moving with a velocity v⃗ = 5.0 × 107 m/s i^ enters a region of space where perpendicular electric and a magnetic fi
stiv31 [10]

Answer:

Magnetic field, B=2\times 10^{-4}\ T

Explanation:

Given that,

Velocity of electron, v=5\times 10^7\ m/s

It enters  a region of space where perpendicular electric and a magnetic fields are present.

Magnitude of electric field, E=10^4\ V/m

We need to find the magnetic field will allow the electron to go through the region without being deflected.

Magnetic force on the electron, F_m=qvB\ sin\theta.......(1)

Electric force on the electron, F = q E........(2)

From equation (1) and (2) we get:

qvB\ sin\theta=qE

B=\dfrac{E}{v}

B=\dfrac{10^4\ V/m}{5\times 10^7\ m/s}

B = 0.0002 T

or

B=2\times 10^{-4}\ T

Hence, this is the required solution.

3 0
3 years ago
The end point of a spring oscillates with a period of 2.0 s when a block with mass m is attached to it. When this mass is increa
Nostrana [21]

Answer: The end point of a spring oscillates with a period of 2.0 s when a block with mass m is attached to it. When this mass is increased by 2.0 kg, the period is found to be 3.0 s.  Then the mass m is 0.625kg.

Explanation: To find the answer, we need to know more about the simple harmonic motion.

<h3>What is simple harmonic motion?</h3>
  • A particle is said to execute SHM, if it moves to and fro about the mean position under the action of restoring force.
  • We have the equation of time period of a SHM as,

                                          T=2\pi \sqrt{\frac{m}{k} }

  • Where, m is the mass of the body and k is the spring constant.
<h3>How to solve the problem?</h3>
  • Given that,

               T_1=2s\\m_1=m\\m_2=m+2kg\\T_2=3s

  • We have to find the value of m,

              T_1=2\pi \sqrt{\frac{m}{k} } \\T_2=2\pi \sqrt{\frac{m+2}{k} } \\\frac{T_1}{T_2} =\sqrt{\frac{m}{m+2} }\\\frac{2}{3} =\sqrt{\frac{m}{m+2} }\\\\

               m=\frac{5}{8} =0.625kg

Thus, we can conclude that, the mass m will be 0.625kg.

Learn more about simple harmonic motion here:

brainly.com/question/28045110

#SPJ4

3 0
2 years ago
A 12 volt car battery is connected to a 3 ohm brake light. What is the current carrying energy to the lights?
zloy xaker [14]

Answer:

4 A

Explanation:

V = IR, where V=voltage, I=current, R=resistance. This is Ohm's Law. (remember that for units V = volts, Ω = ohms, A = amperes.)

V = IR

12 V = I * 3 Ω

12/3 = I

<u>I = 4 A</u>

8 0
3 years ago
How much energy is needed to change 100 g of 0o C ice to 0o C water? The latent heat of fusion for water L=335,000 J/kg. Hint: T
pychu [463]

Answer:

33.5 kJ

Explanation:

here there is no difference is made in the temperature. Only thing happens here is the conversion of the ice in to water of 0 degree. The heat energy taken from the outside is spent for this conversion.

we have ice 100g =0.1 kg

Appplying Q=mL

                  Q= 0.1 kg * 335 000 J kg^{-1}

                   Q = 33 500 J

                   Q = 33.5 kJ

3 0
3 years ago
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