The red light emitted by a helium–neon laser has a wavelength of 632.8 nm. What is the frequency of the light waves?

A thin disk of mass 2.2 kg and radius 61.2 cm is suspended by a horizonal axis perpendicular to the disk through its rim. The di

Alinara [238K]

Answer: The period of the subsequent simple harmonic motion is 1.004 sec.

Explanation:

The given data is as follows.

Mass of disk (m) = 2.2 kg, radius of the disk (r) = 61.2 cm,

Formula to calculate the moment of inertia around the center of mass is as follows.

$I_{cm} = \frac{1}{2}mr^{2}$

= $\frac{1}{2} \times 2.2 kg \times (61.2)^{2}$

= 0.412 $kg m^{2}$

Also,

Distance between center of mass and axis of rotation (d) = r = 0.612 m

Moment of inertia about the axis of rotation (I)

I = $I_{cm} + md^{2}$

I = $0.412 + (2.2) \times (0.612)^{2}$

= 0.339 $kgm^{2}$

Now, we will calculate the time period as follows.

T = $2\pi sqrt{\frac{I}{mgd})$

T = $2 \times 3.14 sqrt{(\frac{0.339}{2.2 \times 9.8 \times 0.612}$

T = 1.435 sec

T = $2 \times 3.14 \times 0.16$

= 1.004 sec

Thus, we can conclude that the period of the subsequent simple harmonic motion is 1.004 sec.

7 0

2 years ago

How far does a cyclist travel if she's going 6 m/s in 10 seconds?

RideAnS [48]

the answer is (b) 0.6

3 0

2 years ago

A charge of 35.0 μC is placed on conducting sphere A of radius 8.00 cm. Another identical conducting sphere B (radius 8.00 cm) c

Wewaii [24]

Answer:

a) 50μC

b) 37.45 m/s

Explanation:

a) If the spheres are connected the charge in both spheres tends to be equal. This because is the situation of minimum energy.

Thus, you have:

$Q_T=35\mu C+65\mu C=100\mu C\\\\Q_s=\frac{Q_T}{2}=50\mu C$

Hence, each sphere has a charge of 50μC.

b) You use the fact that the total work done by the electric force is equal to the change in the kinetic energy of the sphere. Then, you use the following equations:

$\Delta W=\Delta K\\\\\int_{0.4}^\infty Fdr=\frac{1}{2}m[v^2-v_o^2]\\\\F=k\frac{Q^2}{r^2}\\\\v_o=0m/s\\\\m=0.08kg\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=kQ^2[-\frac{1}{r}]_{0.4}^{\infty}=\frac{kQ^2}{0.4m}=\frac{(8.98*10^9Nm^2/C^2)(50*10^{-6}C)^2}{0.4m}\\\\kQ^2\int_{0.4}^{\infty} \frac{dr}{r^2}=56.125J$