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STALIN [3.7K]
3 years ago
8

A mass m0 is attached to a spring and hung vertically. The mass is raised a short distance in the vertical direction and release

d. The mass oscillates with a frequency f0 . If the mass is replaced with a mass nine times as large, and the experiment was repeated, what would be the frequency of the oscillations in terms of f0
Physics
1 answer:
iragen [17]3 years ago
6 0

Answer:

The frequency of the oscillations in terms of fo will be f2=fo/3

E xplanation:

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

 \frac {{f2}/times {fo}}=1:3

⇒f2=fo\3

Here frequency f is inversely poportional to square root of mass m.

so the value of remainder of frequency f2 and fo is equal to 1:3.

⇒\frac{f2} {f1} = \frac sqrt{m1}[m2}

⇒\frac{f2}{fo} = 1:3

⇒f2=\frac{fo} {3}

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The critical angle for a substance is measured at 53.7 degrees. light enters from air at 45.0 degrees. at what angle it will con
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When light travels from a medium with greater refractive index n_1 to a medium with smaller refractive index n_2, there exists an angle (called critical angle) above which the light is totally reflected, and the value of this angle is given by
\theta_c = \arcsin ( \frac{n_2}{n_1} )
In this problem, we know that the critical angle is\theta_c = 53.7^{\circ}, so we can find the ratio between the refractive indices of the two mediums:
\frac{n_2}{n_1} = \sin \theta_c = \sin 53.7^{\circ} =0.81
and since the second medium is air (n=1.00), the refractive index of the first medium is
n_1=  \frac{n_2}{0.81}= \frac{1.00}{0.81}=1.23

In the second part of the problem, we have light entering from air (n_i = 1.00) at angle of incidence of \theta_i = 45.0 ^{\circ}, into the second medium with n_r = 1.23. By using Snell's law, we can find the angle of refraction of the light inside the medium:
n_i \sin \theta_i = n_r \sin \theta_r
\sin \theta_r =  \frac{n_i}{n_r}  \sin \theta_i = \frac{1.00}{1.23} \sin 45^{\circ}=0.574
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3 0
3 years ago
A 0.25-kg ball attached to a string is rotating in a horizontal circle of radius 0.5 m. If the ball revolves twice every second,
Aloiza [94]

Answer:

19.74 N

Explanation:

mass of ball (m) = 0.25 kg

radius (r) = 0.5 m

time (t) = 2 revolutions per seconds = 1/2 = 0.5 second per revolution

find the tension in the string

tension (T) = \frac{mv^{2} }{r}

  • where velocity (v) = \frac{2πr}{t}

tension now becomes (T) = \frac{m}{r} x (\frac{2πr}{t})^{2}

tension (T) = \frac{4π^{2}rm }{t^{2} }

  • now substituting the values of mass (m), time (t) and radius (r) into the equation above we have

tension (T) = \frac{4π^{2}x0.5x0.25}{0.5^{2} }

tension (T) =  2π^{2} =   2x3.142^{2} = 19.74 N

4 0
3 years ago
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