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lidiya [134]
3 years ago
8

The transverse standing wave on a string fixed at both ends is vibrating at its fundamental frequency of 250 Hz. What would be t

he fundamental frequency on a piece of the same string that is twice as long and has four times the tension
Physics
1 answer:
hodyreva [135]3 years ago
7 0

Answer:

Explanation:

fundamental frequency, f = 250 Hz

Let T be the tension in the string and length of the string is l ans m be the mass of the string initially.

the formula for the frequency is given by

f=\frac{1}{2l}\sqrt{\frac{Tl}{m}}    .... (1)

Now the length is doubled ans the tension is four times but the mass remains same.

let the frequency is f'

f'=\frac{1}{2\times 2l}\sqrt{\frac{4T\times 2l}{m}}    .... (2)

Divide equation (2) by equation (1)

f' = √2 x f

f' = 1.414 x 250

f' = 353.5 Hz

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In which medium does light travel faster: one with a critical angle of 27.0° or one with a critical angle of 32.0°? Explain. (Fo
Eddi Din [679]

Answer:

Among those two medium, light would travel faster in the one with a reflection angle of 32^{\circ} (when light enters from the air.)

Explanation:

Let v_{1} denote the speed of light in the first medium. Let v_{\text{air}} denote the speed of light in the air. Assume that the light entered the boundary at an angle of \theta_{1} to the normal and exited with an angle of \theta_{\text{air}}. By Snell's Law, the sine of \theta_{1}\! and \theta_{\text{air}}\! would be proportional to the speed of light in the corresponding medium. In other words:

\displaystyle \frac{v_{1}}{v_{\text{air}}} = \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})}.

When light enters a boundary at the critical angle \theta_{c}, total internal reflection would happen. It would appear as if the angle of refraction is now 90^{\circ}. (in this case, \theta_{\text{air}} = 90^{\circ}.)

Substitute this value into the Snell's Law equation:

\begin{aligned}\frac{v_{1}}{v_{\text{air}}} &= \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})} \\ &= \frac{\sin(\theta_{c})}{\sin(90^{\circ})} \\ &= \sin(\theta_{c})\end{aligned}.

Rearrange to obtain an expression for the speed of light in the first medium:

v_{1} = v_{\text{air}} \cdot \sin(\theta_{1}).

The speed of light in a medium (with the speed of light slower than that in the air) would be proportional to the critical angle at the boundary between this medium and the air.

For 0 < \theta < 90^{\circ}, \sin(\theta) is monotonically increasing with respect to \theta. In other words, for \!\theta in that range, the value of \sin(\theta)\! increases as the value of \theta\! increases.

Therefore, compared to the medium in this question with \theta_{c} = 27^{\circ}, the medium with the larger critical angle \theta_{c} = 32^{\circ} would have a larger \sin(\theta_{c}). such that light would travel faster in that medium.

4 0
3 years ago
A small 18 kilogram canoe is floating downriver at a speed of 1 m/s. What is the canoe's kinetic energy?
In-s [12.5K]

Kinetic energy = (1/2) (mass) (speed²).

A Physicist in the canoe, or on a raft floating downriver next to the canoe, will say that the canoe's kinetic energy is zero.

A Physicist on the riverbank, watching the canoe drift by at 1 m/s, will say that its kinetic energy is 9 Joules.

They're both correct.

8 0
3 years ago
Agustin visits Panama City, Florida, during the month of May. He feels a shore breeze blowing from theocean onto the beach. What
Pavel [41]

Answer: A.The ocean is colder than the land

Explanation:

Based on the information provided in the question, we are informed that Agustin visits Panama City, Florida, during the month of May and that he feels a shore breeze blowing from the ocean onto the beach.

The reason for the shore breeze is simply due to the fact that the ocean is colder than the land. Since the ocean is colder, anyone who goes to the beach will feel the breeze.

7 0
3 years ago
A 55.4 g sample of water at 99.61 °C is placed in a constant pressure calorimeter. Then, 23.4 g of zinc metal at 21.6 °C is adde
Zolol [24]

Answer:

The specific heat capacity of the zinc metal measured in this experiment is 0.427 J/g.°C

Explanation:

From the experimental data, the water loses heat because its initial temperature is greater than the final temperature of the mixture. On the other hand, the zinc metal gains heat because its initial temperature is less than the final temperature of the mixture

Heat loss by water = Heat gain by zinc metal

m1C1(T1 - T3) = m2C2(T3 - T2)

m1 is mass of water = 55.4 g

C1 is specific heat capacity of water = 4.2 J/g.°C

m2 is mass of zinc metal = 23.4 g

C2 is specific heat capacity of zinc metal

T1 is the initial temperature of water = 99.61 °C

T2 is the initial temperature of zinc metal = 21.6 °C

T3 is the final temperature of the mixture = 96.4 °C

55.4×4.2(99.61 - 96.4) = 23.4×C2(96.4 - 21.6)

746.9028 = 1750.32C2

C2 = 746.9028/1750.32 = 0.427 J/g.°C

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4 years ago
Which of the following statements is true?
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My guess is A. I'm not 100% positive but i'm pretty sure.
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