A particular string resonates in four loops at a frequency of 320 Hz . Name at least three other (smaller) frequencies at which it will resonate. Express your answers using two significant figures separated by commas.

Answers

Answer 1
Answer:

Answer:

160 Hz  ,  240 Hz  , 400 Hz

Explanation:

Given that

Frequency of forth harmonic is 320 Hz.

Lets take fundamental frequency = f₁

f_1=(320)/(4)\ Hz

f₁=80 Hz

Frequency of first harmonic = f₂

f₂=2 f₁

f₂ =2 x 80 = 160 Hz

Frequency of second harmonic = f₃

f₃= 3 f₁=3 x 80 = 240 Hz

Frequency of fifth harmonic = f₅

f₅=  5 f₁= 5 x 80 = 400 Hz

Three frequencies are as follows

160 Hz  ,  240 Hz  , 400 Hz

Answer 2
Answer:

Final answer:

The resonant frequencies of a string depend on its length, tension, and linear mass density. For a string resonating in four loops at 320 Hz, three possible smaller frequencies could be 160 Hz, 106.7 Hz, and 80 Hz.

Explanation:

When a string resonates, it vibrates at certain frequencies called its resonant frequencies. The resonant frequencies of a string depend on factors such as its length, tension, and linear mass density. In this case, the string resonates in four loops at a frequency of 320 Hz.

Three other possible resonant frequencies at which the string could vibrate with smaller loops include:

  1. 160 Hz: This is half the frequency of the given resonant frequency, which means the string vibrates with twice the number of loops.
  2. 106.7 Hz: This is one third of the given resonant frequency, which means the string vibrates with three times the number of loops.
  3. 80 Hz: This is one fourth of the given resonant frequency, which means the string vibrates with four times the number of loops.

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Which is true about inelastic collisions: a. An inelastic collision does not obey conservation of energy. b. An inelastic collision conserves kinetic energy. c. Objects will stick together upon collision. d. Momentum is not conserved in inelastic collisions..

Answers

Answer:

Option c is correct

Explanation:

There are two types of collisions-elastic collision and inelastic collision.

In elastic collision, both kinetic energy and total momentum are conserved. On the other hand, in inelastic collision, total momentum is conserved but kinetic energy is not conserved. Thus, option b and d are incorrect.

Total energy is always conserved in both types. Thus, option a is incorrect.

In a perfectly inelastic collision, objects stick together. This happens because maximum kinetic energy is dissipated and used in bonding of the two objects. Thus, correct option is c.

Answer:

i believe its a?

Explanation:

In an inelastic collision, momentum is conserved

What explains why a prism separates white light into a light spectrum?A. The white light, on encountering the prism, undergoes both reflection and refraction; some of the reflected rays re-enter the prism merging with refracted rays changing their frequencies.
B. The white light, on entering a prism, undergoes several internal reflections, forming different colors.
C. The different colors that make up a white light have different refractive indexes in glass.
D. The different colors that make up a white light are wavelengths that are invisible to the human eye until they pass through the prism.
E. The different rays of white light interfere in the prism, forming various colors.

Answers

Answer :

QUESTION①)

What explains why a prism separates white light into a light spectrum ?

C. The different colors that make up a white light have different refractive indexes in glass.

✔ Indeed, depending on the radiation (and therefore colors), which each have different wavelengths, the refraction index varies: the larger the wavelength (red) the less the reflection index is important and vice versa (purple).

✔ That's why purple is more deflected so is lower than red radiation.  

Answer:

I think the answer probably be B

Consider as a system the Sun with Saturn in a circular orbit around it. Find the magnitude of the change in the velocity of the Sun relative to the center of mass of the system during the time Saturn completes half an orbit. Assume the mass of the Sun is 5.68 x10^29 kg, the mass of Saturn is 5.68 x10^26 kg, its period is 9.29 x10^8 s, and the radius of its orbit is 1.43 x 10^12 m. Ignore the influence of other celestial objects.?

Answers

Answer:

v_(su) = 19.44 m/s

Explanation:

m_(su)=5.68x10^(29)kg\nm_(sa)=5.68x10^(26)kg

T=9.29x10^8\nr_(o)=1.43x10^(12)

If the sun considered as x=0 on the axis to put the center of the mass as a:

m_(su)*r_(o)=(m_(sa)+m_(su))*r_(1)

solve to r1

r_1=(m_(sa)*r_(o))/(m_(sa)+m_(su))=(5.68x10^(26)*1.43x10^(12))/(5.68x10^(26)+5.68x10^(26))

r_1=1.428x10^9m

Now convert to coordinates centered on the center of mass.  call the new coordinates x' and y' (we won't need y').  Now since in the sun centered coordinates the angular momentum was  

L = (m_(sa)*2*pi*r_1^2)/(T)

where T = orbital period

then L'(x',y') = L(x) by conservation of angular momentum.  So that means

L_(sun)=(m_(sa)*2*\pi *( 2r_(o)*r_1 -r_1^2))/(T)

Since

L_(su)= m_(su)*v_(su)*r_1

then

v_(su)=(m_(sa)*2*pi*(2r_(o)*r_(1)-r_(1)^2))/(T*m_(sa)*r_1)

v_(su) = 19.44 m/s

Final answer:

In a two-body system such as the Sun-Saturn system, both bodies orbit around their mutual center of mass, or barycenter. Given the Sun's significantly larger mass, this barycenter is near the center of the Sun, and hence the Sun's change in velocity relative to the center of mass of the system as Saturn completes half an orbit is effectively zero.

Explanation:

The problem here is asking for the change in velocity of the Sun relative to the center of mass of the Sun-Saturn system as Saturn completes half an orbit. This is a situation involving orbital physics and center of mass systems.

However, in an isolated two-body orbit system like this, the center of mass does not change velocity - it would remain constant, not unless acted upon by an outside force, which the problem instructs us to ignore.

Saturn and the Sun both orbit around their common center of mass (their barycenter). Given that the Sun is immensely more massive than Saturn, this center of mass is located very close to the center of the Sun.

So, while the Sun does indeed move a little due to Saturn's influence, the change in velocity of Sun relative to the center of mass of the system during the time Saturn completes half an orbit, for all intents and purposes, is zero.

This is especially true unless the problem specifically mentions that the Sun is initially at rest with respect to the center of mass. In any other case, the relative velocity remains constant and hence the change is zero.

Learn more about Orbital Physics here:

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A person's body is producing energy internally due to metabolic processes. If the body loses more energy than metabolic processes are generating, its temperature will drop. If the drop is severe, it can be life-threatening. Suppose that a person is unclothed and energy is being lost via radiation from a body surface area of 1.36 m2, which has a temperature of 34° C and an emissivity of 0.700. Also suppose that metabolic processes are producing energy at a rate of 122 J/s. What is the temperature of the coldest room in which this person could stand and not experience a drop in body temperature

Answers

To solve this problem it is necessary to apply the concepts related to the Stefan-Boltzman law that is responsible for calculating radioactive energy.

Mathematically this expression can be given as

P = \sigma Ae\Delta T^4

Where

A = Surface area of the Object

\sigma = Stefan-Boltzmann constant

e = Emissivity

T = Temperature (Kelvin)

Our values are given as

A = 1.36m^2

\Delta T^4 = T_2^4 -T_1^4 = 307^4-T_1^4

\sigma = 5.67*10^(-8) J/(s m^2 K^4)

P = 122J/s

e = 0.7

Replacing at our equation and solving to find the temperature 1 we have,

P = \sigma Ae\Delta T^4

P = \sigma Ae (T_2^4 -T_1^4)

122 = (5.67*10^(-8))(1.36)(0.7)(307^4-T_1^4)

T_1 = 285.272K = 12.122\°C

Therefore the the temperature of the coldest room in which this person could stand and not experience a drop in body temperature is 12°C

The electric field at the distance of 3.5 meters from an infinite wall of charges is 125 N/C. What is the magnitude of the electric field 1.5 meters from the wall?

Answers

Explanation:

It is given that,

Distance, r = 3.5 m

Electric field due to an infinite wall of charges, E = 125 N/C

We need to find the electric field 1.5 meters from the wall, r' = 1.5 m. Let it is equal to E'. For an infinite wall of charge the electric field is given by :

E=(\lambda)/(2\pi \epsilon_o r)

It is clear that the electric field is inversely proportional to the distance. So,

(E)/(E')=(r')/(r)

E'=(Er)/(r')

E'=(125* 3.5)/(1.5)  

E' = 291.67 N/C

So, the magnitude of the electric field 1.5 meters from the wall is 291.67 N/C. Hence, this is the required solution.

The heating element of a coffeemaker operates at 120 V and carries a current of 4.50 A. Assuming the water absorbs all of the energy converted by the resistor, calculate how long it takes to heat 0.525 kg of water from room temperature (23.0°C) to the boiling point.

Answers

Answer:

It will take 313.376 sec to raise temperature to boiling point

Explanation:

We have given that potential difference V = 120 Volt

Current i = 4.50 A

So resistance R=(V)/(i)=(120)/(4.50)=26.666ohm

Heat flow in resistor will be equal to Q=i^2Rt

It is given that this heat is used for boiling the water

Mass of the water = 0.525 kg = 525 gram

Specific heat of water 4.186 J/gram/°C

Initial temperature is given as 23°C

Boiling temperature of water = 100°C

So change in temperature = 100-23 = 77°C

Heat required to raise the temperature of water Q=mc\Lambda T

So 4.50^2* 26.666* t=525* 4.186* 77

t = 313.376 sec

So it will take 313.376 sec to raise temperature to boiling point

Answer:

Explanation:

Voltage, V = 120 V

Current, i = 4.5 A

mass of water, m = 0.525 kg

initial temperature of water, T1 = 23°C

Final temperature of water, T2 = 100 °C

specific heat of water, c = 4.18 x 1000 J/kg °c

let the time taken is t.

Heat given by the heater = heat gain by the water

V x i x t = m x c x (T2 - T1)

120 x 4.5 x t = 0.525 x 4.18 x 1000 x (100 - 23)

540 t = 47701.5

t = 88.34 s