Answer:

**Answer:**

** 160 Hz , 240 Hz , 400 Hz**

**Explanation:**

Given that

Frequency of forth harmonic is 320 Hz.

Lets take fundamental frequency = f₁

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:
### Final answer:

### Explanation:

### Learn more about Resonant frequencies here:

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.

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:

**160 Hz:**This is half the frequency of the given resonant frequency, which means the string vibrates with twice the number of loops.**106.7 Hz:**This is one third of the given resonant frequency, which means the string vibrates with three times the number of loops.**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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Why does an astronaut in a spacecraft orbiting Earthexperience a feeling of weightlessness?

Find the force necessary to start the crate moving, given that the mass of the crate is 32 kg and the coefficient of static friction between the crate and the floor is 0.57. Express your answer using two significant figures.

Convert this measurement664.2 km=____cm

A popular physics lab involves a hand generator and an assortment of wires with different values of resistance. In the lab, the leads of the generator are connected across each wire in turn. For each wire, students attempt to turn the generator handle at the same constant rate. Students must push harder on the handle when the leads of the generator are connected__________. This is because turning the handle at a given constant rate produces__________ , regardless of what is connected to the leads. So, when turning the handle at a constant rate, lab students must push harder in cases where there is________

A revolutionary war cannon, with a mass of 2260 kg, fires a 21 kg ball horizontally. The cannonball has a speed of 105 m/s after it has left the barrel. The cannon carriage is on a flat platform and is free to roll horizontally. What is the speed of the cannon immediately after it was fired?

Find the force necessary to start the crate moving, given that the mass of the crate is 32 kg and the coefficient of static friction between the crate and the floor is 0.57. Express your answer using two significant figures.

Convert this measurement664.2 km=____cm

A popular physics lab involves a hand generator and an assortment of wires with different values of resistance. In the lab, the leads of the generator are connected across each wire in turn. For each wire, students attempt to turn the generator handle at the same constant rate. Students must push harder on the handle when the leads of the generator are connected__________. This is because turning the handle at a given constant rate produces__________ , regardless of what is connected to the leads. So, when turning the handle at a constant rate, lab students must push harder in cases where there is________

A revolutionary war cannon, with a mass of 2260 kg, fires a 21 kg ball horizontally. The cannonball has a speed of 105 m/s after it has left the barrel. The cannon carriage is on a flat platform and is free to roll horizontally. What is the speed of the cannon immediately after it was fired?

**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**

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.

**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

**Answer:**

**Explanation:**

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

solve to r1

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

where T = orbital period

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

Since

then

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.

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.

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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

Where

A = Surface area of the Object

Stefan-Boltzmann constant

e = Emissivity

T = Temperature (Kelvin)

Our values are given as

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

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

**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 :**

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

**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.**

**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

Heat flow in resistor will be equal to

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

So

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