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Masja [62]
3 years ago
8

Consider a standing wave in a one dimensional ideal medium of length "D" (like a vibrating string).

Physics
1 answer:
stira [4]3 years ago
5 0

Answer:

a) 20 nodes    b) zero nodes

Explanation:

When we have standing waves in a bend we have nodes at the ends and the equation describes the number of possible waves in the string is

         L = n λ/2

Where λ is the wavelength, L is the length of the string, in our case it would be D and n is an entered. We can strip the wavelength of this expression

       

       λ = 2L / n

Let's calculate what value of n we have for a wavelength equal to D/10

       λ = 2D / n

       λ = D / 10

We match and calculate

       2D / n = D / 10

       2 / n = 1/10

       n = 20

Perform them for  λ = D / 20

       λ = 2D / n

       2D / n = D / 20

       n = 2 20 = 40

Since n is an inter there should be a wavelength for each value of n in the bone period there should be 20 different wavelengths

B) for La = 10D

       2D / n = 10D

       1 / n = 5

       n = 1/5 = 0.2

 

La = 20D

       2D / n = 20D

       1 / n = 10

      n = 1/10 = 0.1

These numbers are not entered so there can be no wave in this period

You might be interested in
two masses are kept 2 metre apart there is gravitational force of 2 Newton what is the gravitational force when they are kept at
sukhopar [10]

Answer: 0.5N

Explanation:

Gravitational force is calculated using the formula :

F = Gm1m2/r^2

Where G is the gravitational constant (6.67 × 10^-11)

At a distance 'r' of 2metres apart:

Mass of objects are m1 and m2

Gravitational force 'F1' = 2N

Inputting values into the formula :

2 = Gm1m2 / 2^2 - - - - - (1)

At a distance 'r' of 4meters apart:

Mass of objects are m1 and m2

Gravitational force 'F2' = y

Inputting values

F2 = Gm1m2 / 4^2 - - - - - (2)

Dividing equations 1 and 2

2 = Gm1m2 / 2^2 ÷ F2 = Gm1m2 / 4^2

2 / F2 = (Gm1m2 / 4) / (Gm1m2 / 16)

2 / F2 = (Gm1m2 / 4) × (16 / Gm1m2)

2/F2 = 16 / 4

Cross multiply

2 × 4 = 16 × F2

8 = 16F2

F2 = 8/16

F2 = 0.5N

7 0
3 years ago
Date
sergiy2304 [10]

Answer:

Because Kinetic Energy(KE) is not the same as Momentum(P)

Kinetic Energy is a scalar(has magnitude only). For a body of mass M, velocity V:

KE = 0.5MV^2

The units of KE: Joules.

Energy is the ability to do work.

Momentum is not a form of energy.

Momentum is a vector(has magnitude and direction).

P = MV

Units of momentum: kg m/s

If you have rifles of mass 2, 4, 8, 16 kg, using the same cartridge, with the same load, barrel length(remember momentum of projectile is proportional to velocity), they all have the same recoil momentum.

But the kinetic energy of recoil would be inversely proportional to the mass of the gun.

Thus the 2kg gun(possible even in large powerful calibers due to modern materials like titanium etc), would have 8 times the recoil ENERGY of the 16kg gun.

A lot of confusion exists in America because of retention of old units, namely Foot Pounds(force) for KE, and Pounds(mass) Feet Per Second for Momentum(P). Because of the more awkward momentum units, a lot of old books had a bad habit of calling the momentum units Pounds Feet, leaving out the rest. Naturally this created confusion with Foot Pounds. Multiplication being commutative and all that:).

Remember that the momentum of the rifles is the same. But the ones with the highest recoil energy hurt the most.

Speaking of hurt:

If momentum killed, then consider two dinosaur killer asteroids with the same masses and velocities, striking vertically at the same time antipodal points on the Earth’s surface. Total momentum delivered would be Zero. That would not make us safe at all:)

Similarly, being shot simultaneously at close range from opposite sides with a 5 round burst from each from two M4 assault rifles(by definition must be able to fire full auto) delivered in 0.3 seconds, would deliver zero momentum. But not zero harm.

Also, the recoil momentum of any firearm is equal to the mass of projectile x velocity + mass of propellant x exit velocity of propellant. This is obviously greater, often much greater, depending on range, than the striking momentum of the projectile at the target.

The recoil kinetic energy is vastly less than the kinetic energy of the bullet/projectile. Neglecting propellant contribution:

recoil Momentum = bullet momentum

BUT:

recoil KE/bullet KE = projectile mass/gun mass

This is a very small fraction.

If we consider the M4 carried by American military:

M855(SS109 equivalent) 5.56 bullet of mass 0.004kg(62 grains)is fired from M4 assault rifle of mass, with optic and full mag 4kg, a thousand times as much!

Even allowing for the 0.0015kg powder charge, and the higher velocity of the powder(approx 1400=1500 m/s vs approx 900 m/s muzzle velocity of the bullet), the recoil energy is hundreds of times less than the muzzle energy of the bullet.

That’s why you want to be behind the gun, and not in front.

Explanation:

7 0
3 years ago
If, in
STatiana [176]

Answer:

a)   λ = 121.5 nm , b) 102.6, 97, 91.1 nm

Explanation:

Bohr's model describes the energy of the hydrogen atom

      E_{n} = k² e² / 2m (1 / n²)

A transition occurs when the electron passes from n level to a lower one

       E_{i} -  E_{n} = k² e² / 2m (1 / n_{i}² - 1 /  n_{f}²)

Planck's relationship is

        E = h f = h c / lam

        hc /λ =  k² e²/ 2m(1 / n_{i}² - 1 /  n_{f}²)

        1 / λ = [k² e² / 2m h c] (1 / n_{i}² - 1 /  n_{f}²)

        1 /λ = Ry] (1 / n_{i}² - 1 /  n_{f}²)

a) the first element of the series occurs for n_{f} = 2

        1 / λ = 1.097 10⁷ (1- 1/2²)

        1 / λ = 1.097 10⁷ (1- 0.25)

        1 / λ = 0.82275 10⁷

        λ = 1.215 10⁻⁷ m

        λ = 1,215 10⁻⁷ m (10⁹nm / m)

        λ = 121.5 nm

b) the next elements of the series occur to

n_{f}       n_{i}    1 /λ                            λ (10-7m)       λ (nm)

3        1     1,097 10⁷ (1-1 / 9)     1,0255           102.6

4        1     1,097 10⁷ (1-1 / 16)   0.9723            97.2

∞       1      1,097 10⁷ (1 - 0)     0.91158           91.1

5 0
3 years ago
g Two long parallel wires are a center-to-center distance of 2.50 cm apart and carry equal anti-parallel currents of 2.70 A. Fin
schepotkina [342]

Answer:

864 mT

Explanation:

The magnetic field due to a long straight wire B = μ₀i/2πR where μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, i = current in wire, and R = distance from center of wire to point of magnetic field.

The magnitude of magnetic field due to the first wire carrying current i = 2.70 A at distance R which is mid-point between the wires is B = μ₀i/2πR.

Since the other wire also carries the same current at distance R, the magnitude of the magnetic field is B = μ₀i/2πR.

The resultant magnetic field at B is B' = B + B = 2B = 2(μ₀i/2πR) = μ₀i/πR

Now R = 2.50 cm/2 = 1.25 cm = 1.25 × 10⁻² m and i = 2.70 A.

Substituting these into B' = μ₀i/πR, we have

B' = 4π × 10⁻⁷ H/m × 2.70 A/π(1.25 × 10⁻² m)

B = 10.8/1.25 × 10⁻⁵ T

B = 8.64 × 10⁻⁵ T

B = 864 × 10⁻³ T

B = 864 mT

4 0
3 years ago
What is dispersion, and how is it related to diffraction and refraction?
iragen [17]
Refraction is simply the bending of light when it moves from one material into another. If light is beamed at 90° to a surface, no bending happens. But if you shine the light at an angle it will bend one way or another.
   
    Diffraction is a process in which a beam of light travels through a gap or around a barrier, and spreads out as a result.

<span>   Dispersion is the property that the speed of light in a transparent material is different for different wavelengths. Thus the index of refraction is likewise different.</span>
8 0
3 years ago
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