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noname [10]
4 years ago
6

A periodic wave can be mathematically written as ???? = ???? sin(???????? − ????????) 1. What part of the wave does A describe?

2. If we take a snapshot of the wave, what is the distance between two crests (or troughs) called? How is this related to k? 3. The source that is generating the wave oscillates with a frequency of f . How is this related to ????? 4. Write down the relationship between wave speed, wave frequency and wavelength.
Physics
1 answer:
Veronika [31]4 years ago
8 0

A periodic wave can be written as:

y(x,t)=A sin (kx-\omega t)

where

A is the amplitude, k is the wave number, x is the position, \omega is the angular frequency and t is the time.

1. A represents the amplitude

Explanation: the term A inside the wave equation represents the amplitude, which is the maximum displacement of the wave (along the y-axis) with respect to tis equilibrium position. Basically, when the sine part of the wave is equal to 1, y=A, and the wave has the maximum displacement.

2. Wavelength

Explanation: the wavelength of a wave is defined as the distance between two successive crests (or between succesive throughs) of the wave.

The wavelength (indicated with \lambda) is related to the wave number, k, by the equation

k=\frac{2\pi}{\lambda}

3. \omega = 2 \pi f

The frequency at which the source of the wave is oscillating, f, is related to the angular frequency \omega by the relationship

\omega = 2 \pi f

basically, f expresses the frequency in units of Hertz (1/s), while the angular frequency expresses the frequency in units of radians per second.

4. v=f \lambda

The relationship between wave speed, wave frequency and wavelength is

v=f \lambda

where

v is the wave speed

f is the frequency

\lambda is the wavelength

We can observed that if the speed of the wave is constant, then the frequency and the wavelength are inversely proportional to each other.

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3 years ago
What is the speed vfinal of the electron when it is 10.0 cm from charge 1?
fgiga [73]

Answer:

Two stationary positive point charges, charge 1 of magnitude 3.45 nC and charge 2 of magnitude 1.85 nC, are separated by a distance of 50.0 cm. An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges. What is the speed v(final) of the electron when it is 10.0 cm from

The answer to the question is

The speed v_{final} of the electron when it is 10.0 cm from charge Q₁

= 7.53×10⁶ m/s

Explanation:

To solve the question we have

Q₁ = 3.45 nC = 3.45 × 10⁻⁹C

Q₂ = 1.85 nC = 1.85 × 10⁻⁹ C

2·d = 50.0 cm

a = 10.0 cm

q = -1.6×10⁻¹⁹C

Also initial kinetic energy = 0 and

Initial electric potential energy = k\frac{qQ_1}{d} + k\frac{qQ_2}{d} = kq(\frac{Q_1+Q_2}{d})

Final kinetic energy due to motion = 0.5·m·v²

Final electric potential energy = k\frac{qQ_1}{a} + k\frac{qQ_2}{2d-a} = kq(\frac{Q_1}{a}+\frac{ Q_2}{2d-a})

From the energy conservation principle we have

0+ kq(\frac{Q_1+Q_2}{d})=0.5mv^2+  kq(\frac{Q_1}{a}+\frac{ Q_2}{2d-a})

Solving for v gives

v=\sqrt{\frac{kq(\frac{Q_1+Q_2}{d})-   kq(\frac{Q_1}{a}+\frac{ Q_2}{2d-a})}{0.5m}}

where k = 9.0×10⁹ and m = 9.109×10⁻³¹ kg

gives v =7528188.32769 m/s or 7.53×10⁶ m/s

v_{final} = 7.53×10⁶ m/s

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4 years ago
What is an organism's specific role in an ecosystem
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3 years ago
Read 2 more answers
A merry go round exerts a force of 1000 N on a rider on the
SVEN [57.7K]

Answer:

The radius of the circle made by the person on the merry go round is 74.55 meters

Explanation:

The given parameters are;

The force the merry go round exerts on the rider = 1000 N

The time it takes the merry go round to make one complete revolution = 15 seconds

The weight of the person = 750 N

The radius of the circle made by the person on the merry go round = r

We have;

F_c = \dfrac{m \cdot v^2}{r} = m \cdot \omega ^2 \cdot r

Where;

m = The mass of the person

v = The velocity of the person

F_c = The centrifugal force acting on the person = 1,000 N

r = The radius of the circle made by the person on the merry go round

ω = Angular velocity = 2·π/15 rad/s

We have;

The mass of the person = The weight/(The acceleration due to gravity, g)

∴ The mass of the person = 750/9.81 ≈ 76.45 kg

By substituting the calculated and known values into the equation for  the centripetal force, we have;

F_c = m × ω² × r

1000 = 76.45 × (2·π/15)² × r

r = 1000/(76.45 × (2·π/15)²) = 74.55 m

The radius of the circle made by the person on the merry go round = r = 74.55 m.

3 0
3 years ago
A metal ball attached to a spring moves in simple harmonic motion. The amplitude of the ball's motion is 11.0 cm, and the spring
german

Answer:

a)0.674 kg b) 2.2 s c) 0.9 m/s²

Explanation:

The amplitude of the ball (xo) = 11.0cm, half way between its equilibrium point its maximum displacement x = 11 cm / 2 = 5.5 cm = 5.5 / 100 in meters = 0.055 meters, speed at this point = 27.2 cm /s = (27.2 / 100) in m/s = 0.272 m/s,

spring constant K = 5.5 N/m

a) The mass of the ball (m) can be calculated using the formula below

v =√ (x²o - x²)K/m

make m subject of the formula

v² = (xo² - x²) K/m

m = K ( xo² - x²) / v²

m = 0.674kg

b) The period of the oscillation can be calculated by the following formula

T = 2π√ (m /K)

substitute the values into the formula

T = 2 × 3.142 × √ (0.674/ 5.5) = 2.2s

c) The maximum acceleration of the ball which occurs at the maximum displacement of the ball can be calculated by the following formula

a = K / m × x ( maximum displacement of the body) = 5.5 / 0.674 × 0.11 = 0.9 m/s²

7 0
3 years ago
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