If the period of the wave is 0.34 seconds what would be the frequency of the wave

A car moved 40 km north and 90 km south. What is the displacement of the car?

il63 [147K]

Answer:

A. 50 km south

Explanation:

hope this helps!

8 0

4 years ago

A sinusoidal wave of wavelength 2.00m and amplitude 0.100 m travels on a string with a speed of 1.00 m/s to the right. At t = 0

Shkiper50 [21]

The frequency and angular frequency are 0.500 Hertz and 3.142 rad/s. respectively.
The angular wave number is equal to 3.142 rad/m.
The wave function for this wave is given by: y = Asin(kx - ωt + Φ).
The equation of motion for the left end of the string is given by: y = 0.100sin(3.142x - 3.142t + 0).
The equation of motion for the left end of the string at x = 1.50 m to the right is equal to y = 0.100sin(4.71 - 3.142t + 0).The maximum speed of any point on the string is 0.3142 m/s.

<h3>How to calculate the frequency and angular frequency?</h3>

First of all, we would determine the frequency of this wave by using this formula:

Frequency = wavelength/speed

Frequency = 0.100/2.00

Frequency = 0.500 Hertz.

For the angular frequency, we have:

Angular frequency, ω = 2πf

Angular frequency, ω = 2 × 3.142 × 0.500

Angular frequency, ω = 3.142 rad/s.

<h3>How to determine the angular wave number?</h3>

Angular wave number, k = 2π/∧

Angular wave number, k = (2 × 3.142)/2.00

Angular wave number, k = 3.142 rad/m.

<h3>How to determine the wave function for this wave?</h3>

Mathematically, the wave function for this wave is given by:

y = Asin(kx - ωt + Φ)

For the equation of motion for the left end of the string, we have:

y = 0.100sin(3.142x - 3.142t + 0)

For the equation of motion for the left end of the string at x = 1.50 m to the right, we have:

y = 0.100sin(3.142x - 3.142t + 0)

y = 0.100sin(3.142(1.5) - 3.142t + 0)

y = 0.100sin(4.71 - 3.142t + 0)

<h3>What is the maximum speed of any point on the string?</h3>

Vy = 0.100sin(- 3.142)cos(3.142x - 3.142t)

Vy ≤ 0.3142 m/s (since cosine varies +1 and -1).

Read more on wave function here: brainly.com/question/11181093

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Complete Question:

A sinusoidal wave of wavelength 2.00 m and amplitude 0.100 m travels on a string with a speed of 1.00 m/s to the right. Initially, the left end of the string is at the origin. Find:

(a) the frequency and angular frequency,

(b) the angular wave number, and

(c) the wave function for this wave.

Determine the equation of motion in SI units for

(d) the left end of the string, and

(e) the point on the string at x = 1.50 m to the right of the left end.

(f) What is the maximum speed of any point on the string?

5 0

2 years ago

A child in a boat throws a 5.3 kg package out horizontally with a speed of 10.0m/s. calculate the velocity of the boat immediate

seropon [69]

<span>(Momentum is Zero) 0 = (26.0 kg + 55.0 kg)v + (5.40 kg)(+10.0 m/s)
</span>velocity = -.667 m/s

5 0

3 years ago

Read 2 more answers

Three cars (car F, car G, and car H) are moving with the same speed and slam on their brakes. The most massive car is car F, and

Crazy boy [7]

To solve this problem it is necessary to apply the concepts related to Normal Force, frictional force, kinematic equations of motion and Newton's second law.

From the kinematic equations of motion we know that the relationship of acceleration, velocity and distance is given by

$v_f^2=v_i^2+2ax$

Where,

$v_f =$ Final velocity

$v_i =$ Initial Velocity

a = Acceleration

x = Displacement

Acceleration can be expressed in terms of the drag coefficient by means of

$F_f = \mu_k (mg) \rightarrow$ Frictional Force

$F = ma \rightarrow$ Force by Newton's second Law

Where,

m = mass

a= acceleration

$\mu_k =$ Kinetic frictional coefficient

g = Gravity

Equating both equation we have that

$F_f = F$

$\mu_k mg=ma$

$a = \mu_k g$

Therefore,

$v_f^2=v_i^2+2ax$

$0=v_i^2+2(\mu_k g)x$

Re-arrange to find x,

$x = \frac{v_i^2}{2(-\mu_k g)}$

The distance traveled by the car depends on the coefficient of kinetic friction, acceleration due to gravity and initial velocity, therefore the three cars will stop at the same distance.

3 0

3 years ago

As a recreational boat operator, what actions must you take when in a narrow channel?

alexandr402 [8]

One should never anchor in a narrow channel, until unless required very importantly. One should stay to the starboard side, and use a prolonged blast. The announcement must be done to alarm the nearby vessels, about your approach. The vessel should be kept at the outer limit of the starboard side.

6 0

3 years ago

Read 2 more answers