Two loudspeakers 42.0 m apart and facing each other emit identical 115 Hz sinusoidal sound waves in a room where the sound speed

Response:

3.5 N

Reasoning:

Taking the 0 cm position as the pivot point, to achieve balance, the total moment calculated around this point must equal zero. We will analyze the moment generated at each point, moving from 0 to 100 cm:

- Tension from the string at the 0 cm mark is 0, since the moment arm is nonexistent.

- A 2 N weight at the 10 cm point creates a moment of 20 Ncm moving clockwise.

- Another 2 N weight at the 50 cm position produces a moment of 100 Ncm also clockwise.

- The weight of the 1 N stick located at its center of mass (50 cm) has a moment of 50 Ncm, clockwise.

- A 3 N weight at the 60 cm position generates a moment of 180 Ncm, clockwise.

- The tension T (N) from the string at the 100 cm end contributes a moment of 100T Ncm, moving counter-clockwise.

Total clockwise moments = 20 + 100 + 50 + 180 = 350 Ncm.

Total counter-clockwise moment = 100T.

To achieve balance, we set 100 T = 350, leading to T = 350 / 100 = 3.5 N.

Answer:

The equivalent distance in kilometers is 4012 × km.

Explanation:

It's known that 1 millimeter converts to meters. Then, 1 meter converts to kilometers. Therefore, the conversion for 1 millimeter to kilometers can be stated as

1 mm = m

1 m = km

Thus, 1 mm = × km = km.

Given the distance of 4012 mm, the corresponding distance in kilometers will be

4012 mm = 4012 × km.

The distance therefore is 4012 × km.

Answer

Given:

Wavelength = λ = 18.7 cm

                  = 0.187 m

Amplitude, A = 2.34 cm

Velocity, v = 0.38 m/s

A)  Calculate the angular frequency.

     

     

     

Angular frequency,

ω = 2π f

ω = 2π x 2.03

ω = 12.75 rad/s

B) Calculate the wave number:

     

     

   

C)

Since the wave is traveling in the -x direction, the sign is positive between x and t

y (x, t) = A sin(k x - ω t)

y (x, t) = 2.34 sin(33.59 x - 12.75 t)

Refer to the diagram below.

Ignoring air resistance, use gravitational acceleration g = 9.8 m/s².

The pole vaulter drops with an initial vertical speed u = 0.
At impact with the pad, velocity v satisfies:
v² = 2 × (9.8 m/s²) × (4.2 m) = 82.32 (m/s)²
v = 9.037 m/s

As the pad compresses by 0.5 m to bring the vaulter to rest,
let the average acceleration (deceleration) be a m/s². Then:
0 = (9.037 m/s)² + 2 × a × 0.5 m
Solving for a gives:
a = - 82.32 / (2 × 0.5) = -82 m/s²

Thus, the deceleration magnitude is 82 m/s².

Answer:

Explanation:

To approach this problem, we need to understand two key concepts.

First, the gravitational force on an object in orbit equals its mass multiplied by centripetal acceleration.

Secondly, Newton's law of universal gravitation defines the force between two masses: Fg = mMG/r², where Fg denotes gravitational force, m and M signify the masses, G represents the gravitational constant, and r indicates the distance separating the two masses.

Thus:

Fg = m v²/r

mMG/r² = m v²/r

v² = MG/r

Potential energy for each planet is expressed as:

PE = mgr = m (MG/r²) r = mMG/r

Kinetic energy for each planet is computed as:

KE = 1/2 mv² = 1/2 m (MG/r) = 1/2 mMG/r

Total mechanical energy is calculated as:

ME = PE + KE = 3/2 mMG/r

Since both planets share the same mass, the only variable is their orbital radius. Consequently, Planet A, with a smaller radius, possesses greater potential, kinetic, and mechanical energy.