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4 years ago

5

Having established that a sound wave corresponds to pressure fluctuations in the medium, what can you conclude about the directi

on in which such pressure fluctuations travel

1 answer:

DaniilM [7]4 years ago

4 0

Answer:

the pressure fluctuation is LONGITUDINAL

Explanation:

Sound waves are an oscillating movement of air particles, this can be analyzed in two different, but equivalent ways, as an air oscillation and with a pressure wave due to these oscillations.

The expression for the wave is

ΔP = Δo sin (kx - wt)

Therefore, the pressure variation is in the same direction as the displacement variation, consequently the pressure fluctuation is LONGITUDINAL

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A tennis player receives a shot with the ball (0.0600 kg) traveling horizontally at 50.4 m/s and returns the shot with the ball

jonny [76]

(a) The impulse delivered to the ball by the racket is 5.24 kg.m/s

(b) The work that the racket does on the ball is -35.1 Joule

<h3>Further Explanation</h3>

<u>Given :</u>

mass of ball = m = 0.06 kg

initial velocity = v₁ = -50.4 m/s

final velocity = v₂ = 37.0 m/s

<u>Unknown :</u>

(a) Impulse = I = ?

(b) Work = W = ?

<u>Solution :</u>

<h2>Question (a) :</h2>

In this question , we could use the formula from Second Law of Newton :

$I = \Delta p$

$I = p_2 - p_1$

$I = m \times v_2 - m \times v_1$

$I = m \times (v_2 - v_1)$

$I = 0.06 \times (37.0 - (-50.4))$

$I = 0.06 \times (87.4)$

$I = 5.244~kg.m/s$

$\large { \boxed {I \approx 5.24~kg.m/s} }$

<h2>Question (b) :</h2>

$W = F \times d$

$W = (\frac{I}{\Delta t})(\frac {v_1 + v_2}{2} \Delta t)$

$W = \frac{I(v_1 + v_2)}{2}$

$W = \frac{5.244(-50.4 + 37)}{2}$

$W = \frac{5.244(-13.4)}{2}$

$W = -35.1348~Joule$

$\large { \boxed {W \approx -35.1~Joule}}$

<h3>Learn more</h3>

Newton's Law of Motion: brainly.com/question/10431582

Example of Newton's Law: brainly.com/question/498822

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

Keywords: Newton, Law, Impulse, Work

6 0

4 years ago

Ask Your Teacher The tub of a washer goes into its spin-dry cycle, starting from rest and reaching an angular speed of 2.0 rev/s

adell [148]

Answer:

$\theta = 20\ rev$

Explanation:

Case 1

Given,

initial angular speed = 0 rev/s

final angular speed = 2 rev/s

time, t = 7 s

angular acceleration of the washer

$\alpha = \dfrac{\omega_f-\omega_i}{t}$

$\alpha = \dfrac{2-0}{7}$

$\alpha = 0.286\ rev/s^2$

using equation of rotational motion

$\omega_f^2 = \omega_i^2 + 2 \alpha \theta_1$

$2^2 = 0^2 + 2\times 0.286 \times \theta_1$

$\theta_1 = 7 rev$

Case 2

Given,

initial angular speed = 2 rev/s

final angular speed = 0 rev/s

time, t = 12 s

angular acceleration of the washer

$\alpha = \dfrac{\omega_f-\omega_i}{t}$

$\alpha = \dfrac{0-2}{13}$

$\alpha = -0.154\ rev/s^2$

using equation of rotational motion

$\omega_f^2 = \omega_i^2 + 2 \alpha \theta_2$

$0^2 = 2^2 + 2\times (-0.154) \times \theta_2$

$\theta_2 = 13 rev$

total revolution in this case

$\theta = \theta_1 + \theta_2$

$\theta =7 +13$

$\theta = 20\ rev$

total revolution of the washer is equal to 20 rev.

6 0

3 years ago

Electromagnetic radiation from a 8.25 mW laser is concentrated on a 1.23 mm2 area. Suppose a 1.12 nC static charge is in the bea

mylen [45]

Answer:

The maximum magnetic force is 2.637 x 10⁻¹² N

Explanation:

Given;

Power, P = 8.25 m W = 8.25 x 10⁻³ W

charge of the radiation, Q = 1.12 nC = 1.12 x 10⁻⁹ C

speed of the charge, v = 314 m/s

area of the conecntration, A = 1.23 mm² = 1.23 x 10⁻⁶ m²

The intensity of the radiation is calculated as;

$I = \frac{P}{A} \\\\I = \frac{8.25 \times 10^{-3} \ W}{1.23 \ \times 10^{-6} \ m^2} \\\\I = 6,707.32 \ W/m^2$

The maximum magnetic field is calculated using the following intensity formula;

$I = \frac{cB_0^2}{2\mu_0} \\\\B_0 = \sqrt{\frac{2\mu_0 I}{c} } \\\\where;\\\\c \ is \ speed \ of \ light\\\\\mu_0 \ is \ permeability \ of \ free \ space\\\\B_0 \ is \ the \ maximum \ magnetic \ field\\\\B_0 = \sqrt{\frac{2 \times 4\pi \times 10^{-7} \times 6,707.32 }{3\times 10^8} } \\\\B_0 = 7.497 \times 10^{-6} \ T$

The maximum magnetic force is calculated as;

F₀ = qvB₀

F₀ = (1.12 x 10⁻⁹) x (314) x (7.497 x 10⁻⁶)

F₀ = 2.637 x 10⁻¹² N

5 0

3 years ago

He drives 150 meters in 18 seconds. Assuming constant speed, what is his speed in meters per second?

ira [324]

Answer:

Explanation:

Givens

d = 150 meters

t = 18 seconds

r (rate) = ?

Formula

r = d/t

Solution

r = 150/18

r = 8.33 m/s

4 0

3 years ago

This same car gets pulled over for speeding, and goes from 68 m/s to 0 m/s in 14

Harrizon [31]

Answer:

the acceleration of the car is -4.9m/s2.

the direction is opposite to the actual direction, since the acceleration is negative.

3 0

3 years ago