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4 years ago

In which part of the wave are the particles of the medium closer together?

Physics

2 answers:

yuradex [85]4 years ago

8 0

B.compressions

Explanation:

This is a characteristics of the "longitudinal waves".

Longitudinal waves are waves where the oscillation of the particles of the medium occurs in a direction parallel to the direction of propagation of the wave (an example of longitudinal waves are sound waves). As a consequence, this oscillation creates some regions where particles are closer together, and other regions where particles are more spread apart. The former are called "compressions", while the latter are called "rarefactions".

Lesechka [4]4 years ago

6 0

B.) Compressions
HOPPE IT HELPED

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Si se aplic una fuerza de 150N en un área de de 0.4m2¿cual será la preciosa ejercida?

natali 33 [55]

Answer:

P = 375 Pa

Explanation:

The question says that,"If a force of 150N was applied in an area of 0.4m², what will be the precious exerted?"

We have,

Force, F = 150 N

Area, A = 0.4m²

We need to find the pressure exerted. We know that,

Pressure = forece/area

So,

$P=\dfrac{150\ N}{0.4\ m^2}\\\\P=375\ Pa$

So, the required pressure is equal to 375 pa.

7 0

3 years ago

(i) a force of 35.0 n is required to start a 6.0-kg box moving across a horizontal concrete floor, (a) what is the coefficient o

Serjik [45]

a. The force applied would be equal to the frictional force.

F = us Fn

where, F = applied force = 35 N, us = coeff of static friction, Fn = normal force = weight

35 N = us * (6 kg * 9.81 m/s^2)

us = 0.595

b. The force applied would now be the sum of the frictional force and force due to acceleration

F = uk Fn + m a

where, uk = coeff of kinetic friction

35 N = uk * (6 kg * 9.81 m/s^2) + (6kg * 0.60 m/s^2)

uk = 0.533

4 0

3 years ago

What concept or principle best explains why

andreev551 [17]

Answer:

d. Newton's first law

I hope this helps you

5 0

3 years ago

Compare the gravitational acceleration on the following objects compared to the Sun using:

arsen [322]

The gravitational acceleration of White dwarf compared to Sun is 13,675.86.

The gravitational acceleration of Neutron star compared to Sun is 6.79 x 10⁻²⁴.

The gravitational acceleration of Star Betelgeuse compared to Sun is 8.5 x 10¹⁰.

<h3>Mass of the planets</h3>

Mass of sun = 2 x 10³⁰ kg

Mass of white dwarf = 2.765 x 10³⁰ kg

Mass of Neutron star = 5.5 x 10¹² kg

Mass of star Betelgeuse = 2.188 x 10³¹ kg

<h3>Radius of the planets</h3>

Radius of sun = 696,340 km

Radius of white dwarf = 7000 km

Radius of Neutron star = 11 km

Radius of star Betelgeuse = 617.1 x 10⁶ km

<h3>Gravitational acceleration of White dwarf compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.765 \times 10^{30}}{2\times 10^{30}} \times [\frac{696,340,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 13,675.86$

<h3>Gravitational acceleration of Neutron star compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{5.5 \times 10^{12}}{2\times 10^{30}} \times [\frac{11,000}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 6.79\times 10^{-24}$

<h3>Gravitational acceleration of Star Betelgeuse compared to Sun</h3>

$\frac{g(star)}{g(sun)} = \frac{M(star)}{M(sun)} \times [\frac{R(sun)}{R(star)} ]^2\\\\\frac{g(star)}{g(sun)} = \frac{2.188 \times 10^{31}}{2\times 10^{30}} \times [\frac{617.1 \times 10^9}{7,000,000} ]^2\\\\\frac{g(star)}{g(sun)} = 8.5\times 10 ^{10}$

Learn more about acceleration due to gravity here: brainly.com/question/88039

3 0

3 years ago

At 7.0°c, the volume of a gas is 49 ml. at the same pressure, its volume is 74 ml at what temperature?

Rashid [163]

So i converted everything first;
<span>7.0 C ---> 280 K </span>
<span>49 mL---> 0.049 L </span>
<span>74mL---> 0.074 L </span>
<span>THEN I tried setting it up by the combined law formula which is P1V1/T1=P2V2/T2 </span>

8 0

3 years ago