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3 years ago

12

Explain what happens to the motion of a particle as a wave passes through a medium. How is the motion of the particles like the

motion of a mass on a spring

1 answer:

bezimeni [28]3 years ago

3 0

As they vibrate, they pass the disturbance of energy to the particles next to them, which passes the energy to the particles next to them, and so on.
Energy travels down the spring and reflected like the motion of a mass on a spring.

<u>Explanation</u>:

Particle motion is the superposition of a large component due to the fluid drag toward the filter barrier and a random component due to Brownian motion.

The particles of the medium just vibrate in place. As they vibrate, they pass the unsettling influence of energy to the particles beside them, which passes the energy to the particles by them, etc.

Energy goes down the spring and reflected like the movement of a mass on a spring.

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Difference between on pitch and frequency

shepuryov [24]

Answer:

A high pitch sound corresponds to a high frequency sound wave and a low pitch sound corresponds to a low frequency sound wave. I hope I got it correct !!

8 0

3 years ago

Force → F = ( − 8.0 N ) ˆ i + ( 6.0 N ) ˆ j acts on a particle with position vector → r = ( 3.0 m ) ˆ i + ( 4.0 m ) ˆ j . What a

andrey2020 [161]

Explanation:

It is given that,

Force, $F=(-8\ N)i+(6\ N)j$

Position vector, $r=(3i+4j)\ m$

(a) The torque on the particle about the origin is given by :

$\tau=F\times r\\\\\tau=(-8i+6j)\times (3i+4j)\\\\\tau=(-50k)\ N-m$

(b) To find the angle between r and F use dot product formula as :

$F{\cdot} r=|F||r|\ \cos\theta\\\\\cos\theta=\dfrac{F{\cdot} r}{|F| |r|}\\\\\cos\theta=\dfrac{(-8i+6j){\cdot} (3i+4j)}{\sqrt{(-8)^2+6^2} \times \sqrt{3^2+4^2} }\\\\\cos\theta=\dfrac{-24+24}{\sqrt{(-8)^2+6^2} \times \sqrt{3^2+4^2} }\\\\\cos\theta=0\\\\\theta=90^{\circ}$

Hence, this is the required solution.

8 0

3 years ago

Read 2 more answers

Over a short interval the coordinate of a car in the meters isgiven by x(t) = 27t - 4.0 t3 where time t is in seconds,at the end

wariber [46]

Answer:

5. -24 m/s²

Explanation:

Acceleration: This can be defined as the rate of change of velocity.

The S.I unit of acceleration is m/s².

mathematically,

a = dv/dt ............................ Equation 1

Where a = acceleration, dv/dt = is the differentiation of velocity with respect to time.

But

v = dx(t)/dt

Where,

x(t) = 27t-4.0t³...................... Equation 2

Therefore, differentiating equation 2 with respect to time.

v = dx(t)/dt = 27-12t²............. Equation 3.

Also differentiating equation 3 with respect to time,

a = dv/dt = -24t

a = -24t .................... Equation 4

from the question,

At the end of 1.0 s,

a = -24(1)

a = -24 m/s².

Thus the acceleration = -24 m/s²

The right option is 5. -24 m/s²

4 0

3 years ago

Consider a highway composed of concrete. If the road is 4.5 km long, 30 m wide, and 0.70 m thick, what is its mass?

Aloiza [94]

Answer:

Explanation:

We need to assume that the density of the concrete is about 2350 Kg/m^3. And using the dimensions of the highway we can calculate the volume of the highway.

$vh=4500 * 30 * 0,70\\vh=94500 m^3\\den=m/v\\m=den*v\\m=2350*94500\\\\m=222075 ton$

5 0

3 years ago

A 1.0 kg piece of copper with a specific heat of cCu=390J/(kg⋅K) is placed in 1.0 kg of water with a specific heat of cw=4190J/(

Valentin [98]

Answer:

The temperature change of the copper is greater than the temperature change of the water.

Explanation:

deltaQ = mc(deltaT)

Where,

delta T = change in the temperature

m =mass

c = heat capacity

$\frac{(deltaT)_{Cu}}{(deltaT)_{w}}=\frac{4190J/kg.K}{390J/kg.K}\\ \\(deltaT)_{Cu}=10.74(deltaT)_{w}$

The temperature change in the copper is nearly 11 times the temperature change in the water.

So, the correct option is,

The temperature change of the copper is greater than the temperature change of the water.

Hope this helps!

7 0

3 years ago