Ask question

Ask question

All categories

4 years ago

9

A wave with a frequency of 190 hz and a wavelength of 28.0 cm is traveling along a cord. the maximum speed of particles on the c

ord is the same as the wave speed. part a what is the amplitude of the wave

1 answer:

raketka [301]4 years ago

8 0

Check the attached file for the answer.

You might be interested in

Does the sun fuse hydrogen into helium or vice versa?

NikAS [45]

Hydrogen has the smaller, lighter, simpler nucleus.
Helium has the larger, heavier, more complex nucleus.
Hydrogen nuclei are fused to form helium nuclei.
When that happens, energy is released.

6 0

3 years ago

If The density of this stainless steel is7.85 g/cm3,specific heatis 0.5 J/g.K, melting pointis 1673K, heat of fusion s0.260J/kg.

kodGreya [7K]

Answer:

$\Delta H=687.4 J$

Explanation:

Hello!

In this case, for this melting process, we can identify two sub-processes in order to take the stainless steel from solid to liquid:

1. Heat up from 298.15 K to 1673 K.

2. Undergo the phase transition.

Both process have an associated enthalpy as shown below:

$\Delta H_1=1g*0.5\frac{J}{g*K} (1673K-298.15K)=687.4J$

$\Delta H_2=0.001kg*\frac{0.260J}{kg} =0.00026J$

Therefore, the required heat is:

$\Delta H=\Delta H_1+\Delta H_2\\\\\Delta H=687.4J+0.00026J\\\\\Delta H=687.4J$

Notice the problem is not providing neither the mass or volume, that is why we assumed the mass is 1 g; however, it can be changed to the mass you are given.

Best regards!

4 0

3 years ago

What is the force required to accelerate a baseball with a mass of 0.145 kilograms and an acceleration of 35.00 meters/second sq

Paladinen [302]

Answer:5.075N

Explanation:

Mass=0.145kg

Acceleration=35m/s^2

Force=mass x acceleration

Force=0.145 x 35

Force=5.075N

3 0

3 years ago

You wiggle a string,that is fixed to a wall at the other end, creating a sinusoidalwave with a frequency of 2.00 Hz and an ampli

FinnZ [79.3K]

Answer:

Explanation:

A general wave function is given by:

$f(x,t)=Acos(kx-\omega t)$

A: amplitude of the wave = 0.075m

k: wave number

w: angular frequency

a) You use the following expressions for the calculation of k, w, T and λ:

$\omega = 2\pi f=2\pi (2.00Hz)=12.56\frac{rad}{s}$

$k=\frac{\omega}{v}=\frac{12.56\frac{rad}{s}}{12.0\frac{m}{s}}=1.047\ m^{-1}$

$T=\frac{1}{f}=\frac{1}{2.00Hz}=0.5s\\\\\lambda=\frac{2\pi}{k}=\frac{2\pi}{1.047m^{-1}}=6m$

b) Hence, the wave function is:

$f(x,t)=0.075m\ cos((1.047m^{-1})x-(12.56\frac{rad}{s})t)$

c) for x=3m you have:

$f(3,t)=0.075cos(1.047*3-12.56t)$

d) the speed of the medium:

$\frac{df}{dt}=\omega Acos(kx-\omega t)\\\\\frac{df}{dt}=(12.56)(1.047)cos(1.047x-12.56t)$

you can see the velocity of the medium for example for x = 0:

$v=\frac{df}{dt}=13.15cos(12.56t)$

7 0

3 years ago

The density of gold is 19 300kg/m cube. what is the mass of gold cube with the length 0.2015m?

Sergeeva-Olga [200]

Answer:

157.9 kg

Explanation:

Density: This can be defined as the ratio of the mass of a body and it's volume.

The S.I unit of density is kg/m³.

From the question,

Density = Mass/volume

D = m/v............................ Equation 1

Where D = Density of gold, m = mass of gold, v = volume of gold.

make m the subject of the equation

m = Dv.................... Equation 2

Since the gold is a cube,

v = l³................... Equation 3

Where l = length of the gold cube.

Substitute equation 3 into equation 2

m = Dl³............... Equation 4

Given: D = 19300 kg/m³, l = 0.2015 m

Substitute into equation 4

m = 19300(0.2015)³

m = 157.9 kg.

4 0

3 years ago