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2 years ago

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20. GP A sinusoidal wave traveling in the negative x direction (to the left) has an amplitude of 20.0cm, a wavelength of 35.0cm,

and a frequency of 12.0Hz. The transverse position of an element of the medium at t=0, x=0 is y= 3.00 cm, and the element has a positive velocity here. We wish to find an expression for the wave function describing this wave. the wave Find (d) the angular frequency omega and

1 answer:

umka21 [38]2 years ago

7 0

The angular frequency of the wave is determined as 75.4 rad/s.

<h3>What is wave function?</h3>

A wave function is a mathematical equation for the motion of the wave.

y(x, t) = A sin(kx + ωt + Φ)

where;

ω is angular speed
k is angular wavenumber
Φ is phase angle

<h3>What is angular frequency?</h3>

The angular frequency is the angular displacement of any wave element per unit of time or the rate of change of the waveform phase.

<h3>Angular frequency</h3>

ω = 2πf

ω = 2π(12)

ω = 75.4 rad/s

Thus, the angular frequency of the wave is determined as 75.4 rad/s.

Learn more about angular frequency here: brainly.com/question/3654452

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Use the table below to answer the following questions. Substance Specific Heat (J/g•°C) water 4.179 aluminum 0.900 copper 0.385

lbvjy [14]

1. $-8.78 \cdot 10^5 J$

The energy lost by the water is given by:

$Q=m C_s \Delta T$

where

m = 3.0 kg = 3000 g is the mass of water

Cs = 4.179 J/g•°C is the specific heat

$\Delta T=10.0C-80.0C=-70.0 C$ is the change in temperature

Substituting,

$Q=(3000 g)(4.179 J/gC)(-70.0 C)=-8.78 \cdot 10^5 J$

2. $3.24 \cdot 10^4 J$

The energy added to the aluminium is given by:

$Q=m C_s \Delta T$

where

m = 0.30 kg = 300 g is the mass of aluminium

Cs = 0.900 J/g•°C is the specific heat

$\Delta T=150.0 C-30.0C =120.0 C$ is the change in temperature

Substituting,

$Q=(300 g)(0.900 J/gC)(120.0 C)=3.24 \cdot 10^4 J$

3a. $-5.6^{\circ}C$

The temperature change of the water is given by

$\Delta T=\frac{Q}{m C_s}$

where

$Q = -232 kJ=-2.32\cdot 10^5 J$ is the heat lost by the water

$m=10.0 kg=10000 g$ is the mass of water

Cs = 4.179 J/g•°C is the specific heat

Substituting,

$\Delta T=\frac{-2.32\cdot 10^5 J}{(10000g)(4.179 J/gC)}=5.6^{\circ}C$

3b. $+10.2^{\circ}C$

The temperature change of the copper is given by

$\Delta T=\frac{Q}{m C_s}$

where

$Q = 1.96 kJ=1960$ is the heat added to the copper

$m= 500 g$ is the mass of copper

Cs = 0.385 J/g•°C is the specific heat

Substituting,

$\Delta T=\frac{1960 J}{(500g)(0.385 J/gC)}=10.2^{\circ}C$

4. 42.9 g

The mass of the water sample is given by

$m=\frac{Q}{C_S \Delta T}$

where

$Q=4300 J$ is the heat added

$\Delta T=39 C-15 C=24C$ is the temperature change

Cs = 4.179 J/g•°C is the specific heat

Substituting,

$m=\frac{4300 J}{(4.179 J/gC)(24 C)}=42.9 g$

5. 115.5 J

The heat used to heat the copper is given by:

$Q=m C_s \Delta T$

where

m = 5.0 g is the mass of copper

Cs = 0.385 J/g•°C is the specific heat

$\Delta T=80.0 C-20.0C =60.0 C$ is the change in temperature

Substituting,

$Q=(5.0 g)(0.385 J/gC)(60.0 C)=115.5 J$

6. 0.185 J/g•°C

The specific heat of iron is given by:

$C_s = \frac{Q}{m \Delta T}$

where

Q = -47 J is the heat released by the iron

m = 10.0 g is the mass of iron

$\Delta T=25.0-50.4 C=-25.4 C$ is the change in temperature

Substituting,

$C_s = \frac{-47 J}{(10.0 g)(-25.4 C)}=0.185 J/gC$

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Two hundred and thirsty years

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A singly charged positive ion has a mass of 3.46 × 10−26 kg. After being accelerated through a potential difference of 215 V the

jasenka [17]

Answer:

1.8 cm

Explanation:

$m$ = mass of the singly charged positive ion = 3.46 x 10⁻²⁶ kg

$q$ = charge on the singly charged positive ion = 1.6 x 10⁻¹⁹ C

$\Delta V$ =Potential difference through which the ion is accelerated = 215 V

$v$ = Speed of the ion

Using conservation of energy

Kinetic energy gained by ion = Electric potential energy lost

$(0.5) m v^{2} = q \Delta V\\(0.5) (3.46\times10^{-26}) v^{2} = (1.6\times10^{-19}) (215)\\(1.73\times10^{-26}) v^{2} = 344\times10^{-19}\\v = 4.5\times10^{4} ms^{-1}$

$r$ = Radius of the path followed by ion

$B$ = Magnitude of magnetic field = 0.522 T

the magnetic force on the ion provides the necessary centripetal force, hence

$qvB = \frac{mv^{2} }{r} \\qB = \frac{mv}{r}\\r =\frac{mv}{qB}\\r =\frac{(3.46\times10^{-26})(4.5\times10^{4})}{(1.6\times10^{-19})(0.522)}\\r = 0.018 m \\r = 1.8 cm$

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3 years ago