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2 years ago

What happens to the energy of gas particles when an elastic collision takes place?

Physics

2 answers:

seraphim [82]2 years ago

8 0

The third option because the person above me said so

Gnesinka [82]2 years ago

7 0

Answer:

answer 3

Explanation:

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A 13 kg hanging sculpture is suspended by a 95-cm-long, 5.0 g steel wire. When the wind blows hard, the wire hums at its fundame

Artyom0805 [142]

Answer:

$f=81.96 \ Hz$

Explanation:

Givens

$L=95cm$

$m_{sculpture} =13kg$

$m_{wire}=5g$

The frequency is defined by

$f=\frac{v}{\lambda}$

Where $v$ is the speed of the wave in the string and $\lambda$ is its wave length.

The wave length is defined as $\lambda = 2L = 2(0.95m)=1.9m$

Now, to find the speed, we need the tension of the wire and its linear mass density

$v=\sqrt{\frac{T}{\mu} }$

Where $\mu=\frac{0.005kg}{0.95m}= 5.26 \times 10^{-3}$ and the tension is defined as $T=m_{sculpture} g=13kg(9.81 m/s^{2} )=127.53N$

Replacing this value, the speed is

$v=\sqrt{\frac{127.53N}{5.26 \times 10^{-3} } }=155.71 m/s$

Then, we replace the speed and the wave length in the first equation

$f=\frac{v}{\lambda}\\f=\frac{155.71 m/s}{1.9m}\\ f=81.96Hz$

Therefore, the frequency is $f=81.96 \ Hz$

3 0

2 years ago

Read 2 more answers

How much heat do you need to raise the temperature of 150 g of oxygen from -30c to -15c?

Naddika [18.5K]

The amount of heat needed to raise the temperature of a substance by $\Delta T$ is given by
$Q=m C_s \Delta T$
where
m is the mass of the substance
Cs is its specific heat capacity
$\Delta T$ is the increase in temperature

For oxygen, the specific heat capacity is approximately
$C_s = 0.92 J/(g K)$
The variation of temperature for the sample in our problem is
$\Delta T= -15^{\circ}C-(-30^{\circ} C)=+15^{\circ}C=15 K$
while the mass is m=150 g, so the amount of heat needed is
$Q=m C_s \Delta T=(150 g)(0.92 J/g K)(15 K)=2070 J$

4 0

3 years ago

Read 2 more answers

A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's in

german

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

$\Sigma F = -f = m\cdot a$ (Eq. 1)

Where:

$f$ - Kinetic friction force, measured in newtons.

$m$ - Mass of the box, measured in kilograms.

$a$ - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight ( $W = m\cdot g$ ) and uniform accelerated motion ( $v = v_{o}+a\cdot t$ ), we expand the previous expression:

$-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$

And the magnitude of the friction force exerted on the box is calculated by this formula:

$f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$ (Eq. 1b)

Where:

$W$ - Weight, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$t$ - Time, measured in seconds.

If we know that $W = 52.4\,N$ , $g = 9.807\,\frac{m}{s^{2}}$ , $v_{o} = 1.37\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ and $t = 2.8\,s$ , the magnitud of the kinetic friction force exerted on the box is:

$f = -\left(\frac{52.4\,N}{9.807\,\frac{m}{s^{2}} } \right)\cdot \left(\frac{0\,\frac{m}{s}-1.37\,\frac{m}{s} }{2.8\,s} \right)$

$f = 2.614\,N$

The magnitude of the friction force exerted on the box is 2.614 newtons.

5 0

3 years ago

A very long string (linear density 0.7 kg/m ) is stretched with a tension of 70 N . One end of the string oscillates up and down

rewona [7]

To develop this problem it is necessary to apply the concepts related to Wavelength, The relationship between speed, voltage and linear density as well as frequency. By definition the speed as a function of the tension and the linear density is given by

$V = \sqrt{\frac{T}{\rho}}$

Where,

T = Tension

$\rho =$ Linear density

Our data are given by

Tension , T = 70 N

Linear density , $\rho = 0.7 kg/m$

Amplitude , A = 7 cm = 0.07 m

Period , t = 0.35 s

Replacing our values,

$V = \sqrt{\frac{T}{\rho}}$

$V = \sqrt{\frac{70}{0.7}$

$V = 10m/s$

Speed can also be expressed as

$V = \lambda f$

Re-arrange to find \lambda

$\lambda = \frac{V}{f}$

Where,

f = Frequency,

Which is also described in function of the Period as,

$f = \frac{1}{T}$

$f = \frac{1}{0.35}$

$f = 2.86 Hz$

Therefore replacing to find $\lambda$

$\lambda = \frac{10}{2.86}$

$\lambda = 3.49m$

Therefore the wavelength of the waves created in the string is 3.49m

3 0

3 years ago

Un alumno menciona que al abrir la ventana de su casa sintió cómo el frío ingresaba a su cuerpo. Menciona cuál es la verdadera r

stepan [7]

Answer:

My believe the answer is

A.) or B.)

Explanation:

Here is why I think A is the answer.

If we use the process of elimination, it would look like this,

a) Porque el aire tiene una temperatura menor que la de su cuerpo; por eso se propaga más rápido.

<em>This makes sense because we all know in winter the weather is very cold and freezing.</em>

b) Porque la temperatura de su cuerpo, siente el aire frio que entra por la ventana.

<em>I feel like this answer is the question, but it could also be an answer, sorry, I'm a little uncertain.</em>

c) Porque el calor de su cuerpo se propaga al medio ambiente, al ser la temperatura del niño mayor que la del aire exterior.

<em>This answer has nothing to do with the question, plus it is very false, our body heat is not enough to overcome the very cold temperature from outside.</em>

d) Porque la temperatura del aire es igual a la temperatura del cuerpo.

<em>This is false because again our body heat is not even compared to the freezing cold temperatures from the winter.</em>

<em />

<h2>Well, have a nice rest of the day!</h2><h3>ba baiii!</h3>

3 0

2 years ago