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lakkis [162]
3 years ago
12

Neon atoms at 245 K pass through a fan that gives each mole of neon gas an additional kinetic energy of 16.0 J. Part A What is t

he average temperature of the neon atoms immediately after coming through the fan
Physics
1 answer:
hjlf3 years ago
7 0

Answer:

246.28 K

Explanation:

The total energy of one mole of gas molecules can be calculated by the formula given below

E = \frac{3}{2}\times R\times T

Where R is gas constant and T is absolute temperature.

Put the value of R as 8.314 and temperature as 245 , we get

E = \frac{3}{2}\times 8.314\times 245

= 3055.4 J

Add 16 j to it

Total energy of gas molecules = 3055.4 + 16 = 3071.4 J.

If T be the temperature after addition of energy then

\frac{3}{2}\times 8.314\times T = 3071.4

T =\frac{2\times 3071.4}{3\times 8.314}

T = 246.28 K

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A 15 kg cart is pushed on a frictionless surface from rest horizontally by a 30 N force. What is the cart's acceleration?
Rainbow [258]

Answer:

<em>a. The cart's acceleration is 2 m/s^2</em>

<em>b. The cart will travel 100 m</em>

<em>c. The speed is 20 m/s</em>

Explanation:

a. The acceleration of the cart can be calculated using Newton's second law:

F = m.a

Solving for a:

\displaystyle a=\frac{F}{m}

The cart has a mass of m=15 Kg and is applied a net force of F=30 N, thus:

\displaystyle a=\frac{30}{15}

a=2\ m/s^2

b.

Now we use kinematics to find the distance and speed:

\displaystyle x = v_o.t+\frac{at^2}{2}

The cart starts from rest (vo=0). The distance traveled in t=10 seconds is:

\displaystyle x = 0*10+\frac{2*10^2}{2}

x = 100\ m

The cart will travel 100 m

c.

The final speed is calculated by:

v_f=0+2*10=20\ m/s

The speed is 20 m/s

5 0
3 years ago
Please help...........................
Marianna [84]
Hi!

Neutrons are neutral, which means they don't exactly have an electrical charge. It's because of this neutral charge that it is represented with a '0'. 

On the other hand, protons and electrons <em>do </em>have electrical charges. Electrons flow around the outside of the nucleus, with a negative charge.

Protons are stored in the nucleus with the neutrons, holding a positive charge.

Hopefully, this helps! =)
6 0
4 years ago
Read 2 more answers
Please help
miv72 [106K]

Answer: The verb phrase can be found in the last sentence(?

3 0
2 years ago
A sound wave with a waveenght of 2.5 meters traves 660 meters in 2 seconds calculate the frequemcy of the wave to cacuation the
finlep [7]

The frequency of the wave is 132 Hz

Explanation:

To calculate the speed of the wave, we can use the following formula:

v = \frac{d}{t}

where

d is the distance travelled by the wave

t is the time elapsed

For the sound wave in this problem, we have:

d = 660 m is the distance travelled

t = 2 s is the time interval considered

Substituting and solving for v, we find the speed of the sound wave:

v=\frac{660}{2}=330 m/s

Now we can calculate the frequency of the wave by using the wave equation:

v=f\lambda

where

v = 330 m/s is the speed of the wave

\lambda=2.5 m is the wavelength

f is the frequency

Solving for f, we find:

f=\frac{v}{\lambda}=\frac{330}{2.5}=132 Hz

Learn more about wavelength and frequency:

brainly.com/question/5354733

brainly.com/question/9077368

#LearnwithBrainly

7 0
3 years ago
A 1-meter-long wire consists of an inner copper core with a radius of 1.0 mm and an outer aluminum sheathe, which is 1.0 mm thic
Mazyrski [523]

Answer:

The total resistance of the wire is = 1.917\times10^{-3}

Explanation:

Since the wires will both be in contact with the voltage source at the same time and the current flows along in their length-wise direction, the two wires will be considered to be in parallel.

Hence, for resistances in parallel, the total resistance, R_{Total}

\frac{1}{R_{Total}}  =\frac{1}{R_{cu}  }+\frac{1}{R_{al}}

Parameters given:

Length of wire = 1 m

Cross sectional area of copper A_{cu}= \pi r^{2}= \pi \times (1\times 10^{-3}  )^{2} =3.142\times10^{-6} m^{2}

Cross sectional area of aluminium wire  

A_{al}= \pi( R^{2}-r^{2})\\\\ = \pi \times [ (2\times 10^{-3}  )^{2}-(1\times 10^{-3}  )^{2}] =9.42\times10^{-6} m^{2}\\

Resistivity of copper \rho _{cu}=1.7\times 10^{-8}  \Omega .m

Resistivity of Aluminium \rho _{al}=2.8\times 10^{-8}  \Omega .m

Resistance of copper R_{cu}= \frac{\rho_{cu} \times l}{A_{cu} }  =\frac{1.7\times 10^{-8} \times 1}{3.142\times10^{-6} } =5.41\times 10^{-3}\Omega

Resistance of aluminium R_{al}= \frac{\rho_{al} \times l}{A_{al} }  =\frac{2.8\times 10^{-8} \times 1}{9.42\times10^{-6} } =2.97\times 10^{-3}\Omega

The total resistance of the wire can be obtained as follows;

\frac{1}{R_{Total}}  =\frac{1}{5.41\times10^{-3}  }+\frac{1}{2.97\times10^{-3}}=521.52\frac{1}{\Omega}

R_{Total}= 1.917\times 10^{-3}\Omega

∴ The total resistance of the wire = 1.917\times 10^{-3}\Omega

4 0
3 years ago
Read 2 more answers
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