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Alborosie
4 years ago
7

If 41.5 mol of an ideal gas occupies 86.5 L at 27.00 °C, what is the pressure of the gas?

Physics
2 answers:
Dmitry [639]4 years ago
7 0
PV=nRT
(P)(86.5)=(41.5)(.08206)(300.15)
(P)(86.5)=(1022.157824)
P=11.81685345 atm
AlexFokin [52]4 years ago
5 0

YUP!!and I think its important to note that when using this value of R

  • <em>It is crucial to match your units of Pressure, Volume, number of mole, and Temperature with the units of R. If you use the first value of </em>R,<em> which is </em>0.082057 L <u><em>your unit for Pressure must be atm, Volume must be Liter, & Temperature must be Kelvin.</em></u>

<u><em /></u>

Clarity in Chemistry is crucial.

*_^

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What is a reasonable estimate of the average kinetic energy of an athlete during a 100 m race that takes 10s?
arlik [135]
The average weight of an athlete should be around 60kg so from the information that the athlete can run 100m in 10s, we can calculate that their average speed is 10m/s. Using the kinetic energy formula, Ek = 1/2mv^2 we can calculate the kinetic energy using 60kg as the mass.

(1/2)(60)(10^2) = Ek

Ek= 3000J
7 0
3 years ago
A single sports fan is capable of yelling at an intensity level of 80 decibels from a given distance. If 10,000 similar fans wer
AleksandrR [38]

Answer:

The correct answer is part 'c' 160 dB

Explanation:

When noise levels of different intensities are superimposed the resultant intensity is given by the equation

L_{avg}=20log(\frac{1}{N}\sum_{i=1}^{N}(10^{\frac{L_{i}}{20}}))\\

where,

L_{i} is the intensity of a general sound level

Since we have 10000 fans each producing sound of 80dB thus the resultant intensity is given using the above formula as

L_{avg}=20log(\frac{1}{10000}\sum_{i=1}^{10000}(10^{\frac{80}{20}}))\\

\therefore L_{eq}=160dB

7 0
3 years ago
A 20.0-µF capacitor is charged by a 150.0-V power supply, then disconnected from the power and connected in series with a 0.280
natali 33 [55]

Answer

given,

Capacitance of capacitor = 20.0-µF

Voltage =  150.0-V

inductance = 0.280 m H

a) the oscillation frequency of circuit

f = \dfrac{1}{2\pi}\ sqrt{\dfrac{1}{LC}}

f = \dfrac{1}{2\pi}\ sqrt{\dfrac{1}{0.280 \times 10^{-3}\times 20 \times 10^{-6}}}

f = 2126.9 Hz

b) U = \dfrac{1}{2}CV^2

   U = \dfrac{1}{2}(20\times 10^{-6})(150)^2

         U = 0.225 J

c)Current in the inductor

    V = L \dfrac{dI}{dt}

    150 = 0.28 \times 10^{-3} \dfrac{dI}{dt}

    \dfrac{dI}{dt} = 535714

instantaneous rate of change of current is equal to 535714 A/s

 

7 0
3 years ago
You are a baseball coach and watch as your star
drek231 [11]

Answer:

plzz answer

my question plzz it's important l will follow u is you answer and select as brainly least and thank it

5 0
3 years ago
In a football game the running back is running up the field. He starts from rest and runs 4 seconds with an acceleration of 1.3m
Amiraneli [1.4K]

Answer:

3.38m

Explanation:

Given parameters:

Time  = 4s

Acceleration  = 1.3m/s²

Unknown:

Magnitude of the displacement = ?

Solution:

The body starts at rest and the initial velocity is 0m/s. To solve this problem, we have to use the expression below;

    S   = Ut  + \frac{1}{2}at²

 S  = displacement

t is the time

  a is the acceleration

  U is the initial velocity

  V is the final velocity

Insert the parameters and solve;

   S = (0 x 4)  +  \frac{1}{2} x 1.3² x 4  = 3.38m

6 0
3 years ago
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