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2 years ago

A tank is filled with an ideal gas at 400 K and pressure of 1.00 atm.

Physics

1 answer:

bekas [8.4K]2 years ago

6 0

To find the temperature it is necessary to use the expression and concepts related to the ideal gas law.

Mathematically it can be defined as

$PV=nRT$

Where

P = Pressure

V = Volume

n = Number of moles

R = Gas constant

T = Temperature

When the number of moles and volume is constant then the expression can be written as

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

Or in practical terms for this exercise depending on the final temperature:

$T_2 = \frac{P_2T_1}{P_1}$

Our values are given as

$T_1 = 400K\\P_1 = 1atm\\P_2 = 2atm$

Replacing

$T_2 = \frac{(2)(400)}{1}\\T_2 = 800K$

Therefore the final temperature of the gas is 800K

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Find the voltage drop (in volt) along a 93.4 meter long 10 gauge copper wire carrying acurrent of 72.5 A. The diameter of a 10 g

Sophie [7]

Answer: 5.41 V

Explanation:in order to explain this result we have to use the Ohm law given by:

ΔV=R*I where R is the resistance which is equal R= ρ*L/A . ρ is the resistivity, L the length of the wire and A is the cross section. I is the current.

Then we have

ΔV=ρ*L*I/A= 1.68 * 10^-8 Ωm*93.4 m*72.5A/2.1* 10^-5 m^2=5.41 V

3 0

2 years ago

At launch a rocket ship weighs 4.5 million pounds. When it is launched from rest, it takes 8.00 s to reach 161 km/h; at the end

SpyIntel [72]

Answer:

$5.590278\ m/s^2$

$7.74038461538\ m/s^2$

178.888896 m

12790.56 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

$v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{\dfrac{161}{3.6}-0}{8}\\\Rightarrow a=5.590278\ m/s^2$

The acceleration is $5.590278\ m/s^2$

$v=u+at\\\Rightarrow a=\dfrac{v-u}{t}\\\Rightarrow a=\dfrac{\dfrac{1610}{3.6}-\dfrac{161}{3.6}}{60-8}\\\Rightarrow a=7.74038461538\ m/s^2$

The acceleration is $7.74038461538\ m/s^2$

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=0\times t+\dfrac{1}{2}\times 5.590278\times 8^2\\\Rightarrow s=178.888896\ m$

Distance traveled in the first 8 seconds is 178.888896 m

$s=ut+\dfrac{1}{2}at^2\\\Rightarrow s=\dfrac{161}{3.6}\times 52+\dfrac{1}{2}\times 7.74038461538\times 52^2\\\Rightarrow s=12790.56\ m$

Distance traveled during 8-60 second interval is 12790.56 m

6 0

3 years ago

Assume the acceleration of the object is a(t) = −9.8 meters per second per second. (Neglect air resistance.) A baseball is throw

wel

Answer:8.1 m

Explanation:

Given

ball is launched from height of 3 m

initial velocity $u=10 m/s$

considering the ball is thrown vertically upward

Using $v^2-u^2=2 as$

where,

u=initial velocity

v=final Velocity

a=acceleration

s=distance

At maximum height final velocity will be zero

$v^2-u^2=2 gs$

$0-10^2=2(-9.8)\cdot h$

$h=5.10 m$

Therefore maximum height w.r.t ground is $3+5.10=8.10 m$

8 0

2 years ago

A spherical, non-conducting shell of inner radius = 10 cm and outer radius = 15 cm carries a total charge Q = 13 μC distributed

Vaselesa [24]

Answer:

E = 1580594.95 N/C

Explanation:

To find the electric field inside the the non-conducting shell for r=11.2cm you use the Gauss' law:

$\int EdS=\frac{Q_{in}}{\epsilon_o}$ (1)

dS: differential of the Gaussian surface

Qin: charge inside the Gaussian surface

εo: dielectric permittivity of vacuum = 8.85 × 10-12 C2/N ∙ m2

The electric field is parallel to the dS vector. In this case you have the surface of a sphere, thus you have:

$\int EdS=ES=E(4\pi r^2)$ (2)

Qin is calculate by using the charge density:

$Q_{in}=V_{in}\rho=\frac{4}{3}(r^3-a^3)\rho$ (3)

Vin is the volume of the spherical shell enclosed by the surface. a is the inner radius.

The charge density is given by:

$\rho=\frac{Q}{V}=\frac{13*10^{-6}C}{\frac{4}{3}\pi((0.15m)^3-(0.10m)^3)}\\\\\rho=1.30*10^{-3}\frac{C}{m^3}$

Next, you use the results of (3), (2) and (1):

$E(4\pi r^2)=\frac{4}{3\epsilon_o}(r^3-a^3)\rho\\\\E=\frac{\rho}{3\epsilo_o}(r-\frac{a^3}{r^2})$

Finally, you replace the values of all parameters, and for r = 11.2cm = 0.112m you obtain:

$E=\frac{1.30*10^{-3}C/m^3}{3(8.85*10^{-12}C^2/Nm^2)}((0.112m)-\frac{(0.10)^3}{(0.112m)^2})\\\\E=1,580,594.95\frac{N}{C}$

hence, the electric field is 1580594.95 N/C

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2 years ago

PLEASE HELP ME Color corresponds to the ______________ of light waves. wave speed cycles wavelength

lorasvet [3.4K]

Wavelength. Each wavelength is a certain color. For instance, shorter wavelengths (like 470nm) will be blue or violet, while longer wavelengths (like 650nm) will be red. Hope this helps! :)

5 0

3 years ago