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Minchanka [31]
3 years ago
14

Imagine that you have a 5.50 l gas tank and a 4.00 l gas tank. you need to fill one tank with oxygen and the other with acetylen

e to use in conjunction with your welding torch. if you fill the larger tank with oxygen to a pressure of 145 atm , to what pressure should you fill the acetylene tank to ensure that you run out of each gas at the same time? assume ideal behavior for all gases.
Physics
1 answer:
miv72 [106K]3 years ago
5 0

To solve this problem, we must first assume that the two gases acetylene and oxygen to act like an ideal gas so that we can use the ideal gas equation:

P V = n R T

where,

P is the pressure inside the tank, V is the volume of the tank, n is the number of moles of gas, R is the universal gas constant and T is the absolute temperature

 

Now to ensure that we run out of each gas at the same time, we must also assume that n and T are equal for both gases, therefore:

P V = constant

 

So now we can equate the PV for the two gases to find for the pressure of the other tank:

145 atm (5.50 L) = P (4.00 L)

P = 199.375 atm

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when a temparature of a coin is 75°C, the coin's diameter increases. if the original diameter of a coin is 1.8*10^-2 m and its c
andrey2020 [161]

Answer:

ΔD = 2.29 10⁻⁵ m

Explanation:

This is a problem of thermal expansion, if the temperature changes are not very large we can use the relation

          ΔA = 2α A ΔT

the area is

         A = π r² = π D² / 4

we substitute

         ΔA = 2α π D² ΔT/4

as they do not indicate the initial temperature, we assume that ΔT = 75ºC

    α = 1.7 10⁻⁵ ºC⁻¹

we calculate

          ΔA = 2 1.7 10⁻⁵ pi (1.8 10⁻²) ² 75/4

          ΔA = 6.49 10⁻⁷ m²

by definition

           ΔA = A_f- A₀

           A_f = ΔA + A₀

           A_f = 6.49 10⁻⁷ + π (1.8 10⁻²)² / 4

           A_f = 6.49 10⁻⁷ + 2.544 10⁻⁴

           A_f = 2,551 10⁻⁴ m²

the area is

           A_f = π D_f² / 4

           A_f = \sqrt{4  A_f /\pi }

           D_f = \sqrt{4 \ 2.551 10^{-4} /\pi }

           D_f = 1.80229 10⁻² m

the change in diameter is

           ΔD = D_f - D₀

           ΔD = (1.80229 - 1.8) 10⁻² m

           ΔD = 0.00229 10⁻² m

           ΔD = 2.29 10⁻⁵ m

5 0
3 years ago
To calculate the change in kinetic energy, you must know the force as a function of _______. The work done by the force causes t
aliya0001 [1]

To calculate the change in kinetic energy, you must know the force as a function of position. The work done by the force causes the kinetic energy change

Explanation:

The work-energy theorem states that the change in kinetic enegy of an object is equal to the work done on the object:

\Delta E_k = W

where the work done is the integral of the force over the position of the object:

W=\int F(x) dx

As we see from the formula, the magnitude of the force F(x) can be dependent from the position of the object, therefore in order to solve correctly the integral and find the work done on the object, it is required to know the behaviour of the force as a function of the position, x.

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3 years ago
A neutron star and a black hole are 3.34 x 1012 m from each other at a certain point in their orbit. The neutron star has a mass
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Answer:

  F=1.65 x 10²⁶ N

Explanation:

Given that

Distance ,R= 3.34 x 10¹² m

Mass m₁= 2.78 x 10³⁰ kg

Mass ,m₂= 9.94 x 10³⁰ kg

we know that gravitational force F given as

F=G\dfrac{m_1m_2}{R^2}

G=Constant

G=6.67 x 10⁻¹¹ Nm²/kg²

Now by putting the values

F=6.67\times 10^{-11}\times \dfrac{2.78\times 10^{30}\times 9.94\times 10^{30}}{(3.34\times 10^{12})^2}\ N

F=1.65 x 10²⁶ N

Therefore the force between these two mass will be 1.65 x 10²⁶ N.

5 0
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Calculate the energy of a photon having a wavelength in thefollowing ranges.(a) microwave, with λ = 50.00 cmeV(b) visible, with
IgorLugansk [536]

(a) 2.5\cdot 10^{-6}eV

The energy of a photon is given by:

E=\frac{hc}{\lambda}

where

h=6.63\cdot 10^{-34}Js is the Planck constant

c=3\cdot 10^8 m/s is the speed of light

\lambda is the wavelength

For the microwave photon,

\lambda=50.00 cm = 0.50 m

So the energy is

E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{0.50 m}=4.0\cdot 10^{-25} J

And converting into electronvolts,

E=\frac{4.0\cdot 10^{-25}J}{1.6\cdot 10^{-19} J/eV}=2.5\cdot 10^{-6}eV

(b) 2.5 eV

For the energy of the photon, we can use the same formula:

E=\frac{hc}{\lambda}

For the visible light photon,

\lambda=500 nm = 5 \cdot 10^{-7}m

So the energy is

E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5\cdot 10^{-7} m}=4.0\cdot 10^{-19} J

And converting into electronvolts,

E=\frac{4.0\cdot 10^{-19}J}{1.6\cdot 10^{-19} J/eV}=2.5 eV

(c) 2500 eV

For the energy of the photon, we can use the same formula:

E=\frac{hc}{\lambda}

For the x-ray photon,

\lambda=0.5 nm = 5 \cdot 10^{-10}m

So the energy is

E=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{5\cdot 10^{-10} m}=4.0\cdot 10^{-16} J

And converting into electronvolts,

E=\frac{4.0\cdot 10^{-16}J}{1.6\cdot 10^{-19} J/eV}=2500 eV

6 0
3 years ago
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