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3 years ago

What is a change that will not affect the pressure in a container

Physics

2 answers:

Dimas [21]3 years ago

7 0

Answer:

changing the material that the fluids container is made of.

Lubov Fominskaja [6]3 years ago

7 0

<u>A change that will not affect the pressure in a container:</u>

The factors that affect the pressure inside a container are PVT (Pressure, Volume and Temperature). Apart from these three factors, changing or tweaking any other factor like, changing the material using which the container was made, etc would not affect the pressure.

In this case of a container, a change in pressure(adding or removing air from the container), volume (changing the amount of fluid present in the container), temperature (changing the temperature in which the container is kept) would all affect the pressure inside the container while something like, changing the material using which the container was made, would not change the pressure.

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Which equation describes the cosines function for right triangle?

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3 years ago

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Answer:

The Internal energy of the gas did not change

Explanation:

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3 years ago

If you are six feet tall how far back from a 3 foot mirror do you have to stand in order to see yourself completely?

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3 years ago

Suppose that the sound level of a conversation is initially at an angry 70 db and then drops to a soothing 50 db. assuming that

stepan [7]

Angry sound level = 70 db
Soothing sound level = 50 db
Frequency, f = 500 Hz
Assuming speed of sound = 345 m/s
Density (assumed) = 1.21 kg/m^3
Reference sound intensity, Io = 1*10^-12 w/m^2

Part (a): Initial sound intensity (angry sound)
10log (I/Io) = Sound level
Therefore,
For Ia = 70 db
Ia/(1*10^-12) = 10^(70/10)
Ia = 10^(70/10)*10^-12 = 1*10^-5 W/m^2

Part (b): Final sound intensity (soothing sound)
Is = 50 db
Therefore,
Is = 10^(50/10)*10^-12 = 18*10^-7 W/m^2

Part (c): Initial sound wave amplitude
Now,
I (W/m^2) = 0.5*A^2*density*velocity*4*π^2*frequency^2

Making A the subject;
A = Sqrt [I/(0.5*density*velocity*4π^2*frequency^2)]

Substituting;
A_initial = Sqrt [(1*10^-5)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-8 m = 69.7 nm

Part (d): Final sound wave amplitude
A_final = Sqrt [(1*10^-7)/(0.5*1.21*345*4π^2*500^2)] = 6.97*10^-9 m = 6.97 nm

4 0

3 years ago

A 63 gg ice cube can slide without friction up and down a 30∘30∘ slope. The ice cube is pressed against a spring at the bottom o

tatiyna

Given Information:

slope angle = θ = 30°

spring constant = k = 30 N/m

compressed length = x = 10 cm = 0.10 m

mass of ice cube = m = 63 g = 0.063 kg

Required Information:

distance traveled by ice cube = d = ?

Answer:

distance traveled by ice cube = 0.48 m

Explanation:

Using the the principle of conversation of energy, the following relation holds true for this case,

mgh = 1/2*kx²

h = 1/2*kx²/mg

Where h is the height of the slope, m is the mass of ice cube, k is the spring constant and x is the compressed length o the spring and g is gravitational acceleration.

h = 1/2*kx²/mg

h = 1/2*30(0.1)²/0.063*9.8

h = 0.242 m

From trigonometry ratio,

sinθ = h/d

d = h/sinθ

d = 0.242/sin(30)

d = 0.48 m

Therefore, when the ice cube is released, it will travel a total distance 0.48 up the slope before reversing direction.

3 0

3 years ago