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Over [174]
3 years ago
12

consider a sample of gas in a container on a comfortable spring day. the Celsius temperature suddenly doubles, and you transfer

the gas to a container with twice the volume of the first container. If the original pressure was 12 atm, what is a good estimate for the new pressure?
Physics
1 answer:
Tasya [4]3 years ago
7 0

Answer:

6.6 atm

Explanation:

Using the general gas law

P₁V₁/T₁ = P₂V₂/T₂

Let P₂ be the new pressure

So, P₂ = P₁V₁T₂/V₂T₁

Since V₂ = 2V₁ , P₁ = 12 atm and T₁ = 273 + t where t = temperature in Celsius

T₂ = 273 + 2t (since its Celsius temperature doubles).

Substituting these values into the equation for P₂, we have

P₂ = P₁V₁(273 + 2t)/2V₁(273 + t)

P₂ = 12(273 + 2t)/[2(273 + t)]

P₂ = 6(273 + 2t)/(273 + t)]

assume t = 30 °C on a comfortable spring day

P₂ = 6(273 + 2(30))/(273 + 30)]

P₂ = 6(273 + 60))/(273 + 30)]

P₂ = 6(333))/(303)]

P₂ = 6.6 atm

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Answer:

F'= 4F/9

Explanation:

Two small objects each with a net charge of +Q exert a force of magnitude F on each other. If r is the distance between them, then the force is given by :

F=\dfrac{kQ^2}{r^2} ...(1)

Now, if one of the objects with another whose net charge is + 4Q is replaced and also the distance between +Q and +4Q charges is increased 3 times as far apart as they were. New force is given by :

F'=\dfrac{kQ\times 4Q}{(3r)^2}\\\\F'=\dfrac{4kQ^2}{9r^2}.....(2)

Dividing equation (1) and (2), we get :

\dfrac{F}{F'}=\dfrac{\dfrac{kQ^2}{r^2}}{\dfrac{4kQ^2}{9r^2}}\\\\\dfrac{F}{F'}=\dfrac{kQ^2}{r^2}\times \dfrac{9r^2}{4kQ^2}\\\\\dfrac{F}{F'}=\dfrac{9}{4}\\\\F'=\dfrac{4F}{9}

Hence, the correct option is (d) i.e. " 4F/9"

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3 years ago
How much work is done when a 5 N force moves a block 4 m?
katen-ka-za [31]
20N•m or 20J. Work is equal to force•distance, and 5N•4m is 20N•m, or J
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A squirrel jumps into the air with a velocity of 7 m/s at an angle of 20 degrees. What is the maximum height reached by the squi
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Jill applies a force of 250 N to a machine. The machine applies a force of 25 N to an object. What is the mechanical advantage o
Vinvika [58]

Mechanical advantage is defined as the ratio of output load to the input load. The mechanical advantage of the machine will be 0.1.

<h3>What is mechanical advantage?</h3>

Mechanical advantage is a measure of the ratio of output force to input force in a system,

It is used to obtain the efficiency of forces in levers and pulleys. It is an effective way of amplifying the force in simple machines like levers.

The theoretical mechanical advantage is defined as the ratio of the force responsible for the useful work in the system to the applied force.

Given

applied force = 250 N

Output force = 25

Mechanical advantage = work output / work input

\rm{Mechanical advantage}=\frac{F_O}{F_I}

\rm{Mechanical advantage}=\frac{25}{250}

\rm{Mechanical advantage}=0.1

Hence the mechanical advantage of the machine will be 0.1

To learn more about the mechanical advantage refer to the link;

brainly.com/question/7638820

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