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kari74 [83]
3 years ago
15

What is the new pressure of 150 ml of a gas that is compressed to 50 ml when the original pressure was 3.0 atm and the temperatu

re is held constant?
Physics
1 answer:
Firlakuza [10]3 years ago
5 0

Answer: 9.0 atm

Explanation:

To calculate the new pressure, we use the equation given by Boyle's law. This law states that pressure is directly proportional to the volume of the gas at constant temperature.

The equation given by this law is:

P_1V_1=P_2V_2

where,

P_1\text{ and }V_1 are initial pressure and volume.

P_2\text{ and }V_2 are final pressure and volume.

We are given:

P_1=3.0atm\\V_1=150mL\\P_2=?mmHg\\V_2=50mL

Putting values in above equation, we get:

3.0\times 150mL=P_2\times 50mL\\\\P_2=9.0atm

Thus new pressure of 150 ml of a gas that is compressed to 50 ml is 9.0 atm

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Two forces, one of 100 ponds and the other 150 pounds act on the same object, at angles of 20°and 60°, respectively, withthe pos
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<h2>Resultant is 235.54 pounds at an angle 44.16° to X axis.</h2>

Explanation:

Forces are 100 pound and 150 pound and angles with x axis are 20°and 60°.

That is force 1 is 100 pound with x axis at 20°

           F₁ = 100 cos 20 i  +  100 sin 20 j

           F₁ = 93.97 i  +  34.20 j          

That is force 2 is 150 pound with x axis at 60°

           F₂ = 150 cos 60 i  +  150 sin 60 j

           F₂ = 75 i  +  129.90 j  

F₁ +  F₂ =  93.97 i  +  34.20 j + 75 i  +  129.90 j

F₁ +  F₂ =  168.97 i  +  164.10 j

\texttt{Magnitude = }\sqrt{168.97^2+164.10^2}\\\\\texttt{Magnitude = }235.54pounds\\\\\texttt{Angle = }tan^{-1}\left ( \frac{164.10}{168.97}\right )\\\\\texttt{Angle = }44.16^0

Resultant is 235.54 pounds at an angle 44.16° to X axis.

6 0
3 years ago
if somebody swim to the right for 30 meters then swims to left for 15 meters and then swims back to the right for 50 meters what
zubka84 [21]

50m

Explanation:

Displacement is the length of path traveled which is measured from start to the finishing of the path.

    Analysis of the journey;

 Starts from:

 0                                    30m                                              from right

                  15m                                                                       to left

                                                                       50m              to right

                 

The displacement is 50m from the starting point.

Distance is total path traveled and for this problem it is 30+ 15 + 50 = 95m

learn more:

displacement brainly.com/question/5461768

#learnwithBrainly

8 0
3 years ago
Jonathan has six strings that are the same thickness and are all the same material. He cuts the strings to different lengths and
harkovskaia [24]
The shortest string will have the highest pitch.

7 0
3 years ago
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A planet is discovered orbiting around a star in the galaxy Andromeda at four times the distance from the star as Earth is from
Dmitriy789 [7]

Answer:

Tp/Te = 2

Therefore, the orbital period of the planet is twice that of the earth's orbital period.

Explanation:

The orbital period of a planet around a star can be expressed mathematically as;

T = 2π√(r^3)/(Gm)

Where;

r = radius of orbit

G = gravitational constant

m = mass of the star

Given;

Let R represent radius of earth orbit and r the radius of planet orbit,

Let M represent the mass of sun and m the mass of the star.

r = 4R

m = 16M

For earth;

Te = 2π√(R^3)/(GM)

For planet;

Tp = 2π√(r^3)/(Gm)

Substituting the given values;

Tp = 2π√((4R)^3)/(16GM) = 2π√(64R^3)/(16GM)

Tp = 2π√(4R^3)/(GM)

Tp = 2 × 2π√(R^3)/(GM)

So,

Tp/Te = (2 × 2π√(R^3)/(GM))/( 2π√(R^3)/(GM))

Tp/Te = 2

Therefore, the orbital period of the planet is twice that of the earth's orbital period.

5 0
3 years ago
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At t=0 bullet A is fired vertically with an initial (muzzle) velocity of 450 m/s. When 3s. bullet B is fired upward with a muzzl
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Answer:

At time 10.28 s after A is fired bullet B passes A.

Passing of B occurs at 4108.31 height.

Explanation:

Let h be the height at which this occurs and t be the time after second bullet fires.

Distance traveled by first bullet can be calculated using equation of motion

s=ut+0.5at^2 \\

Here s = h,u = 450m/s a = -g and t = t+3

Substituting

h=450(t+3)-0.5\times 9.81\times (t+3)^2=450t+1350-4.9t^2-29.4t-44.1\\\\h=420.6t-4.9t^2+1305.9

Distance traveled by second bullet

Here s = h,u = 600m/s a = -g and t = t

Substituting

h=600t-0.5\times 9.81\times t^2=600t-4.9t^2\\\\h=600t-4.9t^2 \\

Solving both equations

600t-4.9t^2=420.6t-4.9t^2+1305.9\\\\179.4t=1305.9\\\\t=7.28s \\

So at time 10.28 s after A is fired bullet B passes A.

Height at t = 7.28 s

h=600\times 7.28-4.9\times 7.28^2\\\\h=4108.31m \\

Passing of B occurs at 4108.31 height.

6 0
3 years ago
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