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ahrayia [7]
2 years ago
7

If the volume of a container increases what effect will this have on pressure? Why?

Physics
2 answers:
GaryK [48]2 years ago
7 0

Answer:

 In a container of the same volume, a greater number of molecules results in greater pressure. Gas pressure increases as container volume decreases.

Hope this helps you!!

Please Mark me Brainleast if am correct!!

<h2><em>Thankyou for joining Brainly Communicty!!</em></h2>

frozen [14]2 years ago
5 0

As the volume of the container increases the pressure inside will decrease because the atoms have more room to move around in.

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The bottom of cooking utensils are often kept black. Give reason​
stepan [7]

Answer:

absorb more heat and get heated quickly

Explanation:

7 0
3 years ago
Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is
Ronch [10]

Complete Question

A gas gun uses high pressure gas tp accelerate projectile through the gun barrel.

If the acceleration of the projective is : a = c/s m/s​2

Where c is a constant that depends on the initial gas pressure behind the projectile. The initial position of the projectile is s= 1.5m and the projectile is initially at rest. The projectile accelerates until it reaches the end of the barrel at s=3m. What is the value of the constant c such that the projectile leaves the barrel with velocity of 200m/s?

Answer:

The value of the constant is  c = 28853.78 \ m^2 /s^2

Explanation:

From the question we are told that

         The acceleration is  a =  \frac{c}{s}\   m/s^2

         The  initial position of the projectile is s= 1.5m

         The final position of the projectile is s_f =  3 \ m

          The velocity is  v = 200 \ m/s

     Generally  time  =  \frac{ds}{dv}

   and  acceleration is a =  \frac{v}{time }

so

            a = v  \frac{dv}{ds}

 =>        vdv  =  a ds

             vdv  = \frac{c}{s}  ds

integrating both sides

           \int\limits^a_b  vdv  = \int\limits^c_d \frac{c}{s}  ds

Now for the limit

          a =  200 m/s

             b = 0 m/s  

         c = s= 3 m

          d =s_f= 1.5 m

So we have  

           \int\limits^{200}_{0}  vdv  = \int\limits^{3}_{1.5} \frac{c}{s}  ds

              [\frac{v^2}{2} ] \left | 200} \atop {0}} \right.  = c [ln s]\left | 3} \atop {1.5}} \right.

            \frac{200^2}{2}  =  c ln[\frac{3}{1.5} ]

=>           c = \frac{20000}{0.69315}

              c = 28853.78 \ m^2 /s^2

     

5 0
3 years ago
An object weighs 42.53 newtons. What is its mass if a gravitometer indicates that g = 9.83 m/s2? Round to the nearest tenth.
enot [183]
42.53/9.83 = 4.3265.... = 4.3
3 0
3 years ago
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Aneli [31]

Answer:

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Explanation:

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2 years ago
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1st one and fourth one are true.
7 0
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