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Elena L [17]
2 years ago
13

A rigid tank having a volume of 0.100 m3 contains helium gas at 150 atm. how many balloons can be inflated by opening the valve

at the top of the tank? each filled balloon is a sphere 0.300 m in diameter at an absolute pressure of 1.20 atm
Physics
1 answer:
GalinKa [24]2 years ago
4 0
The initial value of P*V = 0.100*150 atm-m³ 
<span> Each balloon has a volume of (4/3)*π*r³ and N balloons have a volume of N*(4/3)*π*r³ 
</span><span> When all the balloons are inflated, the pressure in the tank is the same as the pressure in the balloons, so the final value of P*V is 
</span><span> 1.20*[N*(4/3)*π*r³ + 0.100] 
</span><span> 0.100*150 = 1.20*[N*(4/3)*π*r³ + 0.100] solve for N: 
</span><span> 15/1.2 = N*(4/3)*π*r³ + 0.100
</span><span> 12.5 - 0.100 = N*(4/3)*π*0.150³ 
</span><span> 12.4 = N*0.01414 
</span><span> N = 877</span>
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Answer:

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Explanation:

The work is defined by

           W = F .ds = F ds cos θ

Bold indicate vectors

We create a reference system where the x-axis is parallel to the ramp and the axis and perpendicular, in the attached we see a scheme of the forces

Let's use trigonometry to break down weight

     sin θ = Wₓ / W

     Wₓ = W sin 60

     cos θ = Wy / W

      Wy = W cos 60

X axis

How the body is going at constant speed

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    fr = mg sin 60

    fr = 15 9.8 sin 60

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Y Axis  

    N - Wy = 0

    N = mg cos 60

    N = 15 9.8 cos 60

    N = 73.5 N

Let's calculate the different jobs

a) The work of the force of gravity is

     W = mg L cos θ

Where the angles are between the weight and the displacement is

      θ = 60 + 90 = 150

     W = 15 9.8 2.50 cos 150

     W = - 318.26 J

b) The work of the normal force

     From Newton's equations

          N = Wy = W cos 60

          N = mg cos 60

         W = N L cos 90

        W = 0

c) The work of the friction force

      W = fr L cos 0

      W = 127.31 2.50

      W = 318.275 J

d) as the body is going at constant speed the force of the tape is equal to the force of friction

      W = F L cos 0

      W = 127.31 2.50

       W = 318.275 J

e) the net force

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    W = F ’L cos 0

    W = 0

4 0
2 years ago
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maxonik [38]
First, let's find the speed v_i of the two blocks m1 and m2 sticked together after the collision.
We can use the conservation of momentum to solve this part. Initially, block 2 is stationary, so only block 1 has momentum different from zero, and it is:
p_i = m_1 v_1
After the collision, the two blocks stick together and so now they have mass m_1 +m_2 and they are moving with speed v_i:
p_f = (m_1 + m_2)v_i
For conservation of momentum
p_i=p_f
So we can write
m_1 v_1 = (m_1 +m_2)v_i
From which we find
v_i =  \frac{m_1 v_1}{m_1+m_2}= \frac{(3.5 kg)(6.3 m/s)}{3.5 kg+1.7 kg}=4.2 m/s

The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force \mu (m_1+m_2)g. The work done by the frictional force to stop the two blocks is
\mu (m_1+m_2)g  d
where d is the distance covered by the two blocks before stopping.
The initial kinetic energy of the two blocks together, just before entering the rough path, is
\frac{1}{2} (m_1+m_2)v_i^2
When the two blocks stop, all this kinetic energy is lost, because their velocity becomes zero; for the work-energy theorem, the loss in kinetic energy must be equal to the work done by the frictional force:
\frac{1}{2} (m_1+m_2)v_i^2 =\mu (m_1+m_2)g  d
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2 years ago
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Answer:

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The acceleration is

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v=u+at

And since the final velocity is zero, v=0, and solving for u,

u=-at=-490(0.030)=14.7 m/s

And now we can find the corresponding distance travelled using the equation:

d=ut+\frac{1}{2}at^2 = (14.7)(0.030)+\frac{1}{2}(-490)(0.030)^2=0.22 m

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2 years ago
How many significant figures are in 0.0069
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