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vagabundo [1.1K]
3 years ago
11

My buddy and I are planning a shore dive. we're descending onto a very gradual slope that begins at 5 m/15 ft, so our descent an

d ascent will be a gradual part of swimming out and back underwater. we have similar cylinders filled to 2000 bar/3000 psi. We plan:
- 50 bar/500 psi reserve
- 20 bar/300 psi for our safety stop
- to turn the dive when we've used 1/3 of the air available to use on the dive

This means we should head back when either of our SPGs read
A. 70 bar/800 psi
B. 145 bar/1900 psi
C. 157 bar/2270 psi
D. 170 bar/2500 psi
Physics
1 answer:
Gnom [1K]3 years ago
4 0

Answer:

C. 157 bar/2270 psi

Explanation:

Calculation to determine what we should head back when either of our SPGs read

SPGs=200 bar -[200 bar-(50 bar + 20 bar)]÷1/3]

SPGs=200 bar-[(200 bar-70 bar)÷1/3]

SPGs=200 bar-(130 bar÷1/3)

SPGs=200 bar-43 bar

SPGs=157 bar/2270 psi

Therefore based on the above calculation we should head back when either of our SPGs read 157 bar/2270 psi

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A billiard ball of mass m moving with speed v1=1 m/s collides head-on with a second ball of mass 2m which is at rest. What is th
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Answer:

The velocity of mass 2m is  v_B = 0.67 m/s

Explanation:

From the question w are told that

     The mass of the billiard ball A is =m

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    The mass of the billiard ball B is = 2 m

    The initial speed  of the billiard ball  B = 0

Let the final speed  of the billiard ball A  = v_A

Let The finial speed  of the billiard ball  B = v_B

      According to the law of conservation of Energy

                 \frac{1}{2} m (v_1)^2 + \frac{1}{2} 2m (0) ^ 2 = \frac{1}{2} m (v_A)^2 + \frac{1}{2} 2m (v_B)^2

              Substituting values  

                \frac{1}{2} m (1)^2  = \frac{1}{2} m (v_A)^2 + \frac{1}{2} 2m (v_B)^2

Multiplying through by \frac{1}{2}m

                1 =v_A^2 + 2 v_B ^2 ---(1)

    According to the law of conservation of Momentum

            mv_1 + 2m(0) = mv_A + 2m v_B

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Multiplying through by m

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making v_A subject of the equation 2

            v_A = 1 - 2v_B

Substituting this into equation 1

         (1 -2v_B)^2 + 2v_B^2 = 1

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          6v_B^2 -4v_B =0

Multiplying through by \frac{1}{v_B}

          6v_B -4 = 0

            v_B = \frac{4}{6}

            v_B = 0.67 m/s

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