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oksian1 [2.3K]
3 years ago
5

A bicycle pump contains 200 cm3 of air and is connected to a bicycle tyre. The volume of the tyre is 800 cm3. The pressure of th

e air in the tyre (it is ‘flat’) is 1.0 atmosphere, the same as the air in the pump. (= volume1). What is the total volume of the air initially?
Please help me
Physics
1 answer:
Aleksandr [31]3 years ago
7 0

Answer:

The total volume of the air is 1000 cubic centimeters.

Explanation:

Since the bicycle pump and the bicycle tyre have the same pressure, then the total volume of the air is the sum of the volume of each element, then we translate this into the following artihmetical expression:

V = 200\,cm^{3}+800\,cm^{3}

V = 1000\,cm^{3}

The total volume of the air is 1000 cubic centimeters.

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Which value is NOT equivalent to the others?
pishuonlain [190]

Answer:

A) 5 × 1011nm

I think It's A

8 0
2 years ago
suppose a 51 kg bungee jumper steps off the royal gorge bridge, in colorado. The bridge is situated 321 m above the arkansas riv
Akimi4 [234]

The total potential energy associated with the jumper at the end of his fall is 90,000 J.

The given parameters;

  • <em>mass of the jumper, m = 51 kg</em>
  • <em>height of the bridge. h  = 321 m</em>
  • <em>spring constant of the cord, k = 32 N/m</em>
  • <em>extension of the cord, x = 179 m - 104 m = 75 m</em>

The total potential energy associated with the jumper at the end of his fall is calculated as follows;

U = ¹/₂kx² + mgΔh

where;

<em>Δh is the change in height after falling </em>

U = ¹/₂(32)(75)²  + (51)(9.8)(0)

U = 90,000 J

Thus, the total potential energy associated with the jumper at the end of his fall is 90,000 J.

Learn more here:brainly.com/question/15731149

6 0
2 years ago
4
telo118 [61]
Ik what it is hit me up for it
5 0
3 years ago
A gas expands and does PV work on its surroundings equal to 319 J. At the same time, it absorbs 136 J of heat from the surroundi
LiRa [457]

Answer:

The change in energy of the gas during the process is -1.83\times 10^{2} joules.

Explanation:

We can represent this process by the First Law of Thermodynamics, in which gas does work on its surroundings and absorbs heat from there to describe its change in energy. In other words:

Q_{in} - W_{out} = \Delta E

Where:

Q_{in} - Heat absorbed by the gas, measured in joules.

W_{out} - Work done by the gas, measured in joules.

\Delta E - Change in energy, measured in joules.

If we know that Q_{in} = 1.36\times 10^{2}\,J and W_{out} = 3.19\times 10^{2}\,J, the change in energy of the gas is:

\Delta E = 1.36\times 10^{2}\,J-3.19\times 10^{2}\,J

\Delta E = -1.83\times 10^{2}\,J

The change in energy of the gas during the process is -1.83\times 10^{2} joules.

3 0
3 years ago
A sharp edged orifice with a 60 mm diameter opening in the vertical side of a large tank discharges under a head of 6 m. If the
Ierofanga [76]

Answer:

The discharge rate is Q = 0.0192 \  m^3 /s

Explanation:

From the question we are told that

   The  diameter is  d =  60 \ mm   =  0.06 \ m

    The  head is  h  =  6 \ m

     The  coefficient of contraction is  Cc  =  0.68

     The  coefficient of  velocity is  Cv  =  0.92

The radius is mathematically evaluated as

         r =  \frac{d}{2}

substituting values

        r =  \frac{ 0.06 }{2}

        r =  0.03 \ m

The  area is mathematically represented as

      A =  \pi r^2

substituting values

      A =  3.142 *  (0.03)^2

      A = 0.00283 \ m^2

 The  discharge rate is mathematically represented as

        Q =  Cv *Cc  *  A  *  \sqrt{ 2 * g *  h}

substituting values

       Q = 0.68 *  0.92*   0.00283  *  \sqrt{ 2 * 9.8 *  6}

       Q = 0.0192 \  m^3 /s

6 0
2 years ago
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