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nata0808 [166]
3 years ago
14

What is the final volume of a gas with an initial volume of 75 mL if the pressure decreases from 300 kPa to 200kPa?

Physics
2 answers:
IgorLugansk [536]3 years ago
7 0

I would subtract 300 to 200 and I  got 100

sammy [17]3 years ago
3 0

Answer:

The final volume of the gas is 112.5 mL.              

Explanation:

It is given that,

Initial volume of the gas, V_1=75\ mL

Initial pressure, P_1=300\ kPa

Final pressure, P_2=200\ kPa

We need to find the final volume of the gas. It can be calculated using the Boyle's law as :

p v = constant

or

\dfrac{P_1}{P_2}=\dfrac{V_2}{V_1}, V₂ is the final volume of the gas

V_2=\dfrac{P_1}{P_2}\times V_1

V_2=\dfrac{300\ kPa}{200\ kPa}\times 75\ mL

V_2=112.5\ mL

So, the final volume of the gas is 112.5 mL. Hence, this is the required solution.

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Answer:

5 is the tripoid stand

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4 0
3 years ago
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A tank, shaped like a cone has height 12 meter and base radius 1 meter. It is placed so that the circular part is upward. It is
Temka [501]

Answer:

376966.991 Joules

Explanation:

Given that :

the height = 12  m

Let assume the tank have a thickness  = dh

The radius of the tank by using the concept of similar triangle is :

\dfrac{1}{r} = \dfrac{12}{h}

r = \dfrac{h}{12}

The area of the tank = \mathbf{\pi r^2}

The area of the tank = \mathbf{\pi( \dfrac{h}{12})^2}

The area of the tank = \mathbf{ \dfrac{\pi}{144}h^2}

The volume of the tank is  = area × thickness

= \mathbf{ \dfrac{\pi}{144}h^2 \  dh}

Weight of the element = \rho_ g * volume

where;

\rho_g = density of water ; which is given as 10000 N/m³

So;

Weight of the element = \mathbf{ 10000 *\dfrac{\pi}{144}h^2 \  dh}

Weight of the element = \mathbf{69.44 \ \pi  \ h^2 \  dh}

However; the work required to pump this water = weight × height  rise

where the height rise = 12 - h

the work required to pump this water  = \mathbf{69.44 \ \pi  \ h^2 \  dh}(12 - h)

the work required to pump this water  = \mathbf{69.44 \pi (12h^2-h^3)dh}

We can determine the total workdone by integrating the work required to pump this water

SO;

Workdone = \mathbf{\int\limits^{12}_0 {69.44 \pi(12h^2-h^3)} dh}

= \mathbf{ 69.44 \pi \int\limits^{12}_0 {(12h^2-h^3)} dh}

=  \mathbf{ 69.44 \pi[ \frac{12h^3}{3}-  \frac{h^4}{4}]^{12}}_0} }

= \mathbf{69.44 \pi [ \frac{12^4}{3}-\frac{12^4}{4}]}

= \mathbf{69.44 \pi*12^4 [ \frac{4-3}{12}]}

= \mathbf{69.44 \pi*12^4 *\frac{1}{12}}

= 376966.991 Joules

6 0
3 years ago
An above ground swimming pool of 30 ft diameter and 5 ft depth is to be filled from a garden hose (smooth interior) of length 10
STALIN [3.7K]

This question involves the concepts of dynamic pressure, volume flow rate, and flow speed.

It will take "5.1 hours" to fill the pool.

First, we will use the formula for the dynamic pressure to find out the flow speed of water:

P=\frac{1}{2}\rho v^2\\\\v=\sqrt{\frac{2P}{\rho}}

where,

v = flow speed = ?

P = Dynamic Pressure = 55 psi(\frac{6894.76\ Pa}{1\ psi}) = 379212 Pa

\rho = density of water = 1000 kg/m³

Therefore,

v=\sqrt{\frac{2(379212\ Pa)}{1000\ kg/m^3}}

v = 27.54 m/s

Now, we will use the formula for volume flow rate of water coming from the hose to find out the time taken by the pool to be filled:

\frac{V}{t} = Av\\\\t =\frac{V}{Av}

where,

t = time to fill the pool = ?

A = Area of the mouth of hose = \frac{\pi (0.015875\ m)^2}{4} = 1.98 x 10⁻⁴ m²

V = Volume of the pool = (Area of pool)(depth of pool) = A(1.524 m)

V = [\frac{\pi (9.144\ m)^2}{4}][1.524\ m] = 100.1 m³

Therefore,

t = \frac{(100.1\ m^3)}{(1.98\ x\ 10^{-4}\ m^2)(27.54\ m/s)}\\\\

<u>t = 18353.5 s = 305.9 min = 5.1 hours</u>

Learn more about dynamic pressure here:

brainly.com/question/13155610?referrer=searchResults

7 0
3 years ago
The athlet at point A runs 150m east, then 70m west and then 100 east
denpristay [2]

Answer:

180m to the east

Explanation:

Displacement is the distance traveled in a specific direction. It is a vector quantity with both magnitude and direction. Therefore, the start and finish position is very paramount.

  point A runs 150m east,

   70m west

  100m east

                                       150m

                --------------------------------------------------------→

                                                                   70m

                                                         ←---------------------

                                                                   100m

                                                         -----------------------------------→

The displacement of the athlete  = 150 - 80 + 100  = 180m to the east

6 0
3 years ago
Two flat 4.0 cm × 4.0 cm electrodes carrying equal but opposite charges are spaced 2.0 mm apart with their midpoints opposite ea
Olenka [21]

Answer:

q =1.07 \times 10^{-7}\ C

Explanation:

given.

Two flat 4.0 cm × 4.0 cm electrodes carrying equal charge

space between the charges = 2 mm

Electric field strength = 7.6 x 10⁶ N/C

ε ₀ = 8.85 × 10⁻¹² C²/N · m²

magnitude of charge =?

Electric field strength between to two plates

E = \dfrac{q}{A\epsilon_0}

q = E A \epsilon_0

q = 7.6 \times 10^6 \times 0.04 \times 0.04 \times 8.85 \times 10^{-12}

q =1.07 \times 10^{-7}\ C

7 0
4 years ago
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