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Alexxx [7]
3 years ago
15

An ideal gas is at a pressure 1.00 Ã 105 n/m2 and occupies a volume 2.00 m3. if the gas is compressed to a volume 1.00 m3 while

the temperature remains constant, what will be the new pressure in the gas?
Physics
1 answer:
blsea [12.9K]3 years ago
4 0
The behavior of an ideal gas at constant temperature obeys Boyle's Law of
p*V = constant
where
p = pressure
V = volume.

Given:
State 1:  
  p₁ = 10⁵ N/m² (Pa)
  V₁ = 2 m³
State 2:
  V₂ = 1 m³

Therefore the pressure at state 2 is given by
p₂V₂ = p₁V₁
or
p₂ = (V₁/V₂) p₁
    = 2 x 10⁵ Pa

Answer: 2 x 10⁵ N/m² or 2 atm.
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The force of gravity depends on the mass of objects and the distance between them. TRUE OR FALSE?
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4 years ago
S Problem Set<br> 2.) 6.4 x 109 nm to cm
anyanavicka [17]

Answer:

6.4\cdot 10^2 cm

Explanation:

First of all, let's convert from nanometres to metres, keeping in mind that

1 nm = 10^{-9} m

So we have:

6.4\cdot 10^9 nm \cdot 10^{-9} m/nm = 6.4 m

Now we can convert from metres to centimetres, keeping in mind that

1 m = 10^2 cm

So, we find:

6.4 m \cdot 10^2 cm/m = 6.4\cdot 10^2 cm

8 0
3 years ago
The National Aeronautics and Space Administration (NASA) studies the physiological effects of large accelerations on astronauts.
m_a_m_a [10]

Answer:

v = 23.95 m/s

Explanation:

As we know that when astronaut is revolving in circular path then the acceleration of the astronaut is due to centripetal acceleration

so it is given as

a_c = \frac{v^2}{R}

here we know that

a_c = 4.50 g

also we know that

R = 13 m

now we have

4.50 \times 9.81 = \frac{v^2}{13}

v = 23.95 m/s

3 0
4 years ago
2. Is the original mixture homogeneous or heterogeneous? Why? .
myrzilka [38]

Answer:

There is no chemical combination of the substances in a mixture, so they retain their physical properties. There is the same composition throughout a homogeneous mixture. The structure of a heterogeneous mixture differs.

Explanation:

4 0
3 years ago
A rod 14.0 cm long is uniformly charged and has a total charge of -22.0 μC. Determine the magnitude and direction of the net ele
lesya692 [45]

Explanation:

It is given that,

Length of the rod, l = 14 cm = 0.14 m

Charge on the rod, q=-22\ \mu C=-22\times 10^{-6}\ C

We need to find the magnitude and direction of the net electric field produced by the charged rod at a point 36.0 cm to the right of its center along the axis of the rod, z = 36 cm = 0.36 m

Electric field at the axis of the rod is given by :

E=\dfrac{\lambda}{2\pi \epsilon_o z}

Where

\lambda is the linear charge density of the rod,

\lambda=\dfrac{q}{l}=\dfrac{-22\times 10^{-6}\ C}{0.14\ m}=-0.00015\ C/m

E=\dfrac{-0.00015}{2\pi\times 8.85\times 10^{-12}\times 0.36}

E = -7493170.57 N/C

or

E=-7.49\times 10^6\ N/C

Negative sign shows that the electric field is acting in inwards direction. Hence, this is the required solution.

4 0
3 years ago
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