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Ivahew [28]
3 years ago
5

When a container decreases in size, what will happen to an amount of gas in the container??

Physics
1 answer:
alexgriva [62]3 years ago
7 0
The gas would also decrease in size since the container lost gas to decrease the size of the container.
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What are the divisions within a shell called
blsea [12.9K]
The divisions within an atom's shell are called subshells. This means that each shell consists of several subshells that are made up of orbitals. Each orbital consists of 1 or 2 electrons. The outermost shell of an atom is what we call the valence electrons, and they are what participate in chemical bonding.
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Myoncedyret
konstantin123 [22]

Answer:

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Explanation:

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An electron is pushed into an electric field where it acquires a 1-V electrical potential. Suppose instead that two electrons ar
sleet_krkn [62]

Answer:

0.5 V

Explanation:

The electric potential distance between different locations in an electric field area is unaffected by the charge that is transferred between them. It is solely dependent on the distance. Thus, for two electrons pushed together at the same distance into the same field, the electric potential will remain at 1 V. However, the electric potential of one of the two electrons will be half the value of the electric potential for the two electrons.

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3 years ago
A mass m = 3.9 kg hangs from a massless string wrapped around a uniform cylinder with mass Mp = 10.53 and radius R= 0.96 m. The
luda_lava [24]

Answer:

the rotational inertia of the cylinder = 4.85 kgm²

the mass moved 7.942 m/s

Explanation:

Formula for calculating Inertia can be expressed as:

I =\frac{1}{2}mR^2

For calculating the rotational inertia of the cylinder ; we have;

I = \frac{1}{2}m_pR^2

I = \frac{1}{2}*10.53*(0.96)^2

I=5.265*(0.96)^2

I=4.852224

I ≅ 4.85 kgm²

mg - T ma and RT = I ∝

T = \frac{Ia}{R^2}

a = \frac{g}{1+\frac{I}{mR^2}}

a = \frac{9.8}{1+\frac{4.85}{3.9*(0.96)^2}}

a = 4.1713 m/s²

Using the equation of motion

v^2 = u^2+2as \\ \\ v^2 = 2as \\ \\ v = \sqrt{2*a*s} \\ \\ v= \sqrt{2*4.1713*7.56} \\ \\ v = 7.942 \ m/s

3 0
3 years ago
Read 2 more answers
Q7) A box sliding with a velocity of 5 m/s accelerates at 2 m/s^2. How
grigory [225]

Answer:

The box displacement after 6 seconds is 66 meters.

Explanation:

Let suppose that velocity given in statement represents the initial velocity of the box and, likewise, the box accelerates at constant rate. Then, the displacement of the object (\Delta s), in meters, can be determined by the following expression:

\Delta s = v_{o}\cdot t+\frac{1}{2}\cdot a\cdot t^{2} (1)

Where:

v_{o} - Initial velocity, in meters per second.

t - Time, in seconds.

a - Acceleration, in meters per square second.

If we know that v_{o} = 5\,\frac{m}{s}, t = 6\,s and a = 2\,\frac{m}{s^{2}}, then the box displacement after 6 seconds is:

\Delta s = 66\,m

The box displacement after 6 seconds is 66 meters.

5 0
3 years ago
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