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adoni [48]
3 years ago
13

1. When does raising the temperature of a gas increase its pressure? when volume is increased and the number of particles is con

stant when volume is increased and the number of particles is increased when volume and the number of particles are constant
Physics
1 answer:
Neporo4naja [7]3 years ago
3 0

Answer:

when volume and the number of particles are constant

Explanation:

Gay Lussac law states that when the volume of an ideal gas is kept constant, the pressure of the gas is directly proportional to the absolute temperature of the gas.

Mathematically, Gay Lussac's law is given by;

PT = K

\frac{P1}{T1} = \frac{P_{2}}{T_{2}}

The ideal gas law is the equation PV = nRT

Where;

P is the pressure.

V is the volume.

n is the number of moles of substance.

R is the ideal gas constant.

T is the temperature.

Generally, raising the temperature of an ideal gas would increase its pressure when volume and the number of particles are constant.

This ultimately implies that, when volume and the number of particles are held constant, there would be a linear relationship between the temperature and pressure of a gas i.e temperature would be directly proportional to the pressure of the gas. Thus, an increase in the temperature of the gas would cause an increase in the pressure of the gas at constant volume and number of particles.

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An electron of mass 9.11 1031 kg has an initial speed of 3.00 105 m/s. It travels in a straight line, and its speed increases to
elena55 [62]

Explanation:

It is given that,

Mass of an electron, m=9.11\times 10^{-31}\ kg

Initial speed of the electron, u=3\times 10^5\ m/s

Final speed of the electron, v=7\times 10^5\ m/s

Distance, d = 5 cm = 0.05 m

(a) The acceleration of the electron is calculated using the third equation of motion as :

a=\dfrac{v^2-u^2}{2d}

a=\dfrac{(7\times 10^5)^2-(3\times 10^5)^2}{2\times 0.05}

a=4\times 10^{12}\ m/s^2

Force exerted on the electron is given by :

F=m\times a

F=9.11\times 10^{-31}\times 4\times 10^{12}

F=3.64\times 10^{-18}\ N

(b) Let W is the weight of the electron. It can be calculated as :

W=mg

W=9.11\times 10^{-31}\times 9.8

W=8.92\times 10^{-30}\ N

Comparison,

\dfrac{F}{W}=\dfrac{3.64\times 10^{-18}}{8.92\times 10^{-30}}

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Hence, this is the required solution.

8 0
3 years ago
A 500 kg block is attached to a horizontal spring that is at its equilibrium length, and whose force constant is 30 N/m. The blo
m_a_m_a [10]

Answer:

x = 0.396 m

Explanation:

The best way to solve this problem is to divide it into two parts: one for the clash of the putty with the block and another when the system (putty + block) compresses it is   spring

Data the putty has a mass m1 and velocity vo1, the block has a mass m2 .  t's start using the moment to find the system speed.

Let's form a system consisting of putty and block; For this system the forces during the crash are internal and the moment is preserved. Let's write the moment before the crash

    p₀ = m1 v₀₁

Moment after shock

    p_{f} = (m1 + m2) v_{f}

   p₀ = p_{f}

   m1 v₀₁ = (m1 + m2) v_{f}

  v_{f} = v₀₁ m1 / (m1 + m2)

   v_{f}= 4.4 600 / (600 + 500)

  v_{f} = 2.4 m / s

With this speed the putty + block system compresses the spring, let's use energy conservation for this second part, write the mechanical energy before and after compressing the spring

Before compressing the spring

   Em₀ = K = ½ (m1 + m2) v_{f}²

After compressing the spring

   E_{mf} = Ke = ½ k x²

As there is no rubbing the energy is conserved

   Em₀ = E_{mf}

   ½ (m1 + m2) v_{f}² = = ½ k x²

   x = v_{f} √ (k / (m1 + m2))

   x = 2.4 √ (11/3000)

   x = 0.396 m

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