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nlexa [21]
3 years ago
6

A fire breaks out and increases the Kelvin temperature of a cylinder of compressed gas by a factor of 2.4. What is the final pre

ssure of the gas relative to its initial pressure?
Physics
1 answer:
kherson [118]3 years ago
7 0

Answer:

P(final) is 2.4 times P(initial).

Explanation:

Here we can assume that the cylinder did not break and it's volume and number of moles of gas present in the cylinder remains constant.

Given the temperature increases by a factor of 2.4. Let us assume that the initial temperature be T_{1} and the final temperature be T_{2}.

Given that T_{2}=2.4\times T_{1}

Now we know the ideal gas equation is PV=nRT

here V=constant , n=constant , R=gas constant(which is constant).

\frac{P}{T}=constant

\frac{P_{1} }{T_{1}}=\frac{P_{2} }{T_{2} }

P_{2}=(\frac{T_{2} }{T_{1} } )P_{1}

P_{2}=(\frac{2.4T_{1} }{T_{1} } )P_{1}

P_{2}=2.4\times P_{1}

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A 6 kg block initially at rest is pulled to East along a horizontal surface with coefficient of kinetic friction μk=0.15 by a co
ozzi

Answer:

1.8 m/s

Explanation:

Draw a free body diagram of the block.  There are four forces:

Normal force Fn up.

Weight force mg down.

Applied force F to the east.

Friction force Fn μ to the west.

Sum the forces in the y direction:

∑F = ma

Fn − mg = 0

Fn = mg

Sum the forces in the x direction:

F − Fn μ = ma

F − mg μ = ma

a = (F − mg μ) / m

a = (12 N − 6 kg × 9.8 m/s² × 0.15) / 6 kg

a = 0.53 m/s²

Given:

Δx = 3 m

v₀ = 0 m/s

a = 0.53 m/s²

Find: v

v² = v₀² + 2aΔx

v² = (0 m/s)² + 2 (0.53 m/s²) (3 m)

v = 1.8 m/s

8 0
3 years ago
Carbon (C) and hydrogen (H) are pure substances. Each is made of only one type of atom, but carbon atoms are different from hydr
eimsori [14]
C is the correct answer.

all substances found on the periodic table are elements by definition. anything that is created using elements, such as methane, carbon dioxide, or water, are all compounds.
5 0
3 years ago
Read 2 more answers
A block of mass m1 = 3.5 kg moves with velocity v1 = 6.3 m/s on a frictionless surface. it collides with block of mass m2 = 1.7
maxonik [38]
First, let's find the speed v_i of the two blocks m1 and m2 sticked together after the collision.
We can use the conservation of momentum to solve this part. Initially, block 2 is stationary, so only block 1 has momentum different from zero, and it is:
p_i = m_1 v_1
After the collision, the two blocks stick together and so now they have mass m_1 +m_2 and they are moving with speed v_i:
p_f = (m_1 + m_2)v_i
For conservation of momentum
p_i=p_f
So we can write
m_1 v_1 = (m_1 +m_2)v_i
From which we find
v_i =  \frac{m_1 v_1}{m_1+m_2}= \frac{(3.5 kg)(6.3 m/s)}{3.5 kg+1.7 kg}=4.2 m/s

The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force \mu (m_1+m_2)g. The work done by the frictional force to stop the two blocks is
\mu (m_1+m_2)g  d
where d is the distance covered by the two blocks before stopping.
The initial kinetic energy of the two blocks together, just before entering the rough path, is
\frac{1}{2} (m_1+m_2)v_i^2
When the two blocks stop, all this kinetic energy is lost, because their velocity becomes zero; for the work-energy theorem, the loss in kinetic energy must be equal to the work done by the frictional force:
\frac{1}{2} (m_1+m_2)v_i^2 =\mu (m_1+m_2)g  d
From which we can find the value of the coefficient of kinetic friction:
\mu =  \frac{v_i^2}{2gd}= \frac{(4.2 m/s)^2}{2(9.81 m/s^2)(1.85 m)}=0.49
3 0
3 years ago
The unusually bright centers found in some galaxies are called ______.
Nimfa-mama [501]
<span>Active Galactic Nuclei.</span>
4 0
3 years ago
g n diffraction, the formula for minima is given by a times s i n (theta )equals m lambda, where a is the width of the slit, the
ki77a [65]

Answer:

θ = 22.2

Explanation:

This is a diffraction exercise

        a sin θ = m λ

The extension of the third zero is requested (m = 3)

They indicate the wavelength  λ = 630 nm = 630 10⁻⁹ m and the width of the slit  a = 5 10⁻⁶ m

         sin  θ = m λ / a

         sin  θ = 3 630 10⁻⁹ / 5 10⁻⁶

         sin  θ = 3.78 10⁻¹ = 0.378

          θ = sin⁻¹  0.378

         

to better see the result let's find the angle in radians

          θ = 0.3876 rad

let's reduce to degrees

         θ = 0.3876 rad (180º /π rad)

         θ = 22.2º

4 0
3 years ago
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