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2 years ago

Hydrogen is a pure substance represented by the chemical symbol . Which of the following best describes hydrogen?

Physics

1 answer:

butalik [34]2 years ago

8 0

Hydrogen is an element made of two or more kinds of atoms. Hydrogen is a compound made of only one kind of atom.

Explanation:

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What did you notice about the angle of incidence and the angle of reflection?

Readme [11.4K]

Answer:

They are equal

Explanation:

angle of incidence = angle of reflection

6 0

2 years ago

A heat engine with 0.300 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 500 K . The gas goes through the fo

LuckyWell [14K]

Complete Question:

A heat engine with 0.300 mol of a monatomic ideal gas initially fills a 1000 cm3 cylinder at 500 K . The gas goes through the following closed cycle: - Isothermal expansion to 5000 cm3. - Isochoric cooling to 400 K . - Isothermal compression to 1000 cm3. - Isochoric heating to 500 K .

a) what is the work for one cycle

b) what is the thermal efficiency

Answer:

a) Work done for 1 cycle = 402.13

b) Thermal efficiency = 20%

Explanation:

Number of moles, n = 0.300 mol

Initial Volume, V₁ = 1000 cm³

Temperature, T = 500 K

Isothermal expansion to 5000 cm³

Final volume, V₂ = 5000 cm³

R = 8.314 J/ mol.K

Work done, W = nRT ln(V₂/V₁)

W = (0.3 * 8.314 * 500) * ln(5000/1000)

W = 1247.1 * ln5

W₁ = 2007.13 J

Isochoric cooling

In an Isochoric process, volume is constant i.e. V₂ = V₁ = V

W = nRT ln(V/V)

But ln(V/V) = ln 1 = 0

Work done, W₂ = 0 Joules

Isothermal Compression to 1000 cm³

V₂ = 1000 cm³

V₁ = 5000 cm³

W = nRT ln(V₂/V₁)

W = 0.3 * 8.314 * 400 ln(1000/5000)

W₃ = -1605 J

Isochoric heating to 500 K

Since there is no change in volume, no work is done

W₄ = 0 J

a) Work done for 1 cycle

W = W₁ + W₂ + W₃ + W₄

W = 2007.13 + 0 + 0 -1605+0

W = 402.13 Joules

b) Thermal efficiency

Thermal efficiency = (Net workdone for 1 cycle)/(Heat absorbed)

Heat absorbed = Work done due to thermal expansion = 2007.13 J

Thermal efficiency = 402.13/2007.13

Thermal efficiency = 0.2

Thermal efficiency = 0.2 * 100% = 20 %

3 0

3 years ago

Projectile motion in two dimensions cannot be determined by breaking the problem into two connected one-dimensional problems.

andriy [413]

The answer I believe is False

8 0

3 years ago

A 5-m steel beam is lowered by means of two cables unwinding at the same speed from overhead cranes. As the beam approaches the

ArbitrLikvidat [17]

Answer:

a) The angular acceleration of the beam is 0.5 rad/s²CW (direction clockwise due the tangential acceleration is positive)

b) The acceleration of point A is 3.25 m/s²

The acceleration of point E is 0.75 m/s²

Explanation:

a) The relative acceleration of B with respect to D is equal:

$a_{B} =a_{D} +(a_{B/D} )_{n} +(a_{B/D} )_{t}$

Where

aB = absolute acceleration of point B = 2.5 j (m/s²)

aD = absolute acceleration of point D = 1.5 j (m/s²)

(aB/D)n = relative acceleration of point B respect to D (normal direction BD) = 0, no angular velocity of the beam

(aB/D)t = relative acceleration of point B respect to D (tangential direction BD)

$a_{B} =a_{D} +(a_{B/D} )_{t}$

$2.5j=1.5j +(a_{B/D} )_{t}\\(a_{B/D} )_{t}=j=1m/s^{2}$

We have that

(aB/D)t = BDα

Where α = acceleration of the beam

BDα = 1 m/s²

Where

BD = 2

$2\alpha =1\\\alpha =0.5rad/s^{2}CW$

b) The acceleration of point A is:

$a_{A} =a_{D} +(a_{A/D} )_{t}$

(aA/D)t = ADαj

$a_{A} =a_{D} +AD\alpha j\\a_{A}=1.5j+(3.5*0.5)j\\a_{A}=3.25jm/s^{2}$

The acceleration of point E is:

(aE/D)t = -EDαj

$a_{E} =a_{D} -ED\alpha j\\a_{E}=1.5j-(1.5*0.5)j\\a_{E}=0.75jm/s^{2}$

7 0

3 years ago

Radio wave radiation falls in the wavelength region of 10.0 to 1000 meters. What is the energy of radio wave radiation that has

dmitriy555 [2]

Answer:

Explanation:

Given

Wavelength of radiation $\lambda =784\ m$

We know Energy of wave with wavelength $\lambda$ is given by

$E=\frac{hc}{\lambda }$

where h=Planck's constant

c=velocity of light

$\lambda$ =wavelength of wave

$E=\frac{6.626\times 10^{-34}\times 3\times 10^8}{784}$

$E=2.53\times 10^{-28}\ J$

Hence the energy of the wave with wavelength 784 m is $2.53\times 10^{-28}\ J$

7 0

3 years ago