Which of the following is an example of an indirect observation?

A cubical box measuring 1.29 m on each side contains a monatomic ideal gas at a pressure of 2.0 atm How much thermal energy do t

Marrrta [24]

Answer:

a) $U = 652.545\,kJ$ , b) $v \approx 659.568\,\frac{m}{s}$

Explanation:

a) According to the First Law of Thermodinamics, the system is not reporting any work, mass or heat interactions. Besides, let consider that such box is rigid and, therefore, heat contained inside is the consequence of internal energy.

$Q = U$

The internal energy for a monoatomic ideal gas is:

$U = \frac{3}{2} \cdot n \cdot R_{u} \cdot T$

Let assume that cubical box contains just one kilomole of monoatomic gas. Then, the temperature is determined from the Equation of State for Ideal Gases:

$T = \frac{P\cdot V}{n\cdot R_{u}}$

$T = \frac{(202.65\,kPa)\cdot(1.29\,m)^{3}}{(1\,kmole)\cdot(8.314\,\frac{kPa\cdot m^{3}}{kmole\cdot K} )}$

$T = 52.325\,K$

The thermal energy contained by the gas is:

$U = \frac{3}{2}\cdot (1\,kmole)\cdot (8.314\,\frac{kPa\cdot m^{3}}{kmole\cdot K})\cdot (52.325\,K)$

$U = 652.545\,kJ$

b) The physical model for the cat is constructed from Work-Energy Theorem:

$U = \frac{1}{2}\cdot m_{cat} \cdot v^{2}$

The speed of the cat is obtained by isolating the respective variable and the replacement of every known variable by numerical values:

$v = \sqrt{\frac{2 \cdot U}{m_{cat}}}$

$v = \sqrt{\frac{2\cdot (652.545 \times 10^{3}\,J)}{3\,kg} }$

$v \approx 659.568\,\frac{m}{s}$

3 0

3 years ago

An apple dropped from the branch of a tree hits the ground in 0.5 s. If the acceleration of the apple during its motion is 10 ms

Ipatiy [6.2K]

Given that,

Time = 0.5 s

Acceleration = 10 m/s²

(I). We need to calculate the speed of apple

Using equation of motion

$v=u+at$

Where, v = speed

u = initial speed

a = acceleration

t = time

Put the value into the formula

$v=0+10\times0.5$

$v=5\ m/s$

(III). We need to calculate the height of the branch of the tree from the ground

Using equation of motion

$s=ut+\dfrac{1}{2}gt^2$

Put the value into the formula

$s=0+\dfrac{1}{2}\times10\times(0.5)^2$

$s=1.25\ m$

(II). We need to calculate the average velocity during 0.5 sec

Using formula of average velocity

$v_{avg}=\dfrac{\Delta x}{\Delta t}$

$v_{avg}=\dfrac{x_{f}-x_{i}}{t_{f}-t_{0}}$

Where, $x_{f}$ = final position

$x_{i}$ = initial position

Put the value into the formula

$v_{avg}=\dfrac{1.25+0}{0.5}$

$v_{avg}=2.5\ m/s$

Hence, (I). The speed of apple is 5 m/s.

(II). The average velocity during 0.5 sec is 2.5 m/s

(III). The height of the branch of the tree from the ground is 1.25 m.

7 0

3 years ago

1. Una carga Q1 = + 12 μC se coloca a una distancia r = 0.024 m desde una carga Q2 = + 16 μC. a) Determina la magnitud de la fue

lyudmila [28]

Answer:

1. a. 3,000 N

b. Repulsión

2. 46.875 × 10⁶ N/C

3. a. 81,000 J

b. 6.75 × 10⁹ V

Explanation:

1. Los parámetros dados son;

Q₁ = +12 μC, Q₂ = +16 μC

La distancia entre las cargas, r = 0.024

La magnitud de la fuerza electrostática, F, entre cargas se da como sigue;

$F = k \times \dfrac{Q_1 \cdot Q_2}{r^2}$

Donde, k = constante de Coulomb = 9.0 × 10⁹ N · m² / C²

Por lo tanto, obtenemos;

F = 9.0 × 10⁹ × 12 × 10⁻⁶ × 16 × 10⁻⁶ / 0.024² = 3.000

La magnitud de la fuerza electrostática, entre las cargas, F = 3000 N

(b) Dado que tanto Q₁ como Q₂ son cargas positivas, y las cargas iguales se repelen entre sí, la fuerza es la repulsión.

2) La intensidad de un campo eléctrico, E, se da como sigue;

$E = \dfrac{k \cdot Q}{r^2}$

La magnitud de la carga, Q = 24 μC

La distancia donde se mide el campo, r = 48 mm = 0.048 m

Por lo tanto, E = 9.0 × 10⁹ × 12 × 10⁻⁶ / 0.048² = 46,875,000 N / C

La intensidad de un campo eléctrico, E = 46,875,000 N / C = 46.875 × 10⁶ N / C

3. La magnitud de las cargas son;

Q₁ = 24 mC

Q₂ = -12 μC

La distancia entre las cargas, r = 0.032 m

un. El potencial eléctrico de una carga, $U_E$ , se da de la siguiente manera;

$U_E = k \times \dfrac{Q_1 \cdot Q_2}{r}$

Por lo tanto;

$U_E$ = 9.0×10⁹ × 24 × 10⁻³ × (-12) × 10⁻⁶ /0.032 = -81,000

La energía potencial eléctrica entre la carga, Q₁ y Q₂= -81,000 J

b. El potencial eléctrico de Q₁ en Q₂, V₁ = $k \times \dfrac{Q_1 }{r}$

Por lo tanto, V₁ = 9.0×10⁹ × 24 × 10⁻³/0.032 = 6.75 × 10⁹

El potencial eléctrico de Q₁ en Q₂, V₁ = 6.75 × 10⁹ V

3 0

3 years ago

Calculate the volume that 42g of nitrogen gas (N2)occupies at standard temperature and pressure.

luda_lava [24]

Nitrogen gas has a molar mass of about 28.0134 g/mol. Then we have a starting amount of

(42 g) / (28.0134 g/mol) ≈ 1.4993 mol

of N₂.

At standard temperature and pressure, one mole of an ideal gas occupies a volume of about 22.4 L. Then 42 g, or 1.4993 mol, of N₂ takes up

(1.4993 mol) × (22.4 L/mol) ≈ 33.6 L

8 0

3 years ago

1. The names of the compounds FeS, NaCl, NaOH, and Pb(CN)2 all end in the suffix A. -ite. B. -ic. C. -ide. D. -ate.

shtirl [24]

They all end with suffix "-ide"

In short, Your Answer would be Option C

Hope this helps!

5 0

3 years ago