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2 years ago

The number of particles in a gas system inversely affects _ and directly affects _.

1 answer:

5 0

The answer is A.
When a substance is a gas, it fills its container no matter how many particles it's made of, however the number of particles present will directly affect the pressure and temperature of the system.

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A comet is in an elliptical orbit around the Sun. Its closest approach to the Sun is a distance of 4.8 1010 m (inside the orbit

creativ13 [48]

Answer:

Explanation:

From the given information:

Distance $d_i = 4.8 \times 10^{10} \ m$

Speed of the comet $V_i = 9.1 \times 10^{4} \ m/s$

At distance $d_2 = 6 \times 10^{12} \ m$

where;

mass of the sun = $1.98 \times 10^{30}$

$G = 6.67 \times 10^{-11}$

To find the speed $V_f$ :

Using the formula:

$E_f = E_i + W \\ \\ where; \ \ W = 0 \ \ \text{since work done by surrounding is zero (0)}$

$E_f = E_i + 0 \\ \\ K_f + U_f = K_i + U_i \\ \\ = \dfrac{1}{2}mV_f^2 + \dfrac{-GMm}{d^2} = \dfrac{1}{2}mV_i^2+ \dfrac{-GMm}{d_i} \\ \\ V_f = \sqrt{V_i^2 + 2 GM \Big [ \dfrac{1}{d_2}- \dfrac{1}{d_i}\Big ]}$

$V_f = \sqrt{(9.1 \times 10^{4})^2 + 2 (6.67\times 10^{-11}) *(1.98 * 10^{30} ) \Big [ \dfrac{1}{6*10^{12}}- \dfrac{1}{4.8*10^{10}}\Big ]}$

$\mathbf{V_f =53.125 \times 10^4 \ m/s}$

3 0

2 years ago

What is atomic theory and how has it changed over time?

vazorg [7]

<span>In chemistry and physics, the atomic theory explains how our understanding of the atom has changed over time. Atoms were once thought to be the smallest pieces of matter. The first idea of the atom came from the Greek philosopher Democritus. Hope I helped!!</span>

6 0

2 years ago

Read 2 more answers

_______ do not have definite size or always take the shape of their container?

amid [387]

Liquids do not have a definite size and always take the shape of the container they're in.

6 0

2 years ago

Read 2 more answers

Two large non-conducting plates of surface area A = 0.25 m 2 carry equal but opposite charges What is the energy density of the

Stells [14]

Answer:

5.1*10^3 J/m^3

Explanation:

Using E = q/A*eo

And

q =75*10^-6 C

A = 0.25

eo = 8.85*10^-12

Energy density = 1/2*eo*(E^2) = 1/2*eo*(q/A*eo)^2 = [q^2] / [2*(A^2)*eo]

= [(75*10^-6)^2] / [2*(0.25)^2*8.85*10^-12]

= 5.1*10^3 J/m^3

8 0

3 years ago

A light, rigid rod is 55.8 cm long. Its top end is pivoted on a frictionless horizontal axle. The rod hangs straight down at res

VMariaS [17]

To solve this problem we will apply the principle of conservation of energy. For this purpose, potential energy is equivalent to kinetic energy, and this clearly depends on the position of the body. In turn, we also note that the height traveled is twice that of the rigid rod, therefore applying these concepts we will have

$KE = PE$