High-level clouds usually occur above 20,000 ft. Which is the best description of the clouds’ appearance?

[25 points] Imagine two billiard balls on a pool table. Ball A has a mass of 7 kilograms and ball B has a mass of 2 kilograms. T

evablogger [386]

6 meters is left because you subtract 12 meters from 6

5 0

3 years ago

On top of Mount Everest, the temperature is -19 ∘C in July. Being a physicist, you determine by how many degrees Celsius one nee

kondaur [170]

Answer:

254 °C

Explanation:

The average kinetic energy of gas molecules K = 3RT/2N where R = gas constant = 8.314 J/mol-K, N = avogadro's constant = 6.022 × 10²³ atoms/mol

T = temperature in Kelvin.

Let K be its average kinetic energy at t = -19°C = 273 + (-19) = 273 - 19 = 254 K = T. K = 3RT/2N = 3 × 8.314 J/mol-K × 254 K/(2 × 6.022 × 10²³ atoms/mol) = 5.26 × 10⁻²¹ J

When its average kinetic energy doubles, it becomes K₁ = 2K = 2 × 5.26 × 10⁻²¹ = 10.52 × 10⁻²¹ J at temperature T₂. So,

K₁ = 3RT₁/2N

T₁ = 2NK₁/3R

T₁ = 2 × 6.022 × 10²³ atoms/mol × 10.52 × 10⁻²¹ J/3 × 8.314 J/mol-K = 508 K

The temperature difference is thus ΔT = T₁ - T = 508 K - 254 K = 254 K.

Since temperature change in kelvin scale equals temperature change in Celsius scale ΔT = 254 °C

So, we need to change the temperature of the air by 254 °C to double its average kinetic energy.

3 0

3 years ago

Why does the balloon stick to the wall

ryzh [129]

Answer:Magic

Explanation:

You know its true.

5 0

3 years ago

Read 2 more answers

A charge q1 of -5.00X10^-9 C and a charge q2 of -2.00X10^-9 C are separated by a distance of 40.0 cm. Find the equilibrium posit

Alex73 [517]

Answer:

The equilibrium position for the third charge is 69.28 cm

Explanation:

Given;

q₁ = -5.00 x 10⁻⁹ C

q₂ = -2.00 x 10⁻⁹ C

q₃ = 15.00 x 10⁻⁹ C

distance between q₁ and q₂ = 40.0 cm = 0.4 m

(-q₁)--------------------------------------(-q₂)---------------------------------(+q₃)

At equilibrium the repulsive force between q₁ and q₂ must be equal to attractive force between q₂ and q₃

According to Coulomb's law, repulsive or attractive force between charges is calculated as;

$F = \frac{Kq_1q_2}{r_1^2} = \frac{Kq_2q_3}{r_2^2}$

where;

F is repulsive or attractive force between charges

K is Coulomb's constant = 8.99 x 10⁹ Nm²/c²

r₁ is the distance between q₁ and q₂

q₁, q₂ and q₃ are the charge

distance between q₂ and q₃, r₂ is calculated as;

$\frac{Kq_1q_2}{r_1^2} = \frac{Kq_2q_3}{r_2^2}\\\\\frac{q_1q_2}{r_1^2} = \frac{q_2q_3}{r_2^2}\\\\r_2^2= \frac{r_1^2q_2q_3}{q_1q_2}\\\\r_2^2= \frac{r_1^2q_3}{q_1} = \frac{0.4^2*15*10^{-9}}{5*10^{-9}} = 0.48\\\\r_2 = \sqrt{0.48} = 0.6928 \ m$

Therefore, the equilibrium position for the third charge is 69.28 cm

3 0

3 years ago

Why electric field inside the conductor zerqual to zero

Arlecino [84]

In electrostatics free charges in a good conductor<span> reside only on the surface.

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5 0

3 years ago