Why just indirect evidence be used to study the structure of atoms?

Dahasolnce [82]

A/ pressure and collision frequency increase.
think of it this way. there's less room in total when u decrease volume, so particles will hit more often because they travel less

8 0

3 years ago

Another droplet of the same mass falls 8.4 cm from rest in 0.250 s, again moving through a vacuum. Find the charge carried by th

Andrej [43]

Answer:

The charge carried by the droplet is $1.330\times10^{-19}\ C$

Explanation:

Given that,

Distance =8.4 cm

Time = 0.250 s

Suppose tiny droplets of oil acquire a small negative charge while dropping through a vacuum in an experiment. An electric field of magnitude $5.92\times10^4\ N/C$ points straight down and if the mass of the droplet is $2.93\times10^{-15} kg$

We need to calculate the acceleration

Using equation of motion

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

Put the value into the formula

$8.4\times10^{-2}=0+\dfrac{1}{2}\times a\times(0.250)^2$

$a=\dfrac{8.4\times10^{-2}\times2}{(0.250)^2}$

$a=2.688\ m/s^2$

We need to calculate the charge carried by the droplet

Using formula of electric filed

$E=\dfrac{F}{q}$

$q=\dfrac{ma}{E}$

Put the value into the formula

$q=\dfrac{2.93\times10^{-15}\times2.688}{5.92\times10^4}$

$q=1.330\times10^{-19}\ C$

Hence, The charge carried by the droplet is $1.330\times10^{-19}\ C$

8 0

2 years ago

What are the five basic postulates of kinetic-molecular theory?

lesantik [10]

Explanation:

The five basic postulate of kinetic molecular theory includes:

1) All gases consist of large amount and numbers of tiny particles that are far apart from each other and also relative to their size.

2) The collisions between gas particles and gas particles against container walls is refer to as elastic collision.

3) All gas particles are in a continuous random and rapid motion. They possess kinetic energy which is energy of motion.

4) There are no attractive force between gas particles.

5) The temperature of a gas depends on the average kinetic energy of the gas particle.

8 0

2 years ago

An ideal photo-diode of unit quantum efficiency, at room temperature, is illuminated with 8 mW of radiation at 0.65 µm wavelengt

nadya68 [22]

Answer:

I = 4.189 mA V = 0.338 V

Explanation:

In order to do this, we need to apply the following expression:

I = Is[exp^(qV/kT) - 1] (1)

However, as the junction of the diode is illuminated, the above expression changes to:

I = Iopt + Is[exp^(qV/kT) - 1] (2)

Now, as the shunt resistance becomes infinite while the current becomes zero, we can say that the leakage current is small, and so:

I ≅ Iopt

Therefore:

I ≅ I₀Aλq / hc (3)

Where:

I₀A = Area of diode (radiation)

λ: wavelength

q: electron charge (1.6x10⁻¹⁹ C)

h: Planck constant (6.62x10⁻³⁴ m² kg/s)

c: speed of light (3x10⁸ m/s)

Replacing all these values, we can get the current:

I = (8x10⁻³) * (0.65x10⁻⁶) * (1.6x10⁻¹⁹) / (6.62x10⁻³⁴) * (3x10⁸)

I = 4.189x10⁻³ A or 4.189 mA

Now that we have the current, we just need to replace this value into the expression (2) and solve for the voltage:

I = Is[exp^(qV/kT) - 1]

k: boltzman constant (1.38x10⁻²³ J/K)

4.189x10⁻³ = 9x10⁻⁹ [exp(1.6x10⁻¹⁹ V / 1.38x10⁻²³ * 300) - 1]

4.189x10⁻³ / 9x10⁻⁹ = [exp(38.65V) - 1]

465,444.44 + 1 = exp(38.65V)

ln(465,445.44) = 38.65V

13.0508 = 38.65V

V = 0.338 V

6 0

3 years ago

Why is a football firm when it is inflated to its proper pressure

Molodets [167]

Answer:

Proper Inflation and Feel When inflating your ball, you can use either a hand pump or an air pump equipped with a gauge that gives readings in pounds per square inch, also called psi. Footballs used in the NFL are inflated to 13 psi, but a proper range can fall between 12.5 and 13.5 psi, according to Wilson Sporting Goods.

Explanation:

8 0

3 years ago