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Alenkasestr [34]

3 years ago

6

In 1911, Ernest Rutherford and his assistants Geiger and Marsden conducted an experiment in which they scattered alpha particles

(nuclei of helium atoms) from thin sheets of gold. An alpha particle, having charge 2e and mass 6.64 10-27 kg, is a product of certain radioactive decays. The results of the experiment led Rutherford to the idea that most of an atom's mass is in a very small nucleus, with electrons in orbit around it. Assume an alpha particle, initially very far from a stationary gold nucleus, is fired with a velocity of 1.50 107 m/s directly toward the nucleus (charge 79e). What is the smallest distance between the alpha particle and the nucleus before the alpha particle reverses direction?

1 answer:

alexandr402 [8]3 years ago

8 0

Answer:

Explanation:

We shall apply conservation of mechanical energy

kinetic energy of alpha particle is converted into electric potential energy.

1/2 mv² = k q₁q₂/d , d is closest distance

d = 2kq₁q₂ / mv²

= 2 x 9 x 10⁹ x 79e x 2e / 4mv²

= 1422 x2x (1.6 x 10⁻¹⁹)² x 10⁹ /4x 1.67 x 10⁻²⁷ x (1.5 x 10⁷)²

= 3640.32 x 10⁻²⁹ /2x 3.7575 x 10⁻¹³

= 484.4 x 10⁻¹⁶

=48.4 x 10⁻¹⁵ m

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A 1000 kg car moving a 10 m/s collides with a stationary 2000 kg truck. The two vehicles interlock as a result of the collision.

IgorLugansk [536]

Answer:

v₃ = 3.33 [m/s]

Explanation:

This problem can be easily solved using the principle of linear momentum conservation. Which tells us that momentum is preserved before and after the collision.

In this way, we can propose the following equation in which everything that happens before the collision will be located to the left of the equal sign and on the right the moment after the collision.

$(m_{1}*v_{1})+(m_{2}*v_{2})=(m_{1}+m_{2})*v_{3}$

where:

m₁ = mass of the car = 1000 [kg]

v₁ = velocity of the car = 10 [m/s]

m₂ = mass of the truck = 2000 [kg]

v₂ = velocity of the truck = 0 (stationary)

v₃ = velocity of the two vehicles after the collision [m/s].

Now replacing:

$(1000*10)+(2000*0)=(1000+2000)*v_{3}\\v_{3}=3.33[m/s]$

7 0

3 years ago

the pressure of gas in a cylinder is 70 kilopascals. if the volume of the cylinder is reduced from 8.0 liters to 4.0 liters, wha

Olegator [25]

By Boyle's law:

P₁V₁ = P₂V₂

70*8 = P<span>₂*4

</span>P<span>₂*4 = 70*8
</span>
P<span>₂ = 70*8/4 = 140
</span>
P<span>₂ = 140 kiloPascals.</span>

3 0

3 years ago

A spring gun consists of a spring inside a plastic tube with spring constant, k. The spring can be compressed 20 cm from its equ

emmainna [20.7K]

Answer: The spring constant is K=392.4N/m

Explanation:

According to hook's law the applied force F will be directly proportional to the extension e produced provided the spring is not distorted

The force F=ke

Where k=spring constant

e= Extention produced

h=2m

Given that

e=20cm to meter 20/100= 0.2m

m=100g to kg m=100/1000= 0.1kg

But F=mg

Ignoring air resistance

assuming g=9.81m/s²

Since the compression causes the plastic ball to poses potential energy hence energy stored in the spring

E=1/2ke²=mgh

Substituting our values to find k

First we make k subject of formula

k=2mgh/e²

k=2*0.1*9.81*2/0.1²

K=3.921/0.01

K=392.4N/m

5 0

3 years ago

A microwave oven operating at 1.22 × 108 nm is used to heat 165 mL of water (roughly the volume of a teacup) from 23.0°C to 100.

ANTONII [103]

<u>Answer:</u> The number of photons are $3.7\times 10^8$

<u>Explanation:</u>

We are given:

Wavelength of microwave = $1.22\times 10^8nm=0.122m$ (Conversion factor: $1m=10^9nm$ )

To calculate the energy of one photon, we use Planck's equation, which is:

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

where,

h = Planck's constant = $6.625\times 10^{-34}J.s$

c = speed of light = $3\times 10^8m/s$

$\lambda$ = wavelength = 0.122 m

Putting values in above equation, we get:

$E=\frac{6.625\times 10^{-34}J.s\times 3\times 10^8m/s}{0.122m}\\\\E=1.63\times 10^{-24}J$

Now, calculating the energy of the photon with 88.3 % efficiency, we get:

$E=1.63\times 10^{-24}\times \frac{88.3}{100}=1.44\times 10^{-24}J$

To calculate the mass of water, we use the equation:

$Density=\frac{Mass}{Volume}$

Density of water = 1 g/mL

Volume of water = 165 mL

Putting values in above equation, we get:

$1g/mL=\frac{\text{Mass of water}}{165mL}\\\\\text{Mass of water}=165g$

To calculate the amount of energy of photons to raise the temperature from 23°C to 100°C, we use the equation:

$q=mc\Delta T$

where,

m = mass of water = 165 g

c = specific heat capacity of water = 4.184 J/g.°C

$\Delta T$ = change in temperature = $T_2-T_1=100^oC-23^oC=77^oC$

Putting values in above equation, we get:

$q=165g\times 4.184J/g.^oC\times 77^oC\\\\q=53157.72J$

This energy is the amount of energy for 'n' number of photons.

To calculate the number of photons, we divide the total energy by energy of one photon, we get:

$n=\frac{q}{E}$

q = 53127.72 J

E = $1.44\times 10^{-24}J$

Putting values in above equation, we get:

$n=\frac{53157.72J}{1.44\times 10^{-24}J}=3.7\times 10^{28}$

Hence, the number of photons are $3.7\times 10^8$

4 0

3 years ago

4. The flow of electric charge is know as

vredina [299]

Answer:

electrons

Explanation:

An electric current is said to exist when there is a net flow of electric charge through a region. In electric circuits this charge is often carried by electrons moving through a wire. It can also be carried by ions in an electrolyte, or by both ions and electrons such as in an ionized gas (plasma).

5 0

3 years ago