The predictions of Rutherford's scattering formula failed to correspond with experimental data when the energy of the incoming alpha particles exceeded 32MeV32MeV. This can be explained by the fact that the predictions of the formula apply when the only force involved is the electromagnetic force and will break down if the incoming particles make contact with the nucleus. Use the fact that Rutherford's prediction ceases to be valid for alpha particles with an energy greater than 32MeV32MeV to estimate the radius rrr of the gold nucleus.

Answers

Answer 1

Answer:

Answer:

r = 7.1 × 10⁻¹⁵

Explanation:

Related Questions

What are some of the benefits of learned optimism that have been found inresearch?OA. Fewer health problemsOOB. All of theseC. Making more moneyOD. A lower divorce rate
An effect analogous to two-slit interference can occur with sound waves, instead of light. In an open field, two speakers placed 1.30 m apart are powered by a single function generator producing sine waves at 1200-Hz frequency. A student walks along a line 12.5 m away and parallel to the line between the speakers. She hears an alternating pattern of loud and quiet, due to constructive and destructive interference. What is : (a) the wavelength of this sound and (b) the distance between the central maximum and the first maximum (loud) position along this line
A 4 kg bowling ball accelerates 1 m/s2. what is the net force on the ball?
A rocket exhausts fuel with a velocity of 1500m/s, relative to the rocket. It starts from rest in outer space with fuel comprising 80 per cent of the total mass. When all the fuel has been exhausted its speed is:________
A single-turn circular loop of wire of radius 5.0 cm lies in a plane perpendicular to a spatially uniform magnetic field. During a 0.02500.0250-\text{s}s time interval, the magnitude of the field increases uniformly from 200 to 300 mT. Determine the magnitude of the emf induced in the loop

What happens to a black body radiator as it increases in temperature? A. it gives off a range of electromagnetic radiation of shorter wavelengths.
B. It gives off only one wavelength of electromagnetic radiation
C. It releases only ultraviolet waves of electromagnetic radiation
D. It becomes hotter but gives off less electromagnetic radiation

Answers

The black body radiator as it increases in temperature gives off a range of electromagnetic radiation of shorter wavelengths so, the option A is correct.

What is radiation?

Radiation is the movement of atomic and subatomic particles as well as waves, such as those that define X-rays, heat rays, and light rays. Radiation of both types, from cosmic and earthly sources, is constantly being thrown at all matter.

The characteristics and behavior of radiation, as well as the matter it interacts with, are outlined in this article, which also explains how energy is transferred from radiation to its surroundings.

The effects of such an energy transfer to living matter, including the typical effects on numerous biological processes, are given a great deal of attention (e.g., photosynthesis in plants and vision in animals).

Thus, the black body radiator gives off a range of electromagnetic radiation of shorter wavelengths.

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Answer: A

Explanation:

Answer is a hope this helps guys!

A golf ball is dropped from rest from a height of 9.50 m. It hits the pavement, then bounces back up, rising just 9.70 m before falling back down again. A boy catches the ball on the way down when it is 1.20 m above the pavement. Ignoring air resistance calculate the total amount of time the ball is in the air, from drop to catch?

Answers

The ball is in the air for 27.70m because
9.5+9.7=19.2...
9.7-1.2=8.5...
19.2+8.5=27.70...

The acceleration due to gravity on Earth is 9.80 m/s2. If the mass of a honeybee is 0.000100 kilograms, what is the weight of this insect?

Answers

Answer:

0.00098 N

Explanation:

The weight of an object is given by:

$W=mg$

where

m is the mass of the object

g is the gravitational acceleration on the planet

In this problem, we have:

$m=0.0001 kg$ is the mass of the honeybee

$g=9.8 m/s^2$ is the acceleration due to gravity

Substituting into the equation, we find:

$W=mg=(0.0001 kg)(9.8 m/s^2)=0.00098 N$

weight = mg
here m = 0.000100 g = 9.80
hence weight = 0.00980 kgm/s2

The unit of electric potential or electromotive force is the

Answers

That would be called VOLT

A volumetric flask made of Pyrex is calibrated at 20.0°C. It is filled to the 150-mL mark with 34.5°C acetone. After the flask is filled, the acetone cools and the flask warms so that the combination of acetone and flask reaches a uniform temperature of 32.0°C. The combination is then cooled back to 20.0°C. (The average volume expansion coefficient of acetone is 1.50 10-4(°C)−1.) (a) What is the volume of the acetone when it cools to 20.0°C?

Answers

Answer:149.73 ml

Explanation:

Given

$\beta \ of\ acetone=1.50* 10^(-4) ^(\circ)C^(-1)$

change in volume is given by

$\Delta V=V_(final)-V_(initial)$

$\Delta V=\nu_(initial)\beta _(acetone)\left [ T_f-T_i\right ]$

$V_(final)=\nu_(initial)+\nu_(initial)\beta _(acetone)\left [ T_f-T_i\right ]$

$V_(final)=150+150* 1.50* 10^(-4)\left [ 20-32\right ]$

$V_(final)=149.73 ml$

Final answer:

The volume of the acetone when it cools to 20.0°C is approximately 142.39 mL.

Explanation:

In order to determine the volume of the acetone when it cools to 20.0°C, we can use the equation for the volume change caused by a temperature change at constant pressure, known as Charles's law. Charles's law states that the volume of a gas is directly proportional to its temperature in Kelvin. We can use the formula V2 = V1 * (T2 / T1) to calculate the volume of the acetone at the lower temperature.

Given that the initial volume of the acetone is 150 mL at a temperature of 34.5°C, we need to convert this temperature to Kelvin by adding 273.15. Therefore, T1 = 34.5°C + 273.15 = 307.65 K.

Since the final temperature is 20.0°C, the final temperature in Kelvin will be T2 = 20.0°C + 273.15 = 293.15 K. We can now plug these values into the equation to find the volume of the acetone at the lower temperature: V2 = 150 mL * (293.15 K / 307.65 K) = 142.39 mL.

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) A circular coil of diameter 20. cm, with 16. turns is in a 0.13 Tesla field. (a) Find the total flux through the coil when the field is perpendicular to the coil plane. (b) If the coil is rotated in 10. ms so its plane is parallel to the field, find the average induced emf.

Answers

Answer:

(a) 0.0041 weber

(b) 0.41 volt

Explanation:

diameter of coil, d = 20 cm

radius of coil, r = half of diameter = 10 cm = 0.1 m

magnetic field strength, B = 0.13 tesla

(a)

The angle between the normal of the coil and the magnetic field is 0°.

Magnetic flux, Ф = B x A x Cos 0°

Ф = 0.13 x 3.14 x 0.1 x 0.1 x 1

Ф = 0.0041 Weber

(b)

angle between the magnetic field and the normal of the coil is 90°.

time, t = 10 ms = 0.01 s

final flux = B x A x cos 90° = 0

induced emf = rate of change of magnetic flux

e = (0.0041 - 0) / 0.01

e = 0.41 Volt

Answer:

a) $\phi=0.4084\ T.m^2$

b) $emf=653.44\ V$

Explanation:

Given:

diameter of the coil, $d=20\ cm=0.2\ m$

no. of turns in the coil, $N=16$

magnetic field strength to which the coil is subjected, $B=0.13\ T$

time taken by the coil to rotate from normal the field to parallel, $t=10* 10^(-3)\ s$

The flux through the coil can be given as:

$\phi=BA$

where:

$A=$ area enclosed by the section of the coil

$\phi=0.13* \pi* (0.2^2)/(4)$

$\phi=0.4084\ T.m^2$

When the coil is rotated there is change in flux which lead to an induced emf in the coil according to the Faraday's law:

$emf=N(d\phi)/(t)$

where:

$d\phi=$ change in the flux

here the flux changes from maximum value to zero when the coil becomes parallel to the field lines because then there is no field line intercepting the coil area.

$emf=16* (0.4084)/(0.01)$

$emf=653.44\ V$

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