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:


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Two identical small charged spheres are a certain distance apart, and each one initially experiences an electrostatic force of magnitude F due to the other. With time, charge gradually leaks off of both spheres. Calculate the magnitude of the electrostatic force, when each of the spheres has lost half of its initial charge. (Your answer will be a function of F, since no values are giving)

Answers

Answer:

1/4F

Explanation:

We already know thatThe electrostatic force is directly proportional to the product of the charge, from Coulomb's law.

So F α Qq

But if it is now half the initial charges, then

F α (1/2)Q *(1/2)q

F α (1/4)Qq

Thus the resultant charges are each halved is (1/4) and the first initial force experienced at full charge.

Thus the answer will be 1/4F

In an electric circuit, resistance and current are ____A. directly proportional
B. inversely proportional
C. have no effect on each other

Answers

In an electric circuit, resistance and current are ____

A. directly proportional

B. inversely proportional

C. have no effect on each other

Explanation:

A

Consider as a system the Sun with Saturn in a circular orbit around it. Find the magnitude of the change in the velocity of the Sun relative to the center of mass of the system during the time Saturn completes half an orbit. Assume the mass of the Sun is 5.68 x10^29 kg, the mass of Saturn is 5.68 x10^26 kg, its period is 9.29 x10^8 s, and the radius of its orbit is 1.43 x 10^12 m. Ignore the influence of other celestial objects.?

Answers

Answer:

v_(su) = 19.44 m/s

Explanation:

m_(su)=5.68x10^(29)kg\nm_(sa)=5.68x10^(26)kg

T=9.29x10^8\nr_(o)=1.43x10^(12)

If the sun considered as x=0 on the axis to put the center of the mass as a:

m_(su)*r_(o)=(m_(sa)+m_(su))*r_(1)

solve to r1

r_1=(m_(sa)*r_(o))/(m_(sa)+m_(su))=(5.68x10^(26)*1.43x10^(12))/(5.68x10^(26)+5.68x10^(26))

r_1=1.428x10^9m

Now convert to coordinates centered on the center of mass.  call the new coordinates x' and y' (we won't need y').  Now since in the sun centered coordinates the angular momentum was  

L = (m_(sa)*2*pi*r_1^2)/(T)

where T = orbital period

then L'(x',y') = L(x) by conservation of angular momentum.  So that means

L_(sun)=(m_(sa)*2*\pi *( 2r_(o)*r_1 -r_1^2))/(T)

Since

L_(su)= m_(su)*v_(su)*r_1

then

v_(su)=(m_(sa)*2*pi*(2r_(o)*r_(1)-r_(1)^2))/(T*m_(sa)*r_1)

v_(su) = 19.44 m/s

Final answer:

In a two-body system such as the Sun-Saturn system, both bodies orbit around their mutual center of mass, or barycenter. Given the Sun's significantly larger mass, this barycenter is near the center of the Sun, and hence the Sun's change in velocity relative to the center of mass of the system as Saturn completes half an orbit is effectively zero.

Explanation:

The problem here is asking for the change in velocity of the Sun relative to the center of mass of the Sun-Saturn system as Saturn completes half an orbit. This is a situation involving orbital physics and center of mass systems.

However, in an isolated two-body orbit system like this, the center of mass does not change velocity - it would remain constant, not unless acted upon by an outside force, which the problem instructs us to ignore.

Saturn and the Sun both orbit around their common center of mass (their barycenter). Given that the Sun is immensely more massive than Saturn, this center of mass is located very close to the center of the Sun.

So, while the Sun does indeed move a little due to Saturn's influence, the change in velocity of Sun relative to the center of mass of the system during the time Saturn completes half an orbit, for all intents and purposes, is zero.

This is especially true unless the problem specifically mentions that the Sun is initially at rest with respect to the center of mass. In any other case, the relative velocity remains constant and hence the change is zero.

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Only one of three balls A, B, and C carries a net charge q. The balls are made from conducting material and are identical. One of the uncharged balls can become charged by touching it to the charged ball and then separating the two. This process of touching one ball to another and then separating the two balls can be repeated over and over again, with the result that the three balls can take on a variety of charges. Which one of the following distribution of charges could not possibly be achieved in this fashion, even if the process were repeated an infinite number of times?Why the answer is qA = 1/2q, qB=3/8q, qC=1/4q. Explain please.

Answers

Answer:

This is greater than the initial charge, which violates the principle that the charge cannot be created or destroyed, consequently this distribution is impossible to achieve

Explanation:

The metals distribute the charge on all surface when they touch the surface increases so that charge density decreases and when the charge is separated into smaller in each metal.

Let's apply this principle to our case.

One of the spheres is loaded with a charge q, when touching a ball its charge is reduced to 1 / 2q for each ball.

         qA = ½ q

         qB = ½ q

         qC = 0

The total charge is q

we make a second contact

If we touch the ball A again with the other sphere not charged C, the chare is distributed and when separated it is reduced by half

         qA = 1/2 (q / 2) = ¼ q

         qC = ¼ q

         qB = ½ q

At this point all spheres have a charge,

      qA = ¼ q

      qb = ½ q

      qC = ¼ q

The total charge is q

Now let's contact spheres B and one of the other two

       Q = ½ q + ¼ q = ¾ q

When splitting the charge

        qB = ½ ¾ q = 3/8 q

        qC = ½ ¾ q = 3/8 q

        qA = ¼ q

The total charge is q

Note that the total load is always equal to q

Now let's analyze the given configuration

Let's look for the total load

       Q = qA + QB + QC

       Q = ½ q + 3/8 q + ¼ q

        Q = 9/8 q

This is greater than the initial charge, which violates the principle that the charge cannot be created or destroyed, consequently this distribution is impossible to achieve

Final answer:

The principle of charge distribution among conductive materials is violated in qA = 1/2q, qB=3/8q, qC=1/4q, as the sum of charges on B and C doesn't equate to the charge on A and 3/8q isn’t a multiple of halving the original charge.

Explanation:

The answer lies in the fact that balls made of conducting materials when in contact, distribute charges evenly among them. This is due to the free movement of electrons within the conductive material that seeks to equalize potential difference, a property exploited in charge distribution problems of this sort.

Given the scenario, every time a charged ball, A, touches an uncharged ball (B or C) the net charge is evenly split between them. Hence, each subsequent distribution halves the charge of the originating ball (A) and gives the complementary half to the ball it's being touched to (B or C).

In the distribution, qA = 1/2q, qB=3/8q, qC=1/4q, the sum of charges on B and C does not equate to A, which is a violation of the charge conservation principle. Moreover, 3/8q isn’t a multiple of halving the original charge q, which negates the manner in which the charge is distributed (i.e., by halving).

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A hemispherical surface (half of a spherical surface) of radius R is located in a uniform electric field of magnitude E that is parallel to the axis of the hemisphere. What is the magnitude of the electric flux through the hemisphere surface?

Answers

Answer:

π*R²*E

Explanation:

According to the definition of electric flux, it can be calculated integrating the product E*dA, across the surface.

As the electric field E is uniform and parallel to the hemisphere axis,  and no charge is enclosed within it, the net flux will be zero, so, in magnitude, the flux across the opening defining the hemisphere, must be equal to the one across the surface.

The flux across the open surface can be expressed as follows:

\int\ {E} \, dA = E*A = E*\pi  *R^(2)

As E is constant, and parallel to the surface vector dA at any point, can be taken out of the integral, which is just the area of the surface, π*R².

Flux = E*π*R²

How many times will the temperature of oxygen with a mass of 1 kg increase if its volume is increased by 4 times, and the pressure is decreased by 2 times?Round off the answer to the nearest whole number.

Answers

Answer:

9.2 Relating Pressure, Volume,

Figure 1. In 1783, the first (a) hydrogen-filled balloon flight, (b) manned hot air balloon flight, and (c) manned hydrogen-filled balloon flight occurred. When the hydrogen-filled balloon depicted in (a) landed, the frightened villagers of Gonesse reportedly destroyed it with pitchforks and knives. The launch of the latter was reportedly viewed by 400,000 people in Paris.

Explanation:

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