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3 years ago

Mark all the fermions a) Proton b) Electron c)Anti-top d) Gluon e) Tau Neutrino

Physics

1 answer:

aliya0001 [1]3 years ago

5 0

Answer:

(a) Proton, (b) Electron (c) tau neutrino.

Explanation:

Elementary particles are the particles without any sub-structure which means they are not composed of other particles.

The elementary particles are classified into three categories which are discussed below:

(1) Quarks: up, down, top, bottom, strange, and charm.

(2) Leptons: electrons, electron neutrino, muon, muon neutrino, tau, tau neutrino.

(3) Bosons: Guon, photons Z bosons, W bosons, Higgs.

Fermions are particles that follow Fermi-Dirac statistics. Fermions consist of all the quarks and leptons. And fermions can also be categorized in elementary particles such as electrons and protons.

Therefore, the fermions in the given question are proton, electron, and tau neutrino.

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Answer:

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2 years ago

Problem 4: A uniform flat disk of radius R and mass 2M is pivoted at point P A point mass of 1/2 M is attached to the edge of th

brilliants [131]

From the case we know that:

The moment of inertia Icm of the uniform flat disk witout the point mass is Icm = MR².
The moment of inerta with respect to point P on the disk without the point mass is Ip = 3MR².
The total moment of inertia (of the disk with the point mass with respect to point P) is I total = 5MR².

Please refer to the image below.

We know from the case, that:

m = 2M

r = R

m2 = 1/2M

distance between the center of mass to point P = p = R

Distance of the point mass to point P = d = 2R

We know that the moment of inertia for an uniform flat disk is 1/2mr². Then the moment of inertia for the uniform flat disk is:

Icm = 1/2mr²

Icm = 1/2(2M)(R²)

Icm = MR² ... (i)

Next, we will find the moment of inertia of the disk with respect to point P. We know that point P is positioned at the arc of the disk. Hence:

Ip = Icm + mp²

Ip = MR² + (2M)R²

Ip = 3MR² ... (ii)

Then, the total moment of inertia of the disk with the point mass is:

I total = Ip + I mass

I total = 3MR² + (1/2M)(2R)²

I total = 3MR² + 2MR²

I total = 5MR² ... (iii)

Learn more about Uniform Flat Disk here: brainly.com/question/14595971

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8 0

1 year ago

A 5 kgkg sphere having a charge of ++ 8 μCμC is placed on a scale, which measures its weight in newtons. A second sphere having

Mrac [35]

Answer:

F_Balance = 46.6 N ,m' = 4,755 kg

Explanation:

In this exercise, when the sphere is placed on the balance, it indicates the weight of the sphere, when another sphere of opposite charge is placed, they are attracted so that the balance reading decreases, resulting in

∑ F = 0

Fe –W + F_Balance = 0

F_Balance = - Fe + W

The electric force is given by Coulomb's law

Fe = k q₁ q₂ / r₂

The weight is

W = mg

Let's replace

F_Balance = mg - k q₁q₂ / r₂

Let's reduce the magnitudes to the SI system

q₁ = + 8 μC = +8 10⁻⁶ C

q₂ = - 3 μC = - 3 10⁻⁶ C

r = 0.3 m = 0.3 m

Let's calculate

F_Balance = 5 9.8 - 8.99 10⁹ 8 10⁻⁶ 3 10⁻⁶ / (0.3)²

F_Balance = 49 - 2,397

F_Balance = 46.6 N

This is the balance reading, if it is calibrated in kg, it must be divided by the value of the gravity acceleration.

Mass reading is

m' = F_Balance / g

m' = 46.6 /9.8

m' = 4,755 kg

6 0

3 years ago

A student hangs a weight on a newtonmeter. The energy currently stored in the spring in the newton meter is 0.045N. The student

castortr0y [4]

Answer:

5x10^-3

Explanation:

Hooke's Law states that the force needed to compress or extend a spring is directly proportional to the distance you stretch it.

Hooke's Law can be represented as

<em>where F is the force </em>

<em> k is the spring constant</em>

<em> x is the extension of the material </em>

<em />

Plug values in the equation

Step 1 find the original extension

0.045 = (400)x

x = 1.125x 10^-4 m d

Step 2 find the new extension

0.045+2 = 400(x)

2.045 = 400x

x = 5.1125x10^-3

Step 3 subtract the new extension with original

Total extension of the spring = 5.1125x10^-3 - 1.125x 10^-4 m = 5x10^-3

8 0

3 years ago

Calculate the change in total internal energy for a system that releases 2.59 × 104 kJ of heat and does 6.46 × 104 kJ of work on

mrs_skeptik [129]

Answer:

See explanation below

Explanation:

The equation to use for this is the following:

dU = q + w

As the heat is being release, this value is negative, and same here happens with the work done, because it's in the surroundings.

Therefore the change in the energy would be:

dU = -2.59x10^4 - 6.46^4

dU = -9.05x10^4 kJ

8 0

3 years ago