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Alchen [17]
2 years ago
12

An antibaryon composed of two antiup quarks

Physics
1 answer:
Sveta_85 [38]2 years ago
5 0

Answer:

(2) −1 e

Explanation:

A quark is the lightest elementary particles which form hadron such as proton and neutron. A quark has fractional charge.

Up, charm and top quarks have +\frac{2}{3} e charge where as down, strange and bottom quarks have -\frac{1}{3}e charge.

The antiparticle of up quark is antiup quark and has charge -\frac{2}{3}e charge.

The antiparticle of down quark is antidown quark and has charge +\frac{1}{3}e charge.

An antibaryon is composed of two anti-up quark and one anti-down quark.

Net charge of the anti-baryon is:

2\times (-\frac{2}{3} e)+1\times (+\frac{1}{3})e=-1e

Thus, antibaryon has -1e charge.

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Rod A has twice the diameter of rod B, but both are made of iron and have the same initial length. Both rods are now subjected t
Vilka [71]

Answer:

ΔLa/ΔLb = 1

Explanation:

The change in length of a solid is given by the following formula:

ΔL = α L ΔT

where,

ΔL = Change in length

α = coefficient of linear expansion

L = Original Length

ΔT = Change in Temperature

Since, the length and change in temperature for both rods are same. Also, the material of each rod is same, which implies that coefficient of linear expansion for both rods is same. Hence, the ratio of change in length of both rods will be:

<u>ΔLa/ΔLb = 1</u>

4 0
2 years ago
A truck collides with a car on horizontal ground. At one moment during the collision, the magnitude of the acceleration of the t
Mice21 [21]

Answer:

The magnitude of the acceleration of the car is 35.53 m/s²

Explanation:

Given;

acceleration of the truck, a_t = 12.7 m/s²

mass of the truck, m_t = 2490 kg

mass of the car, m_c = 890 kg

let the acceleration of the car at the moment they collided = a_c

Apply Newton's third law of motion;

Magnitude of force exerted by the truck = Magnitude of force exerted by the car.

The force exerted by the car occurs in the opposite direction.

F_c = -F_t\\\\m_ca_c = -m_t a_t\\\\a_c =- \frac{m_ta_t}{m_c} \\\\a_c = -\frac{2490 \times 12.7}{890} \\\\a_c = - 35.53 \ m/s^2

Therefore, the magnitude of the acceleration of the car is 35.53 m/s²

3 0
2 years ago
A 60 kg student in a rowboat on a still lake decides to dive off the back of the boat. The studen'ts horizontal aceleration is 2
TiliK225 [7]
As per the third law of Newton, the force exerted by the boat over the student is equal in magnitude to the force that the student exerted on the boat.

So, calculate the force on the student using the second law of Newton, Force = mass * acceleration.

Force on the student = 60 kg * 2.0 m/s^2 = 120 N.

=> horizontal force exerted by the student on the boat = 120 N

Answer: option d. 120 N. toward the back of the boat.

Of course it is toward the back because that is where the student jumped from..
4 0
2 years ago
A highway curve with radius R = 274 m is to be banked so that a car traveling v = 25.0 m/s will not skid sideways even in the ab
belka [17]

Answer:

A)   θ = 13.1º  , B)  E

Explanation:

A) For this exercise, let's use Newton's second law, let's set a reference frame where the axis ax is in the radial direction and is horizontal, the axis y is vertical.

In this reference system the only force that we must decompose is the Normal one, let's use trigonometry

        sin θ = Nₓ / N

        cos θ = N_{y} / N

        Nₓ = N sin θ

       Ny = N cos θ

x-axis (radial)

        Nₓ = m a

where the acceleration is centripetal

         a = v² / R

we substitute

        -N sin θ = -m v² / R                   (1)

the negative sign indicates that the force and acceleration towards the center of the circle

y-axis (Vertical)

          Ny - W = 0

           N cos θ = mg

           N = mg / cos θ

we substitute in 1

          mg / cos θ  sin θ = m v² / R

          g tan θ = v² / R

          θ = tan⁻¹ (v² / gR)

we calculate

        θ = tan⁻¹ (25² / 9.8 274)

        θ = 13.1º

B) when comparing the equations the correct one is E

6 0
1 year ago
You just saw a magazine ad that shows several people having fun at a football game. They're all wearing a specific shoe brand. T
I am Lyosha [343]
The technique used is most likely Bandwagon. 
So, B. 
7 0
2 years ago
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