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2 years ago

Mark all of the antimatter particlesa) Proton b) Electron c)Anti-top d) Gluon e) Tau Neutrino

Physics

1 answer:

Bond [772]2 years ago

3 0

Answer:

c) Anti-top

Explanation:

Protons are particles which belong to the hadron group of particles. The antimatter equivalent particle is antiproton

Electrons are particles that belong to the fermion group of particles. The antimatter equivalent is antielectron

Antitop quarks are particles that belong to the group family of particles. They are the antimatter equivalent of top quarks.

Gluons are particles that belong to the group of particles called bosons.

Tau neutrinos are particles that belong to the fermion group of particles. The antimatter equivalent is Tau antineutrino.

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Explain your answer. What parts of your hypothesis were strong correct? What parts were weak?

Anastaziya [24]

This question is from quizlet.

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2 years ago

Which of the following has the greatest kinetic energy? A. a mass of 4m at velocity v. B. a mass of 3m at velocity 2v. C. a mass

Artist 52 [7]

Answer:

Option C

Explanation:

Kinetic energy is the energy that the body possesses by virtue of its motion.

The formula for Kinetic energy is given by $\frac{1}{2} mv^2$

Using this formula let us find kinetic energy for the bodies given and find out which is the greatest

A) KE = $\frac{1}{2} (4m)(v^2) = 2mv^2$

B) KE = $\frac{1}{2} (3m)(2v)^2 = 6mv^2$

C) KE = $\frac{1}{2} (2m)(3v)^2 = 9mv^2$

D) KE = $\frac{1}{2} (3)(4v)^2 = 8mv^2$

Comparing these we find that 9mv^2 is the highest.

Hence option C is the answer.

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2 years ago

What is the velocity of the object?

dmitriy555 [2]

<h2>Hey There!</h2><h2>_____________________________________</h2><h2>Answer:</h2><h2 /><h2> $\huge\boxed{\text{V = 9.5 m/s}}$ </h2><h2>_____________________________________</h2>

mass = m = 2kg

Distance = x = 6m

Force = 30N

TO FIND:

Work = W = ?

Velocity = V = ?

<h2>SOLUTION:</h2>

According to the object of mass 2 kg travels a distance when the force was exerted on it. The graph between the Force and position was plotted which shows that 30 N of force was used to push the object till the distance of 6.0m.

To find the work, I will use the method of determining the area of the plotted graph. As the graph is plotted in the straight line between the Force and work, THE PICTURE ATTCHED SHOWS THE AREA COVERED IN BLUE AS WORK DONE AND HEIGHT AS 30m AND DISTANCE COVERED AS 6m To solve for the area(work) of triangle is given as,

${\Longrightarrow}\qquad \qquad \qquad W\ =\ \frac{1}{2}\;(Base)\:(Height)$

Base is the x-axis of the graph which is Position i.e. 6m

Height is the y-axis of the graph which is Force i.e. 30N

So,

$W\ =\ \frac{1}{2}\:6\:x\:30$

W = 90 J

The work done is 90 J.

According to the principle of work and kinetic energy (also known as the work-energy theorem) states that the work done by the sum of all forces acting on a particle equals the change in the kinetic energy of the particle.

${\Longrightarrow}\qquad \qquad \qquad W\quad =\quad K.E\\\\{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad \frac{1}{2}\ m\ V^2 \\\\{\Longrightarrow}\qquad \qquad \qquad W\quad =\quad \frac{1}{2}\ 2\ (V_f-V_i)^2\\\\{V_i\ is\ 0\ because\ the\ object\ was\ initially\ at\ rest}\\\\ {\Longrightarrow}\qquad \qquad \qquad W\quad\ =\ \frac{1}{2}\ x\ 2\ (V_f-0)^2 \\\\{\Longrightarrow}\qquad \qquad \qquad 90\quad = \frac{1}{2}\ x\ 2\ (V_f)^2$

$\\\\{\Longrightarrow}\qquad \qquad \qquad V_f\quad =\ \sqrt{\frac{2\ (90)\ }{2}}\\\\{\Longrightarrow}\qquad \qquad \qquad \boxed {V_f\quad =\ 9.48\ m/s}$

$\boxed{The\ Velocity\ of\ the\ Object\ of\ mass\ 2kg\ at\ 6\ meters\ of\ distance\ was\ 9.48\ m/s}$

<h2>_____________________________________</h2><h2>Best Regards,</h2><h2>'Borz'</h2>

8 0

2 years ago

To analyze the motion of a body that is traveling along a curved path, to determine the body's acceleration, velocity, and posit

DiKsa [7]

To solve this problem we will apply the kinematic equations of linear motion and centripetal motion. For this purpose we will be guided by the definitions of centripetal acceleration to relate it to the tangential velocity. With these equations we will also relate the linear velocity for which we will find the points determined by the statement. Our values are given as

$R = 350ft$