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3 years ago

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

Physics

1 answer:

Bond [772]3 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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Answer:

(A) –14m/s

(B) –42.0m

Explanation:

The complete solution can be found in the attachment below.

This involves the knowledge of motion under the action of gravity.

Check below for the full solution to the problem.

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Why cant people with aids foght out the infection?

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A stream moving with a speed of 7.1 m/s reaches a point where the cross-sectional area of the stream decreases to one half of th

lana66690 [7]

Answer:

14.2 m/s

Explanation:

Given data:

Speed of the stream, v₁ = 7.1 m/s

let the cross section area at initial point be A₁

now area at the second point, A₂ = (1/2)A₁ = 0.5A₁

now, from the continuity equation, we have

A₁v₁ = A₂v₂

where, v₂ is the velocity at the narrowed portion

thus, on substituting the values, we get

A₁ × 7.1 = 0.5A₁ × v₂

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3 years ago

What is the number of the lowest energy level that contains an f sublevel?. . 3. . 4. . 5. . 6

DENIUS [597]

The correct answer among all the other choices is 4. This is the number of the lowest energy level that contains an f sublevel. Thank you for posting your question. I hope that this answer helped you. Let me know if you need more help.

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3 years ago

suppose a car manufacturer tested its cars for front end collsion by hauling them up on a crane and dropping them from a certain

IRINA_888 [86]

Initial height: 66.5 m

Explanation:

The problem can be solved by using the principle of conservation of energy.

If we neglect air resistance, the total mechanical energy of the car is conserved during the fall, therefore we can write:

$K_i + U_i = K_f + U_f$

where :

$K_i = 0$ is the kinetic energy of the car at the top (it starts from rest)

$U_i = mgh$ is the gravitational potential energy of the car at the top, with:

m = the mass of the car

g = the acceleration of gravity

h = the heigth of the car

$K_f = \frac{1}{2}mv^2$ is the kinetic energy of the car just before hitting the ground, with

v = 130 km/h final speed of the car

$U_f = 0$ is the gravitational potential energy of the car at the bottom

Re-arranging the equation, we find

$mgh=\frac{1}{2}mv^2$

and we have:

$g=9.8 m/s^2\\v = 130 km/h = 36.1 m/s$

Solving for h, we find the initial height of the car:

$h=\frac{v^2}{2g}=\frac{36.1^2}{2(9.8)}=66.5 m$

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3 years ago