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3 years ago

Select all that apply. Quarks are thought to be the basic component of _____. protons electrons neutrons

2 answers:

5 0

Answer: protons and neutrons.

Justification:

Contrary to the belief of some years ago, protons and neutrons are not considered elementary or fundamental particles any more.

Electrons remain being considered elementary particles.

All three, protons, electrons and neutrons are subatomic partilces, but the only elemental particle of them is the electron.

That means that protons and neutrons are composite particles (they are composed by other smaller, elementary particles).

Those particles that constitute the protons and neutrons are the quarks, as stated by the question.

KengaRu [80]3 years ago

3 0

Quarks are types of particles and a basic constituent of matter. They <span>form composite particles called hadrons when they combine.
They are fundamental components of Protons and neutrons.</span>

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Tony has a mass of 50 kg, and his friend Sam has a mass of 45 kg. Assume that both friends push off on their rollerblades with t

Citrus2011 [14]

Answer:

Sam

Explanation:

Hes lighter

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4 years ago

A 20 g ball is fired horizontally with speed v0 toward a 100 g ball hanging motionless from a 1.0-m-long string. The balls un-de

sladkih [1.3K]

Answer:7.93 m/s

Explanation:

Given

mass of ball $m_1=20\ gm$

Mass of hanging ball $m_2=100\ gm$

Length of string $L=1\ m$

Maximum angle turned $\theta _{max}=50^{\circ}$

$v_o$ is the initial velocity of ball 1 and 0 is the initial velocity of ball 2

For Perfectly elastic final velocity of ball 1 and 2 is given by

$v_2'=\frac{2m_1}{m_1+m_2}\cdot v_1-\frac{m_1-m_2}{m_1+m_2}\cdot v_2$

$v_1'=\frac{m_1-m_2}{m_1+m_2}\cdot v_1+\frac{2m_2}{m_1+m_2}\cdot v_2$

where $v_1$ and $v_2$ are the velocity of 1 and 2 before collision

thus $v_2'=\frac{2\times 20}{120}v_0-0$

$v_2'=\frac{v_o}{3}$

$v_1'=\frac{20-100}{120}\times v_o+0$

$v_1'=-\frac{2}{3}v_o$

By energy conservation on second ball we get

Kinetic energy=Potential Energy

$\frac{1}{2}m_2(v_2')^2=m_2gL(1-\cos \theta )$

$v_2'=\sqrt{2gl(1-\cos \theta )}$

$v_2'=\sqrt{2\times 9.8\times 1(1-0.642)}$

$v_2'=2.64\ m/s$

thus $v_o=3\times v_2'=7.93\ m/s$

5 0

3 years ago

Just wondering if I did this right

Kisachek [45]

Yeah

All they are all correct

5 0

3 years ago

Consider a positive charge Q and a point B twice as far away from Q as point A. What is the ratio of the electric field strength

Vikentia [17]

Answer:

$\frac{E_{A}}{E_{B}}=4$

Explanation:

The electric field is defined as the electric force per unit of charge, this is:

$E=\frac{F}{q}$ .

The electric force can be obtained through Coulomb's law, which states that the electric force between to electrically charged particles is inversely proportional to the square of the distance between them and directly proportional to the product of their charges. The electric force can be expressed as

$F=\frac{kQq}{r^{2}}$ .

By substitution we get that

$E=\frac{kQq}{qr^{2}}\\\\E=\frac{kQ}{r^{2}}$

Now, letting $E_{A}$ be the electric field at point A, letting $E_{B}$ be the electric field at point B, and letting R be the distance from the charge to A:

$E_{A}=\frac{kQ}{R^{2}}\\\\E_{B}=\frac{kQ}{(2R)^{2}}$ .

The ration of the electric fields is

$\frac{E_{A}}{E_{B}}=\frac{\frac{kQ}{R^{2}}}{\frac{kQ}{(2R)^{2}}}\\\\\frac{E_{A}}{E_{B}}=\frac{\frac{1}{R^{2}}}{\frac{1}{(2R)^{2}}}\\\\\frac{E_{A}}{E_{B}}=\frac{\frac{1}{R^{2}}}{\frac{1}{(4)R^{2}}}\\\\\\\frac{E_{A}}{E_{B}}=\frac{1}{\frac{1}{(4)}}\\\\\frac{E_{A}}{E_{B}}=4$

This means that at half the distance, the electric field is four times stronger.

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3 years ago

What is the speed of a rocket that travels 9,000 meters in 12 seconds?

Mademuasel [1]

9,000/12=750

Speed is equal to distance divided by time

7 0

3 years ago

Read 2 more answers