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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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A 3,220 lb car enters an S-curve at A with a speed of 60 mi/hr with brakes applied to reduce the speed to 45 mi/hr at a uniform

Grace [21]

The magnitude of the total friction force exerted by the road on the tires at B will be 919.46 lb.

<h3>What is the friction force?</h3>

It is a type of opposition force acting on the surface of the body that tries to oppose the motion of the body. Its unit is Newton (N).

Mathematically, it is defined as the product of the coefficient of friction and normal reaction.

The given data in the problem is;

The weight is,W= 3,220 lb

The speed is,u= 60 mi/hr

The reducing speed is,v= 45 mi/hr

The distance traveled is,d= 300 ft

The radius of curvature of the path of the car at B is,R= 600 ft.

1 mile = 5280 ft

From the Newtons' equation of motion;

$\rm v^2 = u^2 +2ad \\\\ \rm (45 \times \frac{5280}{3600} )^2 = (60 \times \frac{5280}{3600} )^2 +2a\times 300 \\\\$

The tangential accelerations are;

$\rm a_t = \frac{66^2 -88^2}{600} \\\\ \rm a_t = -5.65 ft/sec^2 \\\\$

The force is found as;

$\rm \sum F = ma \\\\ \frac{3220}{32.2} \times -5.65 \\\\ F_T= 565 \ lb$

The normal force is;

$\rm F_n = \frac{3220}{32.2} \times \frac{66^2}{600} \\\\ F_N =726 \ lb$

The net or the total friction force exerted by the road on the tires at B. is found as;

$\rm F = \sqrt{(565)^2+(726)^2} \\\ F = 919.946 \ lb$

Hence, the magnitude of the total friction force exerted by the road on the tires at B will be 919.46 lb.

To learn more about the friction force, refer to the link;

brainly.com/question/1714663

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2 years ago

The equation used to predict the theoretical period Ty of a simple pendulum assumes a small amplitude of oscillation. A student

astra-53 [7]

Answer:

The answer is "Choice E".

Explanation:

In this situation the option e is right because its resistance decreases through time, however, the time is the same for the same reason, whereas the sphere deteriorates, somehow it travels shorter distances however if the air resistance becomes are using the amplitude of movement declines, that's why other choices were wrong.

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3 years ago

A battery charger is connected to a dead battery and delivers a current of 8.9 A for 4.7 hours, keeping the voltage across the b

adell [148]

Answer:

1807.56 kJ

Explanation:

Parameters given:

Current, I = 8.9A

Time, t = 4.7hrs = 4.7 * 3600 = 16920 secs

Voltage, V = 12V

Electrical energy is given as:

E = I*V*t

Where I = Current

V = Voltage/Potential differenxe

t = time in seconds.

E = 8.9 * 12 * 16920

E = 1807056 J = 1807.056 kJ

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3 years ago

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A man weighs 1190 n on earth. what would he weigh on jupiter, where the free-fall acceleration is 25.9 m/s 2 ? the acceleration

Elden [556K]

The answer would be 3,145N. Using W=mg solve for the mass of the man on earth. Once you have the mass you can multiply it by the gravity of Jupiter giving you his weight in Newton’s on Jupiter.

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3 years ago

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nikklg [1K]

Answer:

Explanation:

First, we need to calculate linear expansivity, then after finding that value, we can move on to finding the area expansivity.

<u />

=============================================================

Finding Linear Expansivity :

⇒ α = Final length - Original length / (Original length × ΔT)

⇒ α = 9 - 4 / (4 × 70 - 20)

⇒ α = 5 / 5 × 50

⇒ α = <u>0.02</u>

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Finding Area Expansivity :

⇒ Area Expansivity = 2 × Linear Expansivity

⇒ β = 2 × α

⇒ β = 2 × 0.02

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1 year ago

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