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Law Incorporation [45]
3 years ago
15

The ball of a ballpoint pen is 0.5 mm in diameter and has an ASTM grain size of 12. How many grains are there in the ball

Physics
1 answer:
Fudgin [204]3 years ago
8 0

Given :

The ball of a ballpoint pen is 0.5 mm in diameter and has an ASTM grain size of 12.

To Find :

How many grains are there in the ball?

Solution :

Volume of ball of the ballpoint is :

V = \dfrac{4 \pi r^3}{3}\\\\V = \dfrac{4\times 3.14 \times 0.5^3}{3}\ mm^3\\\\V = 0.523\  mm^3

Now, grain size of 12 has about 520000 grains/mm³.

Therefore, number of grains are :

n = 520000\times 0.523\ grains\\\\n = 271960\ grains

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Leptons have +1e, -1e or 0 charge.

Photons have 0 charge.

Only quarks have a charge of +2/3e or -1/3e of an electron charge. 

To be exact, only up-type quarks (Up, Down and Top quarks) have a +2/3e or two thirds of an electron charge.

So the correct answer is D) Quark.
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When a pendulum is swinging, the velocity is highest at which point?
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At the center, when the bob is hanging straight down

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The decomposition of ammonia to the elements is a first-order reaction with a half-life of 200 s at a certain temperature. How l
Feliz [49]

Based on the half-life of the reaction, time taken for the partial pressure of ammonia to decrease from 0.100 atm to 0.00625 atm is 800 seconds.

<h3>What is the half-life of a substance?</h3>

The half-life of a substance is the time taken for half the amount of atoms present in that substance to decay.

The half-life of the reaction is 200 seconds.

After, one half-life, pressure reduces to 0.0500 atm

After, two half-lives, pressure reduces to 0.0250 atm

After, three half-lives, pressure reduces to 0.01250 atm

After, four half-lives, pressure reduces to 0.00625 atm

Time taken = 4 * 200 = 800 seconds

Therefore, time taken for the partial pressure of ammonia to decrease from 0.100 atm to 0.00625 atm is 800 seconds.

Learn more about half-life at: brainly.com/question/26689704

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7 0
2 years ago
A 900-kg car cruising at a constant speed of 60 km/h is to accelerate to 100 km/h in 4 s. The additional power needed to achieve
kiruha [24]

To solve this problem we will apply the concepts related to power as a function of the change of energy with respect to time. But we will consider the energy in the body equivalent to kinetic energy. The change in said energy will be the difference between the two velocity data given by half of the mass. We will first convert the given units into an international system like this

Initial Velocity,

V_i = 60km/h (\frac{1000m}{1km})(\frac{1h}{3600s})

V_i = 16.6667m/s

Final Velocity,

V_f = 100km/h (\frac{1000m}{1km})(\frac{1h}{3600s})

V_f = 27.7778m/s

Now Power is defined as the change of Energy over the time,

P = \frac{E}{t}

But Energy is equal to Kinetic Energy,

P = \frac{\frac{1}{2} m\Delta v^2}{t}

P = \frac{\frac{1}{2} m(v_f^2-v_i^2)}{t}

Replacing,

P = \frac{\frac{1}{2} (900)(27.7778^2-16.6667^2)}{4}

P = 56kW

Therefore the correct answer is A.

8 0
3 years ago
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