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2 years ago

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]2 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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Help ASAP with equation and answer pls. Numbers 2 3 4

scoray [572]

Answer:

2. 200N

3.50kg

4.700N

Explanation:

Weight is another word for the force of gravity

Weight is a force that acts at all times on all objects near Earth.

F=m*g

where g=acceleration due to gravity

2. due to the gravitational fields of the earth , assume gravitational acceleration=10m/s2

F=20*10= 200N

3.same as above

mass=Force/gravitational acceleration

mass=500/10 = 50kg

4.force=mass*gravitational acceleration

force=70*10=700N

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2 years ago

A 3.0-A current is maintained in a simple circuit that consists of a resistor between the terminals of an ideal battery. If the

harina [27]

Answer:

$R=2.78\ \Omega$

Explanation:

Given that,

The current flowing in the circuit, I = 3 A

The power of the battery, P = 25 W

We need to find the resistance of the battery. We know that the power of the battery is given by the formula as follows :

$P=I^2R$

Put all the values to find R.

$R=\dfrac{P}{I^2}\\\\R=\dfrac{25}{(3)^2}\\\\R=2.78\ \Omega$

So, the resistance is equal to $2.78\ \Omega$ .

7 0

2 years ago

The speed of sound in air is around 330 m/s. If a bat emits a single high-pitched ‘click’ of sound in a cave that is 25m wide, c

Bingel [31]

Answer:

0.15 s

Explanation:

From the question given above, the following data were obtained:

Speed of sound (v) = 330 m/s

Distance (x) = 25 m

Time (t) =?

The time taken for the echo of the sound to the bat can be obtained as follow:

v = 2x / t

330 = 2 × 25 / t

330 = 50 / t

Cross multiply

330 × t = 50

Divide both side by 330

t = 50 / 330

t = 0.15 s

Thus, it will take 0.15 s for the echo of the sound to the bat

4 0

2 years ago

I'm a little bit unsure about this question.

slava [35]

Answer:

Option C. 4 Hz

Explanation:

To know the correct answer to the question given above, it is important we know the definition of frequency.

Frequency can simply be defined as the number of complete oscillations or circles made in one second.

Considering the diagram given above, the wave passes through the medium over a period of one second.

Thus, we can obtain the frequency by simply counting the numbers of complete circles made during the period.

From the diagram given above,

The number of circles = 4

Thus,

The frequency is 4 Hz

8 0

2 years ago

A 15.5 kg block is pulled by two forces. The first is 11.8 N at a 53.7 angle and the second is 22.9 at a -15.8 angle. What is th

Ainat [17]

Answer:

$1.88 m/s^2$ at $6.5^{\circ}$

Explanation:

We need to calculate the components of the resultant force on both the x (horizontal) and y (vertical) direction.

Components of the first force F1:

$F_{1x} =(11.8) cos (53.7^{\circ})=7.0 N\\F_{1y} = (11.8) sin (53.7^{\circ})=9.5 N$

Components of the second force F2:

$F_{2x} =(22.9) cos (-15.8^{\circ})=22.0 N\\F_{2y} = (22.9) sin (-15.8^{\circ})=-6.2 N$

So the components of the resultant force are

$R_x = F_{1x}+F_{2x}=7.0+22.0 = 29.0 N\\R_y = F_{1y}+F_{2y} = 9.5+(-6.2)=3.3 N$

So the magnitude of the resultant force is

$F=\sqrt{(29.0)^2+(3.3)^2}=29.2 N$

And the direction is

$\theta = tan^{-1} (\frac{R_y}{R_x})=tan^{-1} (\frac{3.3}{29.0})=6.5^{\circ}$

The magnitude of the acceleration can be found by using Newton's second law:

$a=\frac{F}{m}=\frac{29.2 N}{15.5 kg}=1.88 m/s^2$

while the direction is the same as the resultant force, $6.5^{\circ}$ .

5 0

2 years ago