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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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IgorC [24]

the batteries would heat up due to the over load of power not going into any thing and the screw driver is giving it a boost of energy



8 0
3 years ago
Please solve it ,,,,,,,,,,................................
fredd [130]

Answer:

the distance covered by the body in 5th second =u+a(n- 1/2)=7+4(5- 1/2)

=7+4×9/2

=7+18

=25m

6 0
3 years ago
Two convex lenses are placed 15 cm apart. The left lens has a focal length of 10 cm, and the right lens a focal length of 5 cm.
Minchanka [31]

Answer:

The image distance from right lens is 2.86 cm and image is real.

Explanation:

Given that,

Focal length of left lens = 10 cm

Focal length of right lens = 5 cm

Distance between the lenses d= 15 cm

Object distance = 50 cm

We need to calculate the image distance from left lens

Using formula of lens

\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}

Put the value into the formula

\dfrac{1}{v}=\dfrac{1}{10}-\dfrac{1}{-50}

\dfrac{1}{v}=\dfrac{3}{25}

v=8.33\ cm

We need to calculate the image distance from right lens

The object distance will be

u = 15-8.33 = 6.67\ cm

Using formula of lens

\dfrac{1}{v}=\dfrac{1}{f}-\dfrac{1}{u}

Put the value into the formula

\dfrac{1}{v}=\dfrac{1}{5}-\dfrac{1}{-6.67}

\dfrac{1}{v}=\dfrac{1167}{3335}

v=2.86\ cm

The image is real.

Hence, The image distance from right lens is 2.86 cm and image is real.

4 0
3 years ago
When you lift a bowling ball with a force of 61.1 N, the ball accelerates upward with an acceleration a. If you lift with a forc
gtnhenbr [62]

Answer:

the weight of the ball is w = 51.94 N ( mass = 5.3 kg)

Explanation:

Following Newton's second law:

net force = mass * acceleration = weight/gravity * acceleration

then denoting 1 and 2 as the first and second lift

F₁ - w= w/g *a₁

F₂ -w = w/g *a₂ = w/g * 2.07a

dividing both equations

(F₂- w)/(F₁ -w)= 2.07

(F₂- w) = 2.07 * (F₁ -w)

1.07*w = 2.07*F₁ - F₂

w = (2.07*F₁ - F₂ )/ 1.07

replacing values

w = (2.07*61.1 N - 70.9 N )/ 1.07  = 51.94 N

then the weight of the ball is w = 51.94 N ( mass = 5.3 kg)

7 0
3 years ago
If the timber weighs 670 N, calculate its angle of inclination when the water surface is 2.1 m above the pivot. Above what depth
Ymorist [56]

Answer:

Q = 40.1 degrees

Explanation:

Given:

- The weight of the timber W = 670 N

- Water surface level from pivot y = 2.1 m

- The specific density of water Y = 9810 N / m^3

- Dimension of timber = (0.15 x 0.15 x 0.0036) m

Find:

- The angle of inclination Q that the timber makes with the horizontal.

Solution:

- Calculate the Flamboyant Force F_b acting upwards at a distance x along the timber, which is unknown:

                                   F_b = Y * V_timber

                                   F_b = 9810*0.15*0.15*x

                                   F_b = 226.7*x N

- Take static equilibrium conditions for the timber, and take moments about the pivot:

                                   (M)_p = 0

                                   W*0.5*3.6*cos(Q) - x/2 * F_b*cos(Q) = 0

- Plug values in:

                                   670*0.5*3.6 - x^2 * 0.5*226.7 = 0

                                   x^2 = 1206 / 113.35

                                   x = 3.26 m

- Now use the value of x and vertical height y to compute the angle of inclination to be:

                                   sin(Q) = y / x

                                   sin(Q) = 2.1 / 3.26

                                   Q = sin^-1 (0.6441718)

                                   Q = 40.1 degrees

5 0
3 years ago
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