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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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A factory has a solid copper sphere that needs to be drawn into a wire. The mass of the copper sphere is 76.5 kg. The copper nee
baherus [9]

Answer:

120.125 m

Explanation:

Density = Mass/volume

D = m/v .............................. Equation 1.

Where D = Density of the solid copper sphere, m = mass of the solid copper sphere, v = volume of the solid copper sphere.

Making v the subject of the equation,

v = m/D............................... Equation 2

Given: m = 76.5 kg,

Constant: D = 8960 kg/m .

Substituting into equation 2

v = 76.5/8960

v = 0.0085379 m³

Since the copper sphere is to be drawn into wire,

Volume of the copper sphere = volume of the wire

v = volume of the wire

Volume of wire = πd²L/4

Where d = diameter of the wire, L = length of the wire.

Note: A wire takes the shape of a cylinder.

v = πd²L/4 ........................ equation 3.

making L the subject of the equation,

L = 4v/πd²..................... Equation 4

Given: v = 0.0085379 m³, d = 9.50 mm = 0.0095  and π = 3.14

Substitute into equation 4

L = 4×0.0085379/(3.15×0.0095²)

L = 0.0341516/0.0002843

L = 120.125 m.

L = 120.125 m

Thus the length of the wire produced = 120.125 m

4 0
4 years ago
A deer with a mass of 146kg is running head on toward you with a speed of 17 m/s. You are going north. Find the momentum of the
Pavel [41]
Momentum = Mass * Velocity = 146 * 17 = 2482  kgm/s
6 0
4 years ago
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A heavy piece of hanging sculpture is suspended by a 90 cm-long, 5.0 g steel wire. When the wind blows hard, the wire hums at it
kupik [55]

Answer: The mass of the sculpture is 11.8kg

Explanation:

Using the equation of fundamental frequency of a taut string.

f = (1/2L)*√(T/μ) .... (Eqn1)

Where

f= frequency in Hertz =80Hz

T = Tension in the string = Mg

M represent the mass of the substance (sculpture) =?

g= 9.8m/s^2

L= Length of the string=90cm=0.9m

μ= mass density = mass of string /Length of string

mass of string =5g=0.005kg

L=0.9m

μ=0.005/0.9 = 0.0056kg/m

Using (Eqn1)

80= 1/(2*0.9) √(T/0.0056)

144= √(T/0.0056)

Square both sides

20736= T/0.0056

T= 116.12N

Recall that T =Mg

116.12= M * 9.8

M=116.12/9.8

M= 11.8kg

Therefore the mass of the sculpture is 11.8kg

4 0
3 years ago
What is inertia? explain please!
MrRa [10]
Inertness, especially with regard to effort, motion, action, and the like; inactivity; sluggishness.
3 0
3 years ago
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An oscillator consists of a block of mass 0.373 kg connected to a spring. When set into oscillation with amplitude 33 cm, the os
aniked [119]

Answer:

(a)  T = 0.412s

(b)  f = 2.42Hz

(c)  w = 15.25 rad/s

(d)  k = 86.75N/m

(e)  vmax = 5.03 m/s

Explanation:

Given information:

m: mass of the block = 0.373kg

A: amplitude of oscillation = 22cm = 0.22m

T: period of oscillation = 0.412s

(a) The period is the time of one complete oscillation = 0.412s

The period is 0.412s

(b) The frequency is calculated by using the following formula:

f=\frac{1}{T}=\frac{1}{0.412s}=2.42Hz

The frequency is 2.42 Hz

(c) The angular frequency is:

\omega=2\pi f=2\pi (2.42Hz)=15.25\frac{rad}{s}

The angular frequency is 15.25 rad/s

(d) The spring constant is calculated by solving the following equation for k:

\omega=\sqrt{\frac{k}{m}}\\\\k=m\omega^2=(0.373kg)(15.25rad/s)^2=86.75\frac{N}{m}

The spring constant is 86.75N/m

(e) The maximum speed is:

v_{max}=\omega A=(15.25rad/s)(0.33m)=5.03\frac{m}{s}

(f) The maximum force applied by the spring if for the maximum elongation, that is, the amplitude:

F=kA=(86.75N/m)(0.2m)=17.35N

The maximum force that the spring exerts on the block is 17.35N

3 0
4 years ago
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