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aivan3 [116]
4 years ago
15

What items can be classified as matter?

Physics
1 answer:
Agata [3.3K]4 years ago
7 0
Gas , Liquid , or solid
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A string is stretched and fixed at both ends, 200 cm apart. If the density of the string is 0.015 g/cm, and its tension is 600 N
Gre4nikov [31]

Answer:

<h2>f₀ = 158.12 Hertz</h2>

Explanation:

The fundamental frequency of the string  f₀ is expressed as f₀ = V/4L where V is the speed experienced by the string.

V = \sqrt{\frac{T}{\mu} } where T is the tension in the string and  \mu is the density of the string

Given T = 600N and \mu = 0.015 g/cm  = 0.0015kg/m

V = \sqrt{\frac{600}{0.0015} }\\ \\V =  \sqrt{400,000}\\ \\V = 632.46m/s

The next is to get the length L of the string. Since the string is stretched and fixed at both ends, 200 cm apart, then the length of the string in metres is 2m.

L = 2m

Substituting the derived values into the formula f₀ = V/2L

f₀ = 632.46/2(2)

f₀ = 632.46/4

f₀ = 158.12 Hertz

Hence the fundamental frequency of the string is 158.12 Hertz

5 0
3 years ago
An iron wire has a length of 1.50 m and a cross sectional area of 0.290 mm2. If the resistivity of iron is 10.0 ✕ 10−8 Ω · m and
lapo4ka [179]

Answer:

1.35 A

Explanation:

Applying,

V = IR

I = V/R..................... Equation 1

I = Current, V = Voltage, R = Resistance.

But,

R = Lρ/A............... Equation 2

Where L = Length of the wire, ρ = resistivity, A = Cross-sectional area of the wire.

Sustitute equation 2 into equation 1

V = AV/Lρ............... Equation 3

From the question,

Given: V = 0.7 V, A = 0.290 mm² = 2.9×10⁻⁷ m², L = 1.5 m, ρ = 10×10⁻⁸ Ω.m

Substitute these values into equation 3

I = (0.7× 2.9×10⁻⁷)/(1.5× 10×10⁻⁸ )

I = (2.03×10⁻⁷)/(15×10⁻⁸)

I = 1.35 A

5 0
3 years ago
The atmosphere protects us from all of the following except Meteorites The sun's rays Pollution Wind
denis23 [38]
Meteorites is the answer

6 0
3 years ago
6. A 2-kg ball B is traveling around in a circle of radius r1 = 1 m with a speed (vB)1 = 2 m/s. If the attached cord is pulled d
kipiarov [429]

Answer:

Explanation:

Given that,

Mass of ball m = 2kg

Ball traveling a radius of r1= 1m.

Speed of ball is Vb = 2m/s

Attached cord pulled down at a speed of Vr = 0.5m/s

Final speed V = 4m/s

Let find the transverse component of the final speed using

V² = Vr²+ Vθ²

4² = 0.5² + Vθ²

Vθ² = 4²—0.5²

Vθ² = 15.75

Vθ =√15.75

Vθ = 3.97 m/s.

Using the conservation of angular momentum,

(HA)1 = (HA)2

Mb • Vb • r1 = Mb • Vθ • r2

Mb cancels out

Vb • r1 = Vθ • r2

2 × 1 = 3.97 × r2

r2 = 2/3.97

r2 = 0.504m

The distance r2 to the hole for the ball to reach a maximum speed of 4m/s is 0.504m

The required time,

Using equation of motion

V = ∆r/t

Then,

t = ∆r/Vr

t = (r1—r2) / Vr

t = (1—0.504) / 0.5

t = 0.496/0.5

t = 0.992 second

7 0
3 years ago
If you could live any where except for where you are now where would you live
laila [671]

Answer:

I would live in the Atlantic ocean on a lux liner

Explanation:

:)

4 0
3 years ago
Read 2 more answers
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