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Darya [45]
3 years ago
7

A wire of length 5mm and Diameter 2m extends by 0.25 when a force of 50N was use. calculate the

Physics
1 answer:
bazaltina [42]3 years ago
3 0

Answer and Explanation:

Data provided in the question

Force = 50N

Length = 5mm

diameter = 2.0m = 2\times 10^{-3}

Extended by = 0.25mm = 0.25\times 10^{-3}

Based on the above information, the calculation is as follows

a. The Stress of the wire is

= \frac{force\ applied}{area\ of \ circle}

here area of circle = perpendicular to the are i.e cross-sectional  i.e

= \frac{\pi d^{2}}{4}

= \frac{\pi(2\times 10^{-3})^2}{4}

Now place these above values to the above formula

= \frac{4\times 50}{\pi\times 4 \times 10^{-6}} \\\\ = \frac{50}{\pi}

= 15.92 MPa

As 1Pa = 1 by N m^2

So,

MPa = 10^6 N m^2

b. Now the strain of the wire is

= \frac{Change\ in\ length}{initial\ length} \\\\ = \frac{0.25\times 10^{-3}}{5}

= 5 \times 10^{-5}

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A standing wave pattern is created on a string with mass density μ = 3.4 × 10-4 kg/m. A wave generator with frequency f = 61 Hz
uranmaximum [27]

Answer:

1) λ = 0.413 m , 2)v = 25,213 m / s , 3)  T = 0.216 N , 4) m = 22.04 10-3 kg

Explanation:

1) The resonance occurs when the traveling wave bounces at the ends and the two waves are added, the ends as they are fixed have a node, the wavelength and the length of the string are related

         λ = 2L / n               n = 1, 2, 3 ...

In this case L = 0.62 m and n = 3

Let's calculate

        λ = 2 0.62 / 3

        λ = 0.413 m

2) the velocity related to wavelength and frequency

      v =  λ f

      v = 0.413 61

      v = 25,213 m / s

3) let's use the equation

     v = √T /μ

     T = v² μ

     T = 25,213² 3.4 10⁻⁴

     T = 0.216 N

4) the rope tension is proportional to the hanging weight

      T-W = 0

     T = W

    W = m g

    m = W / g

    m = 0.216 / 9.8

    m = 22.04 10-3 kg

5) n = 2

     λ = 2 0.62 / 2

     λ = 0.62 m

6) v =  λ f

     v = 0.62 61

     v = 37.82 m / s

7) T = v² μ

   T = 37.82² 3.4 10⁻⁴

   T = 0.486 N

8) m = W / g

   m = 0.486 / 9.8

   m = 49.62 10⁻³ kg

9) n = 1

    λ = 2 0.62

    λ = 1.24 m

    v = 1.24 61

    v = 75.64 m / s

    T = v² miu

    T = 75.64² 3.4 10⁻⁴

    T = 2.572 10⁻² N

    m = 2.572 10⁻² / 9.8

    m = 262.4 10⁻³ kg

5 0
3 years ago
Not every lightning strike produces thunder. a. Trueb.False
podryga [215]

Answer: The given statement is false.

Explanation:

Lightening is defined as steady expansion of air within and surrounding path due to the sudden increase in pressure and temperature that leads to emission of lightning.

Basically, lightning is a state of plasma in which the molecules are very rapidly striking with each other as it also contains positively charged ions and negatively charged electrons.

Lightning always leads to the production of thunder.

Therefore, we can conclude that the statement not every lightning strike produces thunder, is false.

7 0
3 years ago
As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a p
Irina-Kira [14]

Answer:

The magnitude of each force is 2.45 x 10⁻¹⁶ N

Explanation:

The charge of proton, +q = 1.603 x 10⁻¹⁹ C

The charge of electron, -q = 1.603 x 10⁻¹⁹ C

Distance between the two charges, r = 971 nm = 971 x 10⁻⁹ m

Apply Coulomb's law;

F = \frac{kq_1q_2}{r^2}

where;

k is Coulomb's constant = 8.99 x 10⁹ Nm²/C²

q₁ and q₂ are the charges of proton and electron respectively

F is the magnitude of force between them

Substitute in the given values and solve for F

F = \frac{(8.99*10^9)(1.603*10^{-19})^2}{(971*10^{-9})^2} \\\\F = 2.45*10^{-16} \ N

Therefore, the magnitude of each force is 2.45 x 10⁻¹⁶ N

4 0
3 years ago
A meter stick swinging from one end oscillates with a frequency f0. What would be the frequency, in terms of f0 , if the bottom
snow_lady [41]

To solve this problem we will proceed to define the Period of a stick, then we will define the frequency, which is the inverse of the period. We will compare the change suffered by the new length and replace that value. The Time period of meter stick is

T = 2\pi\sqrt{ \frac{L}{g}}

Here,

L = Length

g = Gravity

At the same time the frequency is

f = \frac{1}{T}

Therefore the frequency in Terms of the Period is

f_0 = \frac{1}{2\pi}\sqrt{\frac{g}{L}}

If bottom third were cut off then the new length is

L' = \frac{2}{3} L

Replacing this value at the new frequency we have that,

f ' = \frac{1}{2\pi} \sqrt{\frac{3g}{2L}}

f' = \sqrt{\frac{3}{2}}(\frac{1}{2\pi}\sqrt{\frac{g}{L}})

Finally,

\therefore f' = \sqrt{\frac{3}{2}}f_0

3 0
3 years ago
Read 2 more answers
What is the minimum height of a planar mirror placed in a vertical plane 2m distant that will permit a person 2m tall to see the
Taya2010 [7]

Answer:

Height of mirror will be 1 m for complete image of person

Explanation:

We have given distance of person from plane mirror d = 2 m

And height of person = 2 m

We have to find the height of plane mirror which given full image of person

Height of mirror will be equal to half of height of person

So height of mirror =\frac{height\ of\ person}{2}=\frac{2}{2}=1m

So height of mirror will be 1 m for complete image of person

8 0
3 years ago
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