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sweet [91]
3 years ago
14

A wire has a cross sectional area of 4.00 mm2 and is stretched by 0.100 mm by a certain force. How far will a wire of the same m

aterial and length stretch if its cross-sectional area is 8.00 mm2 and the same force is used to stretch it
Physics
1 answer:
Nina [5.8K]3 years ago
5 0

Answer: 0.05\ mm

Explanation:

Given

Cross-sectional area of wire A_1=4\ mm^2

Extension of wire \delta l=0.1\ mm

Extension in a wire is given by

\Rightarrow \delta l=\dfrac{FL}{AE}

where, E=\text{Youngs modulus}

\Rightarrow \delta_1=\dfrac{FL}{A_1E}\quad \ldots(i)

for same force, length and material

\Rightarrow \delta_2=\dfrac{FL}{A_2E}\quad \ldots(ii)

Divide (i) and (ii)

\Rightarrow \dfrac{0.1}{\delta_2}=\dfrac{A_2}{A_1}\\\\\Rightarrow \delta_2=0.1\times \dfrac{4}{8}\\\\\Rightarrow \delta_2=0.05\ mm

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It has been suggested that rotating cylinders several miles in length and several miles in diameter be placed in space and used
stepladder [879]

Answer:

the required revolution per hour is 28.6849

Explanation:

Given the data in the question;

we know that the expression for the linear acceleration in terms of angular velocity is;

a_{c} = rω²

ω² = a_{c} / r

ω = √( a_{c} / r )

where r is the radius of the cylinder

ω is the angular velocity

given that; the centripetal acceleration equal to the acceleration of gravity a a_{c}  = g = 9.8 m/s²

so, given that, diameter = 4.86 miles = 4.86 × 1609 = 7819.74 m

Radius r = Diameter / 2 = 7819.74 m / 2 = 3909.87 m

so we substitute

ω = √( 9.8 m/s² / 3909.87 m )

ω = √0.002506477 s²  

ω = 0.0500647 ≈ 0.05 rad/s  

we know that; 1 rad/s = 9.5493 revolution per minute

ω = 0.05 × 9.5493 RPM

ω = 0.478082 RPM  

1 rpm = 60 rph  

so  

ω = 0.478082 × 60

ω = 28.6849  revolutions per hour  

Therefore, the required revolution per hour is 28.6849

7 0
3 years ago
A vibrating object produces periodic waves with a wavelength of 53 cm and a frequency of 15 Hz. How fast do these waves move awa
DerKrebs [107]

Answer:

v = 7.95 m/s

Explanation:

Given that,

Wavelength of a wave, \lambda=53\ cm=0.53\ m

Frequency of a wave, f = 15 Hz

We need to find the speed of the wave. The speed of a wave is given by :

v=f\lambda\\\\v=15\ Hz\times 0.53\ m\\\\v=7.95\ m/s

So, the wave move with a speed of 7.95 m/s.

5 0
3 years ago
An ocean wave traveling in one direction has a wavelength of 1.0 m and a frequency of 1.25 Hz. Take the direction of wave propag
aliina [53]

Answer:

a) v=1*1.25=1.25\: m/s

b) y(x,t)=2sin(2\pi( x-1.25 t)  

c) y(3,10)=1.73 m    

Explanation:

a) The speed of a wave is given by the following equation:

v=\lambda f

Where:

λ is the wavelength

f is the frequency

v=1*1.25=1.25\: m/s

b) The harmonic wave has the following equation:

y=Asin(kx-\omega t)

A is the amplitude (2 m)

k is the wavenumber (2π/λ)  

ω is the angular frequency (2πf)

y(x,t)=2sin(2\pi x-2\pi*1.25 t)  

y(x,t)=2sin(2\pi( x-1.25 t)  

c) Here we need to find the heigth at x=3 m and t =10 s, so we need to find y(3,10).

y(3,10)=2sin(2\pi(3-1.25*10)

y(3,10)=2sin(2\pi(3-1.25*10)              

y(3,10)=1.73 m              

I hope it helps you!

 

6 0
3 years ago
A clown at a birthday party has brought along a helium cylinder, with which he intends to fill balloons. When full, each balloon
Furkat [3]

Answer:

11

Explanation:

According to Boyle's law:

P\times V=constant

Thus,

P_{cylinder}\times V_{cylinder}=n\times P_{balloon}\times V_{balloon}

Where, n is the number of the balloons

From the question, it is given that:

For balloon:

P = 1.2\times 10^5 Pa

V = 0.040 m³

For cylinder:

P = 1.80\times 10^7 Pa

V = 0.0031 m³

So,

1.80\times 10^7\times 0.0031=n\times 1.2\times 10^5\times 0.040

n = 11.625

<u>So, Maximum number of balloons = 11</u>

7 0
3 years ago
Calculate the heat required, in calories, to raise
Goshia [24]

Answer:Q=4450.74\ J

Explanation:

Given

mass of tin m=100 gm

Also we know that

specific heat of tin c_{tin}=0.21 J/gm-^{\circ}C

Melting Point of Tin T=231.94^{\circ}C

Heat required to raise the temperature of 100 gm tin is given by

Q=mc\Delta T

where Q=heat\ added

c=specific\ heat\ of\ tin

\Delta T=change\ in\ temperature

Q=100\times 0.21\times (231.94-20)

Q=4450.74\ J

5 0
3 years ago
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