Ask question

Ask question

All categories

2 years ago

11

When tension is applied to a metal wire of length L , it stretches by Δ L . If the same tension is applied to a metal wire of th

e same material with the same cross-sectional area but of length 2 L , by how much will it stretch?

2 answers:

hichkok12 [17]2 years ago

7 0

Answer:

It will stretch by 2ΔL

Explanation:

The young modulus for the metals in both cases is the same as it is a property of the metal.

Y1 = Y2

Tensile force and area in both cases are the same.

Let the unknown increase in length be x.

F/A × L/ΔL = F/A × 2L/x

By canceling out like terms and rearranging, x = 2ΔL.

Anna [14]2 years ago

6 0

Answer:

The metal wire will stretch by $2 \delta L$

Explanation:

$T = \frac{kA \delta L}{L}$ ......................................(1)

Where T = Tension applied

ΔL = Extension

L = length

k = constant

T₁ = T₂ = T

A₁ = A₂ =A

L₁ = L

L₂ = 2L

(ΔL)₁ = ΔL

(ΔL)₂ = ?

From equation (1)

$TL/kA = \delta L$ .....................(2)

$TL/kA = (\delta L) ........................(3)\\ 2TL/kA = (\delta L)_{2} ........................(4)$

Divide (4) by (3)

$\frac{(\delta L)_{2} }{\delta L} =\frac{\frac{2TL}{kA} }{\frac{TL}{kA} } \\\frac{(\delta L)_{2} }{\delta L} = 2\\ (\delta L)_{2} = 2\delta L$

You might be interested in

An offshore oil well is 2 kilometers off the coast. The refinery is 4 kilometers down the coast. Laying pipe in the ocean is twi

shusha [124]

Answer:

Rectangular path

Solution:

As per the question:

Length, a = 4 km

Height, h = 2 km

In order to minimize the cost let us denote the side of the square bottom be 'a'

Thus the area of the bottom of the square, A = $a^{2}$

Let the height of the bin be 'h'

Therefore the total area, $A_{t} = 4ah$

The cost is:

C = 2sh

Volume of the box, V = $a^{2}h = 4^{2}\times 2 = 128$ (1)

Total cost, $C_{t} = 2a^{2} + 2ah$ (2)

From eqn (1):

$h = \frac{128}{a^{2}}$

Using the above value in eqn (1):

$C(a) = 2a^{2} + 2a\frac{128}{a^{2}} = 2a^{2} + \frac{256}{a}$

$C(a) = 2a^{2} + \frac{256}{a}$

Differentiating the above eqn w.r.t 'a':

$C'(a) = 4a - \frac{256}{a^{2}} = \frac{4a^{3} - 256}{a^{2}}$

For the required solution equating the above eqn to zero:

$\frac{4a^{3} - 256}{a^{2}} = 0$

$\frac{4a^{3} - 256}{a^{2}} = 0$

a = 4

Also

$h = \frac{128}{4^{2}} = 8$

The path in order to minimize the cost must be a rectangle.

8 0

2 years ago

A 0.9 µF capacitor is charged to a potential difference of 10.0 V. The wires connecting the capacitor to the battery are then di

jeyben [28]

Answer:

3.6μF

Explanation:

The charge on the capacitor is defined by the formula

q = CV

because the charge will be conserved

q₁ = C₁V₂

q₂ = C₂V₂ where C₂ V₂ represent the charge on the newly connected capacitor and the voltage drop across the two capacitor will be the same

q = q₁ + q₂ = C₁V₂ + C₂V₂

CV = CV₂ + C₂V₂

CV - CV₂ = C₂V₂

C ( V - V₂) = C₂V₂

C ( V/ V₂ - V₂ /V₂) = C₂

C₂ = 0.9 ( 10 /2) - 1) = 0.9( 5 - 1) = 3.6μF

7 0

2 years ago

Why in a rollercoaster the height of the tracks is either equal to or less than the first track

umka2103 [35]

Answer:

I really don't know

Explanation:

sorry

6 0

1 year ago

Electromagnetic radiation behaves both as particles (called photons) and as waves. Wavelength (λ) and frequency (ν) are related

inessss [21]

<h2>Answer: 136.363 m</h2>

Explanation:

We can find the wavelength of the radiation produced by the microwave oven by using the following given equation:

$c=\lambda.\nu$ (1)

Clearing $\lambda$ :

$\lambda=\frac{c}{\nu}$ (2)

Knowing $\nu=2.20 GHz=2.20(10)^{6}Hz=2.20(10)^{6}s^{-1}$

$\lambda=\frac{3(10)^{8}m/s}{2.20(10)^{6}s^{-1}}$ (3)

$\lambda=136.363m$ This is the wavelength of the radiation produced by the microwave

8 0

2 years ago

Objects with masses of 235 kg and a 535 kg are separated by 0.330 m. (a) find the net gravitational force exerted by these objec

saul85 [17]

The 61.0 kg object<span> ... F = (300kg)(6.673×10−11 </span>N m<span>^2 </span>kg<span>^−2)(61kg)/(.225m)^2. F = 2.412e-5 </span>N<span> towards the 495 </span>kg<span> block. </span>b. [195kg] ===.45m ... (b<span>) You cannot achieve this </span>position<span>. For the </span>net force<span> to become zero, one or both of the </span>masses<span> must ...</span>

7 0

2 years ago