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maksim [4K]
3 years ago
14

A steel wire of length 4.7 m and cross section 3 x 103 m2 stretches by the same amount as a copper wire of length 3.5 m and cros

s section 4 x 10-5 m2 under a given load. What is the ratio of the Young's modulus of steel to that of copper? (a) 3.83 x 103 (b) 1.46 x 10-2 (d) 5.85 x 10-3 (c) 1.79 x 10-2 2.
Physics
1 answer:
EleoNora [17]3 years ago
5 0

Answer:

The ratio of the young's modulus of steel and copper is 1.79\times10^{-2}

(c) is correct option.

Explanation:

Given that,

Length of steel wire = 4.7 m

Cross sectionA = 3\times10^{-3}\ m^2

Length of copper wire = 3.5 m

Cross sectionA = 4\times10^{-5}\ m^2

We need to calculate the ratio of young's modulus of steel and copper

Using formula of young's modulus for steel wire

Y=\dfrac{\dfrac{F}{A}}{\dfrac{\Delta l}{l}}

Y_{s}=\dfrac{Fl_{s}}{A_{s}\Delta l}....(I)

The young's modulus for copper wire

Y_{c}=\dfrac{Fl_{c}}{A_{c}\Delta l}....(II)

From equation (I) and (II)

The ratio of the young's modulus of steel and copper

\dfrac{Y_{s}}{Y_{c}}=\dfrac{\dfrac{Fl_{s}}{A_{s}\Delta l}}{\dfrac{Fl_{c}}{A_{c}\Delta l}}

\dfrac{Y_{s}}{Y_{c}}=\dfrac{A_{c}\times l_{s}}{A_{s}\times l_{c}}

\dfrac{Y_{s}}{Y_{c}}=\dfrac{4\times10^{-5}\times4.7}{3\times10^{-3}\times3.5}

\dfrac{Y_{s}}{Y_{c}}=1.79\times10^{-2}

Hence, The ratio of the young's modulus of steel and copper is 1.79\times10^{-2}

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Answer:

a) 4.49Hz

b) 0.536kg

c) 2.57s

Explanation:

This problem can be solved by using the equation for he position and velocity of an object in a mass-string system:

x=Acos(\omega t)\\\\v=-\omega Asin(\omega t)\\\\a=-\omega^2Acos(\omega t)

for some time t you have:

x=0.134m

v=-12.1m/s

a=-107m/s^2

If you divide the first equation and the third equation, you can calculate w:

\frac{x}{a}=\frac{Acos(\omega t)}{-\omega^2 Acos(\omega t)}\\\\\omega=\sqrt{-\frac{a}{x}}=\sqrt{-\frac{-107m/s^2}{0.134m}}=28.25\frac{rad}{s}

with this value you can compute the frequency:

a)

f=\frac{\omega}{2\pi}=\frac{28.25rad/s}{2\pi}=4.49Hz

b)

the mass of the block is given by the formula:

f=\frac{1}{2\pi}\sqrt{\frac{k}{m}}\\\\m=\frac{k}{4\pi^2f^2}=\frac{427N/m}{(4\pi^2)(4.49Hz)^2}=0.536kg

c) to find the amplitude of the motion you need to know the time t. This can computed by dividing the equation for v with the equation for x and taking the arctan:

\frac{v}{x}=-\omega tan(\omega t)\\\\t=\frac{1}{\omega}arctan(-\frac{v}{x\omega })=\frac{1}{28.25rad/s}arctan(-\frac{-12.1m/s}{(0.134m)(28.25rad/s)})=2.57s

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x=Acos(\omega t)\\\\A=\frac{0.134m}{cos(28.25rad/s*2.57s )}=0.45m

5 0
2 years ago
A 0.25 kg skeet (clay target) is fired at an angle of 30 degrees to the horizon with a speed of 25 m/s. When it reaches the maxi
kozerog [31]

Answer:

6.51 m and 37.13 m

Explanation:

from the question we were given

mass of skeet = 0.25 kg

speed of skeet = 25 m/s

angle = 30 degrees to the horizon

mass of pellet = 15 g

speed of pellet = 2000 m/s

without being hit by the pellet, the x and y components of the skeet velocity are  

Vx = 25 cos 30 = 21.65

Vy = 25 sin 30 = 12.5

now

V = U + (a x t)

where V = final velocity, U = initial velocity , a = acceleration, t = time and s = distance

-25 sin 30 = 25 sin 30 + (-9.8 x t)

-12.5 = 12.5 - 9.8 t

t = 2.55 s

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V^2 = U^2 + 2as  ( s = vertical distance and V = 0 )

0 = (25 sin 30)^2 + 2 x (-9.8) x Y

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the distance traveled without the pellet hitting the skeet can be gotten using distance = speed x time

distance = 21.65 x 2.55 = 55.2 m

applying the conservation of linear momentum

on the x axis : (Ms x Us) + (Mp x Up) = (Ms x Vx) + (Mp x Vx)  ...equation 1

on the y axis :   (Ms x Us) + (Mp x Up) = (Ms x Vy) + (Mp x Vy) ...equation 2

(0.25 x 25 cos 30) + 0 = (0.25 +0.015) Vx

 Vx = 20.42m/s

0 + (0.015 x 200) = (0.25 + 0.015) Vy

 Vy = 11.32 m/s

now V^2 = U^2 + 2 as

 0 = 11.3^2 + (2 x (-9.8) x s)

s = 6.51 m                          

  • to find the extra distance moved after collision we apply

s = ut + 1/2 at^2

-7.98 = 11.32t + 1/2 (-9.8) t^2

4.9 t^2 - 11.32t + 7.98

t =  3.17 s

recall that distance = speed x time

distance = 20.42 x 3.17 = 64.73 m

the distance of the skeet before being hit would be half of the distance it travels without being hit, this is because the skeet was hit at its maximum height = 55.2 /2

= 27.6 m

therefore the extra distance traveled would be the change in distance = 64.73 -27.6 = 37.13 m

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