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Sliva [168]2 years ago

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Answer:hello

Explanation:

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When the mass is removed, the length of the cable is found to be l0 = 4.66 m. After the mass is added, the length is remeasured

zaharov [31]

Answer:

$\gamma=6.07*10^5\frac{N}{m^2}$

Explanation:

For a linear elastic material Young's modulus is a constant that is given by:

$\gamma=\frac{F/A}{\Delta L/L_0}$

Here, F is the force exerted on an object under tensio, A is the area of the cross-section perpendicular to the applied force, $\Delta L$ is the amount by which the length of the object changes and $L_0$ is the original length of the object. In this case the force is the weight of the mass:

$F=mg\\F=55kg(9.8\frac{m}{s^2})\\F=539N$

Replacing the given values in Young's modulus formula:

$\gamma=\frac{F/\pi r^2}{(L_1-L_0)/L_0}\\\gamma=\frac{539N/\pi(0.045m)^2}{(5.31m-4.66m)/4.66m}\\\gamma=6.07*10^5\frac{N}{m^2}$

6 0

3 years ago

Match the correct term with the statement that is true about it

Sholpan [36]

Do you have an image?

5 0

4 years ago

Based on the reading which statement is FALSE?

zvonat [6]

The answer is the last option.

4 0

3 years ago

Read 2 more answers

An airplane in a holding pattern flies at constant altitude along a circular path of radius 3.38 km. If the airplane rounds half

Andrej [43]

Answer:

a) 6076 m

b) 43.33 m/s

c) 68 m/s

Explanation:

(a) If the airplane rounds half the circle in 156s, its displacement is the circle diameter in 156s, or twice the circle's radius

s = 2r = 2* 3.38km = 6.76 km or 6760 m

(b) The average velocity would be displacement over unit of time

v = s/t = 6760 / 156 = 43.33 m/s

(c) The length of the chord it's swept in 156s is half of the circle perimeter

c = πr = π3.38 = 10.62 km or 10620 m

The airplane average speed is its chord length over a unit of time

c / t = c / 156 = 68 m/s

4 0

3 years ago

An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate

OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 × $10^{-4}$ m

dynamic viscosity = 1.75 × $10^{-5}$ Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ $\frac{du}{dy}$

so

= µ $\frac{v}{h}$ ............1

put here value

= 1.75× $10^{-5}$ × $\frac{v}{10^{-4}}$

= 0.175 v

and

area between air and puck is given by

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

area = $\frac{\pi }{4} 0.1^{2}$

area = 7.85 × $\frac{v}{10^{-3}}$ m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × $10^{-3}$

force = 1.374 × $10^{-3}$ v

and now apply newton second law

force = mass × acceleration

- force = $mass \frac{dv}{dt}$

- 1.374 × $10^{-3}$ v = $0.03 \frac{0.9v - v }{t}$

t = $\frac{0.1 v * 0.03}{1.37*10^{-3} v}$

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0

3 years ago