What is the center of our solar system?

An aluminum bar 600mm long, with diameter 40mm long has a hole drilled in the center of the bar.The hole is 30mm in diameter and

Svetradugi [14.3K]

Answer:

Total contraction on the Bar = 1.22786 mm

Explanation:

Given that:

Total Length for aluminum bar = 600 mm

Diameter for aluminum bar = 40 mm

Hole diameter = 30 mm

Hole length = 100 mm

elasticity for the aluminum is 85GN/m² = 85 × 10³ N/mm²

compressive load P = 180 KN = 180 × 10³ N

Calculate the total contraction on the bar = ???

The relation used in calculating the contraction on the bar is:

$\delta L = \dfrac{P *L }{A*E}$

The relation used in calculating the total contraction on the bar can be expressed as :

Total contraction in the Bar = (contraction in part of bar without hole + contraction in part of bar with hole)

i.e

Total contraction on the Bar = $\dfrac{P *L_1 }{A_1*E} + \dfrac{P *L_2 }{A_2 *E}$

Let's find the area of cross section without the hole and with the hole

Area of cross section without the hole is :

Using A = πd²/4

A = π (40)²/4

A = 1256.64 mm²

Area of cross section with the hole is :

A = π (40²-30²)/4

A = 549.78 mm²

Total contraction on the Bar = $\dfrac{P *L_1 }{A_1*E} + \dfrac{P *L_2 }{A_2 *E}$

Total contraction on the Bar = $\dfrac{180 *10^3 \N }{85*10^3 \ N/mm^2} [\dfrac{500}{1256.64}+ \dfrac{100}{549.78}]$

Total contraction on the Bar = 2.117( 0.398 + 0.182)

Total contraction on the Bar = 2.117*(0.58)

Total contraction on the Bar = 1.22786 mm

5 0

4 years ago

Provide your own example that demonstrates Newton’s 3 rd law of motion?

Sedaia [141]

Answer:

EXAMPLE:when you jump, your legs apply a force to the ground, and the ground applies and equal and opposite reaction force that propels you into the air.

6 0

3 years ago

A small object begins a free-fall from a height of =81.5 m at 0=0 s . After τ=2.20 s , a second small object is launched vertica

-BARSIC- [3]

Answer:

33.2 m

Explanation:

For the first object:

y₀ = 81.5 m

v₀ = 0 m/s

a = -9.8 m/s²

t₀ = 0 s

y = y₀ + v₀ t + ½ at²

y = 81.5 − 4.9t²

For the second object:

y₀ = 0 m

v₀ = 40.0 m/s

a = -9.8 m/s²

t₀ = 2.20 s

y = y₀ + v₀ t + ½ at²

y = 40(t−2.2) − 4.9(t−2.2)²

When they meet:

81.5 − 4.9t² = 40(t−2.2) − 4.9(t−2.2)²

81.5 − 4.9t² = 40t − 88 − 4.9 (t² − 4.4t + 4.84)

81.5 − 4.9t² = 40t − 88 − 4.9t² + 21.56t − 23.716

81.5 = 61.56t − 111.716

193.216 = 61.56t

t = 3.139

The position at that time is:

y = 81.5 − 4.9(3.139)²

y = 33.2

7 0

3 years ago

Suppose that the speed of yellow visible light in a certain transparent medium is 2.01* 105 km/s. What, approximately, is the in

True [87]

Answer:

$1.50$

Explanation:

Index of refraction:

$n=\frac{c}{v}$ where $n$ is the refractive index, $c$ is the speed of light in vacuum and $v$ is the speed of light in medium.

The speed of light in vacuum is $3.00 \times 10^5 \text{km/s}$

Speed of light in medium is $2.01 \times 10^5 \text{km/s}$

Thus,

$n=\frac{3.00\times 10^5}{2.01\times 10^5}$

$n=1.50$

Index of refraction of this substance through yellow light is $1.50$

4 0

3 years ago

Reading glasses with a power of 1.50 diopters make reading a book comfortable for you when you wear them 1.8 cmcm from your eye.

Vlad1618 [11]

Answer:

The near point is $n =44.8 \ cm$

Explanation:

From the question we are told that

The power is $P = 1.50$

The distance from the eye is $k = 1.8 \ cm$

The distance of the book from the eye is $z = -28 \ cm$

Generally the focal length of the glasses is

$f = \frac{1}{P}$

=> $f = \frac{1}{1.50 }$

=> $f = 0.667 \ m$

=> $f = 66.7 \ cm$

The object distance is evaluated as

$u = z + k$

=> $u = -28 + 1.8$

=> $u = -26.2 \ cm$

The image distance is evaluated from lens formula as

$\frac{1}{v} = \frac{1}{f} + \frac{1}{u}$

=> $\frac{1}{v} = \frac{1}{66.7} + \frac{1}{-26.2}$

=> $v=- \frac{1}{0.0232}$

=> $v=- 43 \ cm$

The near point is evaluated as

$n = -v + k$

=> $n =-(-43) + 1.8$

=> $n =44.8 \ cm$

3 0

3 years ago