a cruise took 6 hours to fulfill its trip it travelled at an average speed of 160 km/h for thefirst 2 h 15 min

A single crystal of a metal that has the FCC crystal structure is oriented such that a tensile stress is applied parallel to the

algol13

Stress required to cause slip on in the direction [ 1 1 0 ] is 7.154 MPa

Explanation:

Given -

Stress Direction, A = [1 0 0 ]

Slip plane = [ 1 1 1]

Normal to slip plane, B = [ 1 1 1 ]

Critical stress, Sc = 2.92 MPa

Let the direction of slip on = [ 1 1 0 ]

Let Ф be the angle between A and B

cos Ф = A.B/ |A| |B| = [ 1 0 0 ] [1 1 1] / √1 √3

cos Ф = 1/√3

σ = Sc / cosФ cosλ

For slip along [ 1 1 0 ]

cos λ = [ 1 1 0 ] [ 1 0 0 ] / √2 √1

cos λ = 1/√2

Therefore,

σ = 2.92 / 1/√3 1/√2

σ = √6 X 2.92 MPa = 2.45 X 2.92 = 7.154MPa

Therefore, stress required to cause slip on in the direction [ 1 1 0 ] is 7.154MPa

4 0

3 years ago

If a sled has at the top of 10 m hill had 1000 J of potential energy what would happen to the PE if the sled were to moved to a

nataly862011 [7]

Answer:

500J

Explanation:

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7 0

3 years ago

Sammy squirrel is steering his boat at a heading of 327 degree at 18mph. The current is flowing at 4mph at a heading of 60 degre

Tanya [424]

Answer:

59.97 º at 18.23 mph

Explanation:

To find Sammy's course you have to add the two velocities (vectors), 18 mph 327º and 4 mph 60º.

To add the two vectors analytically you decompose each vector into their vertical and horizontal components.

1. 18 mph 327º

Horizontal component: 18 mph × cos (327º) = 15.10 mph

Vertical component: 18 mph × sin (327º) = - 9.80 mph

Vector notation:

$15.10\hat i-9.80\hat j$

2. 4 mph 60º

Horizontal component: 4 mph × cos (60º) = 2.00 mph

Vertical component: 4 mph × sin (60º) = 3.46 mph

Vector notation:

$2.00\hat i+3.46\hat j$

3. Addition:

You add the corresponding components:

$15.10\hat i-9.80\hat j+2.00\hat i+3.46\hat j\\ \\ 17.10\hat i-6.34\hat j$

To find the magnitude use Pythagorean theorem:

$\sqrt{17.1^2+6.34^2}= 18.23$

4. Direction:

Use the tangent ratio:

$tan(\alpha )=opposite/adjacent=3.46/2.00=1.73$

Find the inverse:

arctan (1.73) ≈ 59.97º

5 0

3 years ago

Can you help me with this please

WARRIOR [948]

Defenition of skeletal muscles: a muscle that is connected to the skeleton to form part of the mechanical system that moves the limbs and other parts of the body.

{GOOGLE~SEARCHED}

6 0

3 years ago

Read 2 more answers

Is the force of friction the sum of all the microscopic contact forces between the bottom of the block and the surface?

kotegsom [21]

The force of friction is that force that tends to oppose the motion of the body with the surface in contact... are you clear with it?

6 0

2 years ago