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Alja [10]
3 years ago
9

ASAP pls answer right I will mark brainiest . All I know is 4. Is A

Physics
1 answer:
Sonja [21]3 years ago
4 0

Answer:

Q1: B.2 Q2: B.Waxing crescent Q3: A.Waxing Gibbous

Explanation:

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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

<u>Explanation:</u>

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
Did i get any of these answers right if not what do i have to change
gregori [183]

Answer:

You got them right

8 0
3 years ago
A circular track has a radius of 50m.What is the displacement?
stepladder [879]

The displacement of a moving object is the straight-line distance between the place it starts from and the place where it stops.

The displacement of anything moving along a circular track depends on how  far around it goes before it stops.  The greatest displacement it can possibly have is the diameter of the track ... 100m on this particular one ... because that's as far apart as two places on a circle can ever be.

The most interesting case is when the object goes around the circle exactly once.  Then it stops at the same place it started from, the distance between the starting point and ending point is zero, and after all that motion, the displacement is zero.

5 0
3 years ago
1. Do alto de uma plataforma com 15m de altura, é lançado horizontalmente um projéctil. Pretende-se atingir um alvo localizado n
sveta [45]

Answer:

(a). The initial velocity is 28.58m/s

(b). The speed when touching the ground is 33.3m/s.

Explanation:

The equations governing the position of the projectile are

(1).\: x =v_0t

(2).\: y= 15m-\dfrac{1}{2}gt^2

where v_0 is the initial velocity.

(a).

When the projectile hits the 50m mark, y=0; therefore,

0=15-\dfrac{1}{2}gt^2

solving for t we get:

t= 1.75s.

Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that

50m = v_0(1.75s)

which gives

\boxed{v_0 = 28.58m/s.}

(b).

The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,

v_x = 28.58m/s.

the vertical component of the velocity is

v_y = gt \\v_y = (9.8m/s^2)(1.75s)\\\\{v_y = 17.15m/s.

which gives a speed v of

v = \sqrt{v_x^2+v_y^2}

\boxed{v =33.3m/s.}

4 0
2 years ago
Help with this one please! Please give correct answers!
Verdich [7]

the answer is C because the air balloon is  going down meaning that negative


5 0
3 years ago
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