A crystalline grain of nickel in a metal plate is situated so that a tensile load is oriented along the [111] crystal direction (a) If the applied stress if 0.45 MPa, what will be the resolved shear stress, tRss, along the [101] direction within the (11 T) plane? (b) What tensile stress is required to produce a critical resolved shear stress, TcRss of 0.242 MPa

Answers

Answer 1

Answer:

The resolved shear stress along the [101] direction is 0.3673 MPa and the tensile stress is required to produce a critical resolved is 0.2964 MPa.

What is a lattice unit cell?

The symmetrical 3-D structural arrangement of the ions, atoms, or molecules inside a crystalline lattice solid as a point.

A crystalline grain of nickel in a metal plate is situated so that a tensile load is oriented along the [111] crystal direction.

(a) If the applied stress is 0.45 MPa. Then the resolved shear stress, along the [101] direction within the (11T) plane will be

The θ be the angle between [101] and (111) will be

$\rm cos\ \theta = \frac{1*1+1*0+1*1}{√(1^2+1^2+1^2)\sqer{1^2+0^2+1^2}}\n\n\ncos\ \theta = \sqrt{ (2)/(3)}$

And ∅ be the angle between [111] and [111], then we have

$\rm cos\ \phi= \frac{1*1+1*1+1*1}{√(1^2+1^2+1^2)\sqer{1^2+1^2+1^2}}\n\n\ncos\ \phi= 1$

Now, for the resolved components along [101], we have

$\rm = \sigma cos \ \theta \ \ cos \ \phi\n\n= 0.45 *10^6*\sqrt{(2)/(3)}*1= 0.3673 \ \ MPa$

(b) The tensile stress is required to produce a critical resolved shear stress of 0.242 MPa will be

$\rm \sigma _1 = (shear\ stress)/(cos\ \theta \ cos \ \phi)\n\n\n\sigma _1 = \frac{0.242*10^6}{\sqrt{(2)/(3)} * 1}\n\n\n\sigma _1 = 0.2964 \ MPa$

More about the crystalline lattice link is given below.

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Answer 2

Answer:

Given:

Applied stress, $\sigma = 0.45 MPa$

Critical Resolved Stress, $T_(cRss)= 0.242 MPa$

Solution:

a). According to the question, orientation of tensile load is along [1 1 1],

$\sigma = 0.45 MPa$

Now, for resolved shear stress, $\t_(Rss)$ along [1 0 1] within (1 1 T)

let ' $\theta$ ' be the angle between [1 1 1] and [1 0 1], then by coordinate formula:

$cos\theta$ = $\frac{1* 1 + 1* 0 + 1* 1}{\sqrt{1^(2)+1^(2)+1^(2)\sqrt{1^(2)+0^(2)+1^(2)}}}$

$cos\theta$ = $\sqrt{(2)/(3)}$

let ' $\phi$ ' be the angle between [1 1 1] and [1 1 1], then by coordinate formula:

$cos\phi = \frac{1* 1 + 1* 1 + 1* 1}{\sqrt{1^(2)+1^(2)+1^(2)\sqrt{1^(2)+1^(2)+1^(2)}}}$

$cos\phi$ = 1

Now, for the resolved components along [1 0 1]

$\t_(Rss)$ = $\sigma cos\theta cos\phi$

$\t_(Rss)$ = $0.45* 10^(6)* \sqrt{(2)/(3)}* 1$ = 0.3673 MPa

b). For required tensile stress to produce $T_(cRss)= 0.242 MPa$ :

$\sigma _(1) = (T_(cRss))/(cos\theta cos\phi )$

$\sigma _(1) = \frac{0.242* 10^(6)}{\sqrt{(2)/(3)}* 1}$

$\sigma _(1) = 0.2964 MPa$

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Find the force necessary to start the crate moving, given that the mass of the crate is 32 kg and the coefficient of static friction between the crate and the floor is 0.57. Express your answer using two significant figures.

Answers

Answer:

178.75 N

Explanation:

The force necessary to start moving the crate must be equal to or more than the frictional force (resistive force) acting on the crate but moving in an opposite direction to the frictional force.

So, we find the frictional force, Fr:

Fr = -μmg

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m = mass

g = acceleration due to gravity

The frictional force is negative because it acts against the direction of motion of the crate.

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Garza travels at a speed of 5 m/s. How long will it take him to travel 640 m?

Answers

Answer:

128 s

Explanation:

The distance, speed and time are related as;

$Distance=Speed* Time$

Given that the speed = 5 m/s

Distance = 640 m

Time = ?

So,

$Distance=Speed* Time$

$640\ m=5\ m/s* Time$

$Time=\frac {640\ m}{5\ m/s}=128\ s$

Thus, Garza takes 128 s to travel 640 m at 5 m/s speed.

The distance in the x direction between two control points on a vertical aerial photograph is 4.5". If the distance between these same two points is 3.6" on another photograph having a scale of 1:24,000, determine the scale of the first vertical aerial photograph. Of the focal length of the camera is 6"and the average elevation at these points is 100 ft, determine the flying height from which each photograph was taken

Answers

Answer:

Use proportions to find the scale of the first photo, then use that scale and other given information to fill in the equation

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Where:

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To calculate the change in kinetic energy, you must know the force as a function of _______. The work done by the force causes the kinetic energy change.'

Answers

Answer: The force is as a function of Distance

Explanation:

The force and distance must be parallel to each other. Only the component of the force in the same direction as the distance traveled does any work. Hence, if a force applied is perpendicular to the distance traveled, no work is done. The equation becomes force times distance times the cosine of the angle between them.

where both the force F and acceleration are vectors. This makes sense since both force and acceleration have a direction.

On the other hand, the kinetic energy

K=12mv2

looks completely different. It doesn't seem to depend on the direction.

Answer:

Distance

Explanation:

dW = F. dx

The physics of wind instruments is based on the concept of standing waves. When the player blows into the mouthpiece, the column of air inside the instrument vibrates, and standing waves are produced. Although the acoustics of wind instruments is complicated, a simple description in terms of open and closed tubes can help in understanding the physical phenomena related to these instruments. For example, a flute can be described as an open-open pipe because a flutist covers the mouthpiece of the flute only partially. Meanwhile, a clarinet can be described as an open-closed pipe because the mouthpiece of the clarinet is almost completely closed by the reed.1. Consider a pipe of length 80.0 cm open at both ends. What is the lowest frequency f of the sound wave produced when you blow into the pipe?
2. A hole is now drilled through the side of the pipe and air is blown again into the pipe through the same opening. The fundamental frequency of the sound wave generated in the pipe is now:______.
a. the same as before.
b. lower than before.
c. higher than before.
3. If you take the original pipe in Part A and drill a hole at a position half the length of the pipe, what is the fundamental frequency of the sound that can be produced in the pipe?
4. What frequencies, in terms of the fundamental frequency of the original pipe in Part A, can you create when blowing air into the pipe that has a hole halfway down its length?
4-1. Recall from the discussion in Part B that the standing wave produced in the pipe must have an antinode near the hole. Thus only the harmonics that have an antinode halfway down the pipe will still be present.
A. Only the odd multiples of the fundamental frequency.
B. Only the even multiples of the fundamental frequency.
C. All integer multiples of the fundamental frequency.
E. What length of open-closed pipe would you need to achieve the same fundamental frequency as the open pipe discussed in Part A?
A. Half the length of the open-open pipe.
B. Twice the length of the open-open pipe.
C. One-fourth the length of the open-open pipe.
D. Four times the length of the open-open pipe.
E. The same as the length of the open-open pipe.
F. What is the frequency of the first possible harmonic after the fundamental frequency in the open-closed pipe described in Part E?
F-1. Recall that possible frequencies of standing waves that can be generated in an open-closed pipe include only odd harmonics. Then the first possible harmonic after the fundamental frequency is the third
harmonic.

Answers

Final answer:

The physics of wind instruments is based on standing waves. The lowest frequency of a sound wave produced in an open-open pipe can be calculated. When a hole is drilled through the pipe, the fundamental frequency is lower than before. Only odd multiples of the fundamental frequency will be present in a pipe with a hole halfway down its length. An open-closed pipe needs to be twice the length of an open-open pipe to achieve the same fundamental frequency. The first possible harmonic after the fundamental frequency in an open-closed pipe is the third harmonic.

Explanation:

The lowest frequency f of the sound wave produced when blowing into an open-open pipe can be calculated using the formula f = v/2L, where v is the speed of sound and L is the length of the pipe. Plugging in the values, we get f = 343/(2*0.8), which equals 214.375 Hz.

When a hole is drilled through the side of the pipe, the fundamental frequency of the sound wave generated in the pipe is lower than before. This is because the effective length of the pipe has been changed, resulting in a lower frequency.

The fundamental frequency of the sound that can be produced in the original pipe with a hole drilled halfway down its length can be calculated as f = v/L, where L is the new effective length of the pipe. Since the hole is halfway down, the effective length becomes half of the original length, resulting in a frequency equal to the original fundamental frequency.

When blowing air into the pipe with a hole halfway down its length, only the odd multiples of the fundamental frequency will be present. Therefore, the frequencies that can be created are only the odd multiples of the fundamental frequency.

The length of an open-closed pipe needed to achieve the same fundamental frequency as an open-open pipe is twice the length of the open-open pipe. This is because an open-closed pipe has only odd harmonics, which are spaced twice as far apart as the harmonics in an open-open pipe.

The first possible harmonic after the fundamental frequency in an open-closed pipe is the third harmonic.

Learn more about standing waves here:

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You are traveling on an interstate highway at the posted speed limit of 70 mph when you see that the traffic in front of you has stopped due to an accident up ahead. You step on your brakes to slow down as quickly as possible. Assume that you to slow down to 30 mph in about 5 seconds. A) With this same average acceleration, how much longer would it take you to stop?B) What total distance would you travel from when you first apply the brakes until the car stops?

Answers

A.The time taken for the car to stop is 8.75 s

B.The distance travelled when the brakes were applied till the car stops is 136.89 m

A. Determination of the time taken for the car to stop.

We'llbegin bycalculatingthedecelerationof thecar

Initial velocity (u) = 70 mph = 0.447 × 70 = 31.29 m/s

Final velocity (v) = 30 mph = 0.447 × 30 = 13.41 m/s

Time (t) = 5 s

Deceleration (a) =?

$a \: = (v \: - u)/(t) \n \n a = (13.41 - 31.29)/(5) \n \n a \: = ( - 17.88)/(5) \n \n$

a = –3.576 m/s²

Finally,we shall determine the time taken for the car to stop.

Initial velocity (u) = 31.29 m/s

Final velocity (v) = 0 m/s

Deceleration (a) = –3.576 m/s²

Time (t) =?

$v \: = u \: + at \n 0 \: = 31.29 \: + \: ( - 3.576 * t) \n 0 \: = 31.29 \: - 3.576 * t \n collet \: like \: terms \n 0 - 31.29 \: = - 3.576 * t \n - 31.29 \: = - 3.576 * t \n divide \: both \: side \: by \: - 3.576 \n t \: = (- 31.29)/(- 3.576) \n$

t = 8.75 s

Thus, the time taken for the car to stop is 8.75 s

B.Determination of the total distance travelled when the brakes were applied.

Initial velocity (u) = 31.29 m/s

Final velocity (v) = 0 m/s

Deceleration (a) = –3.576 m/s²

Distance (s) =?

${v}^(2) = {u}^(2) + 2as \n {0}^(2) = {31.29}^(2) + (2 * - 3.576 * s) \n 0 = 979.0641 - 7.152 s \n collect \: like \: terms \n 0 - 979.0641 = - 7.152 s \n - 979.0641 = - 7.152 s \n divide \: both \: side \: by \: - 7.152 \n s = (- 979.0641)/(- 7.152) \n \n$