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horsena [70]
2 years ago
12

A curve with 0.3 m constant radius contains a bead that is moving on it such that its rotational velocity is 3t2 sec-1. If the b

ead had zero velocity at theta = 0, determine its vel and acc when theta (θ)= 45 degrees.

Engineering
1 answer:
Schach [20]2 years ago
8 0

Answer:

v = 0.5846 m/s

a = 1.1394 m/s^2

Explanation:

See it in the pic.

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A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane-strain fracture tough
jeyben [28]

Complete question:

A structural component in the form of a wide plate is to be fabricated from a steel alloy that has a plane strain fracture toughness of 98.9 MPa root m (90 ksi root in.) and a yield strength of 860 MPa (125,000 psi). The flaw size resolution limit of the flaw detection apparatus is 3.0 mm (0.12 in.). If the design stress is one-half of the yield strength and the value of Y is 1.0, determine whether or not a critical flaw for this plate is subject to detection.

Answer:

Since the flaw 17mm is greater than 3 mm the critical flaw for this plate is subject to detection

so that critical flow is subject to detection  

Explanation:

We are given:

Plane strain fracture toughness K = 98.9 MPa \sqrt{m}

Yield strength Y = 860 MPa

Flaw detection apparatus = 3.0mm (12in)

y = 1.0

Let's use the expression:

oc = \frac{K}{Y \sqrt{pi * a}}

We already know

K= design

a = length of surface creak

Since we are to find the length of surface creak, we will make "a" subject of the formula in the expression above.

Therefore

a= \frac{1}{pi} * [\frac{k}{y*a}]^2

Substituting figures in the expression above, we have:

= \frac{1}{pi} * [\frac{98.9 MPa \sqrt{m}} {10 * \frac{860MPa}{2}}]^2

= 0.0168 m

= 17mm

Therefore, since the flaw 17mm > 3 mm the critical flow is subject to detection  

3 0
2 years ago
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Interpret the assembly program below: MOV R3,R0;
Reika [66]

Answer:

Explanation:

1.  With the operands R0, R1, the program would compute AND operation and ADD operation .

2. The operands could truly be signed 2's complement encoded (i.e Yes) .

3. The overflow truly occurs when two numbers that are unsigned were added and the result is larger than the capacity of the register, in that situation, overflow would occur and it could corrupt the data.

 When the result of an operation is smaller in magnitude than the smallest value represented by the data type, then arithmetic underflow will occur.

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Answer:

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dimaraw [331]

Answer:

all of the above

Explanation:

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