Hydraulic refer to liquids (e.g., water), electric refers to electricity and pneumatic refers to gases (e.g., air). They are different devices that can be applied to a robot drive.
<h3>Hydraulic, electric and pneumatic devices </h3>
Hydraulic devices refer to the technologies that involve the use of liquids such as oil or water.
Pneumatic devices refer to the technologies that involve the use of gases such as air.
Electricity devices refer to the technologies that involve the use of an electric field to produce a given outcome (e.g., motion in an electric motor).
A robot drive may exploit these different devices in order to perform specific functions.
Learn more about pneumatic devices here:
brainly.com/question/9778362
Answer:
hello your question is incomplete attached below is the missing diagram to the question and the detailed solution
Answer : principal stresses : 0.82 MPa, -33.492 MPa
shear stress = 17.157 MPa
∅ = 9.09 ≈ 10°
Explanation:
The principal stress ( б1 ) = 0.82 MPa
( б2 ) = -33.492 MPa
The shear stress = 17.157 MPa
∅ = 9.09 ≈ 10°
attached below is the detailed solution and the Mohr's circle
Answer:
0.34
Explanation:
While talking about Load and Resistance Factor Design (LRFD) method, we can say that it is based on the principle that strength (resistance) of various materials is scaled down by some factors while the applied loads are scaled up by some factors, and thereby the structural elements are designed using reduced strength and increased loads.
Refer to attachment for the step by step solution.
1.25, because change the fractions into decimals and then add it from there so it would be add like, 0.75+0.5=1.25
Answer:
(a) 16.27 Vpk
(b) 48.7%
Explanation:
The transformer is assumed to be an ideal 10:1 voltage divider with no internal impedance. The diode is assumed to be modeled in the forward direction by a perfect 0.7 V voltage drop with no internal impedance. That means the frequency of the supply voltage is irrelevant.
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<h3>(a)</h3>
The peak voltage will be 0.7 V less than the transformer secondary peak voltage:
((120 V)√2)/10 -0.7 V ≈ 16.27 V
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<h3>(b)</h3>
The fraction of the amplitude for which the diode is non-conducting is ...
0.7/(12√2) ≈ 0.041248
The period of conduction is symmetrical about the peak of the waveform, so it is convenient to use the arccos function to find the (half) conduction angle:
arccos(0.041248) ≈ 87.64°
As a fraction of half the cycle, this is ...
conduction fraction ≈ 87.64°/180° ≈ 48.7%