Answer:
The diameter increases
Explanation:
The expansion in the metal is uniform in every dimension
Answer:
total width bandwidth = 8kHz
Explanation:
given data
transmitter operating = 3.9 MHz
frequencies up to = 4 kHz
solution
we get here upper side frequencies that is
upper side frequencies = 3.9 ×
+ 4 × 10³
upper side frequencies = 3.904 MHz
and
now we get lower side frequencies that is
lower side frequencies = 3.9 ×
- 4 × 10³
lower side frequencies = 3.896 MHz
and now we get total width bandwidth
total width bandwidth = upper side frequencies - lower side frequencies
total width bandwidth = 8kHz
Answer:
peak flow and any engineering considerations related thereto
Explanation:
It should be no surprise that a peak flow meter will report peak flow, sometimes with important maximum-value, time-constant, or bandwidth limitations. There are many engineering issues related to flow rates. A peak flow meter can allow you to assess those issues with respect to the flows actually encountered.
Peak flow can allow you to assess adequacy of flow and whether there may be blockages or impediments to flow that reduce peak levels below expected values. An appropriate peak flow meter can help you assess the length of time that peak flow can be maintained, and whether that delivers sufficient volume.
It can also allow you to assess whether appropriate accommodation is made for unexpectedly high flow rates. (Are buffers or overflow tanks of sufficient size? Is there adequate protection against possible erosion? Is there adequate support where flow changes direction?)
Since the applied stress required for failure due to crack propagation is still higher than 550 MPa, the ceramic is expected to fail due to overload and not because of the flaws
Explanation:
<u>Plane -Strain Fracture toughness is calculated as</u>
=
б
F=geometry factor of the flaw
б=Stress applied
=Fracture toughness
a=Flaw size
<u>Given that </u>
Internal Flaw,a=0.001cm
Fracture Toughness
=45MPa
Tensile Strength б=550 MPa
Geometry Factor,
=1
<u>Calculation</u>
An internal Flaw i s 0.001 cm
2a=0.001cm
a=0