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arlik [135]
1 year ago
11

Select the correct answer.

Engineering
1 answer:
andre [41]1 year ago
8 0

Answer:

A. energy transformations

Explanation:

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What is the definition of a duty cycle?
ira [324]

Answer:

D=\frac{PW}{T}*100

Explanation:

In electrical terms, is the ratio of time in which a load or circuit is ON compared to the time in which the load or circuit is OFF.

The duty cycle or power cycle, is expressed as a percentage of the activation time. For example, a 70% duty cycle is a signal that 70% of the time is activated and the other 30% disabled. Its equation can be expressed as:

D=\frac{PW}{T}*100

Where:

D=Duty\hspace{3}Cycle

PW=Pulse\hspace{3}Active\hspace{3}Time

T=Period\hspace{3}of\hspace{3}the\hspace{3}Signal

Here is a picture that will help you understand these concepts.

5 0
3 years ago
An angle is observed repeatedly using the same equipment and procedures producing the data below:35 ∘ 40'00",35 ∘ 40'10",35 ∘ 40
Helen [10]

Answer: (a) +/- 7.5° (b) +/- 3.75°

Explanation:

See attachment

6 0
3 years ago
In a TDM communication example, 15 voice signals are badlimited to 5kHz and transmitted simultaneously using PAM. What is a prel
MA_775_DIABLO [31]

Answer:

Option D

160 kHz

Explanation:

Since we must use at least one synchronization bit, total message signal is 15+1=16

The minimum sampling frequency, fs=2fm=2(5)=10 kHz

Bandwith, BW required is given by

BW=Nfs=16(10)=160 kHz

5 0
3 years ago
Which measuring tool will be used to determine the diameter of a crankshaft journal?
guapka [62]

Answer:

The dial bore gauge measures the inside of round holes, such as the bearing journals. This one tool can measure 2″ up to 6″ diameter holes. Both tools are needed in order to check the interior and exterior dimensions of the crankshaft, rods and engine block journals, as well as the thickness of the bearings themselves.

Hope it's helpful to you

pls mark me as brain list

4 0
3 years ago
A smooth sphere with a diameter of 6 inches and a density of 493 lbm/ft^3 falls at terminal speed through sea water (S.G.=1.0027
Pachacha [2.7K]

Given:

diameter of sphere, d = 6 inches

radius of sphere, r = \frac{d}{2} = 3 inches

density,  \rho} = 493 lbm/ ft^{3}

S.G = 1.0027

g = 9.8 m/ m^{2} = 386.22 inch/ s^{2}

Solution:

Using the formula for terminal velocity,

v_{T} = \sqrt{\frac{2V\rho  g}{A \rho C_{d}}}              (1)

(Since, m = V\times \rho)

where,

V = volume of sphere

C_{d} = drag coefficient

Now,

Surface area of sphere, A = 4\pi r^{2}

Volume of sphere, V = \frac{4}{3} \pi r^{3}

Using the above formulae in eqn (1):

v_{T} = \sqrt{\frac{2\times \frac{4}{3} \pir^{3}\rho  g}{4\pi r^{2} \rho C_{d}}}

v_{T} = \sqrt{\frac{2gr}{3C_{d}}}  

v_{T} = \sqrt{\frac{2\times 386.22\times 3}{3C_{d}}}

Therefore, terminal velcity is given by:

v_{T} = \frac{27.79}{\sqrt{C_d}} inch/sec

3 0
3 years ago
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