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OLga [1]
2 years ago
10

Does an electronic clock use electrical energy?​

Engineering
2 answers:
kati45 [8]2 years ago
5 0
Yes it does because it's using the flow of energy from a conductor to power the clock :)
prohojiy [21]2 years ago
3 0

Answer:

Yes

Explanation:

It uses electrical energy because it is using the flow of energy from a conductor to power the clock.

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Answer:<em>, Locke believed that human nature gave people the chance to be selfish. This is apparent with the introduction of currency. In a natural state, all people were equal and independent and alone at times, and everyone had a natural right to defend his "life, health, liberty, or possessions.."</em>

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How many power station do we have​
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10% A steel beam W18x76 spans 32 feet and is subjected to a Moment of 334 kips-ft. Find the load w on the beam. Determine the de
Lilit [14]

Answer:

w = 10.437 kips

deflection at 1/4 span  20.83\E ft

at mid span = 1.23\E ft

shear stress  7.3629 psi

Explanation:

area of cross section = 18*76

length of span = 32 ft

moment = 334 kips-ft

we know that

moment = load *eccentricity

334 = w * 32

w = 10.437 kips

deflection at 1/4 span

\delta = \frac{wa^2b^2}{3EI}

= \frac{10.4375*8^2 *24^2}{3E \frac{BD^3}{12}}

         =\frac{10.437 *8^2*24^2}{3E \frac{18*16^3}{12}}

         = 20.83\E ft

at mid span

\delta = \frac{wl^3}{48EI}

= \frac{10.43 *32^3}{48 *E*\frac{18*16^3}{12}}

\delta = 1.23\E ft

shear stress

\tau = \frac{w}{A} = \frac{10.43 7*10^3}{18*76} =7.3629 psi

6 0
3 years ago
Consider the 'scope trace showing the input signal to an rc filter (the one having the larger amplitude) and the output of the f
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6 0
2 years ago
A train starts from rest at station A and accelerates at 0.6 m/s^2 for 60 s. Afterwards it travels with a constant velocity for
Aleks [24]

Answer:

The distance between the station A and B will be:

x_{A-B}=55.620\: km  

Explanation:

Let's find the distance that the train traveled during 60 seconds.

x_{1}=x_{0}+v_{0}t+0.5at^{2}

We know that starts from rest (v(0)=0) and the acceleration is 0.6 m/s², so the distance will be:

x_{1}=\frac{1}{2}(0.6)(60)^{2}

x_{1}=1080\: m

Now, we need to find the distance after 25 min at a constant speed. To get it, we need to find the speed at the end of the first distance.

v_{1}=v_{0}+at

v_{1}=(0.6)(60)=36\: m/s

Then the second distance will be:

x_{2}=v_{1}*1500

x_{2}=(36)(1500)=54000\: m        

The final distance is calculated whit the decelerate value:

v_{f}^{2}=v_{1}^{2}-2ax_{3}

The final velocity is zero because it rests at station B. The initial velocity will be v(1).

0=36^{2}-2(1.2)x_{3}

x_{3}=\frac{36^{2}}{2(1.2)}  

x_{3}=540\: m

Therefore, the distance between the station A and B will be:

x_{A-B}=x_{1}+x_{2}+x_{3}  

x_{A-B}=1080+54000+540=55.620\: km  

I hope it helps you!

 

7 0
2 years ago
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