All categories

2 years ago

What type of diagram chart in a service manual can be used to easily service an electronic circuit board with ICs?

1 answer:

3 0

That could be letter B: flowchart. Flowchart diagrams can be used to easily service an electronic circuit board with ICs because they let you see the flow of the current as you could imagine it. You can use different shapes to further your explanation and illustration on how the electricity flows from a source, to its medium to the switches. You could also decide on flowcharts and be able to see where that decision leads you and you could also try to learn different decision methods which is basically illustrated on the chart.

You might be interested in

In hydrogen, the transition from level 2 to level 1 has a rest wavelength of 121.6 nm.1).Find the speed for a star in which this

soldier1979 [14.2K]

Answer:

1). $v = - 2960526m/s$

2). Toward us

3). $v = - 493421m/s$

4). Toward us

5). $v = 1480263m/s$

6). Away from us

7). $v = 3207236m/s$

8). Away from us

Explanation:

Spectral lines will be shifted to the blue part of the spectrum if the source of the observed light is moving toward the observer, or to the red part of the spectrum when it is moving away from the observer (that is known as the Doppler effect).

The wavelength at rest is 121.6 nm ( $\lambda_{0} = 121.6nm$ )

$Redshift: \lambda_{measured} > \lambda_{0}$

$Blueshift: \lambda_{measured} < \lambda_{0}$

Then, for this particular case it is gotten:

Star 1: $\lambda_{measured} = 120.4nm$

Star 2: $\lambda_{measured} = 121.4nm$

Star 3: $\lambda_{measured} = 122.2nm$

Star 4: $\lambda_{measured} = 122.9nm$

Star 1:

$Blueshift: 120.4nm < 121.6nm$

Toward us

Star 2:

$Blueshift: 121.4nm < 121.6nm$

Toward us

Star 3:

$Redshift: 122.2nm > 121.6nm$

Away from us

Star 4:

$Redshift: 122.9nm > 121.6nm$

Away from us

Due to that shift the velocity of the star can be determine by means of Doppler velocity.

$v = c\frac{\Delta \lambda}{\lambda_{0}}$ (1)

Where $\Delta \lambda$ is the wavelength shift, $\lambda_{0}$ is the wavelength at rest, v is the velocity of the source and c is the speed of light.

$v = c(\frac{\lambda_{measured}- \lambda_{0}}{\lambda_{0}})$ (2)

Case for star 1 $\lambda_{measured} = 120.4 nm$ :

$v = (3x10^{8}m/s)(\frac{120.4nm-121.6nm}{121.6nm})$

$v = - 2960526m/s$

Notice that the negative velocity means that is approaching to the observer.

Case for star 2 $\lambda_{measured} = 121.4 nm$ :

$v = (3x10^{8}m/s)(\frac{121.4nm-121.6nm}{121.6nm})$

$v = - 493421m/s$

Case for star 3 $\lambda_{measured} = 122.2 nm$ :

$v = (3x10^{8}m/s)(\frac{122.2nm-121.6nm}{121.6nm})$

$v = 1480263m/s$

Case for star 4 $\lambda_{measured} = 122.9 nm$ :

$v = (3x10^{8}m/s)(\frac{122.9nm-121.6nm}{121.6nm})$

$v = 3207236m/s$

4 0

3 years ago

Samples of different materials, A and B, have the same mass, but the sample

madam [21]

Answer:

A.) The particles that make up material A have more mass than the

particles that make up material B.

HOPE that helps!!! :)

4 0

2 years ago

A projectile is launched upward at an angle of 70⁰ from the horizontal and strikes the ground a certain distance down range. For

OLga [1]

Answer:20°

Explanation:

Recall

Range R of a projectile is given by U^2sin2A/g

We're U = velocity,A= angle of projection and g is acceleration due to gravity

From the question the range R are the same

Hence R1=R2

U1^2sin2A/g=U2^2sin2B/g

But U1=U2 and g=g

Hence sin2A=sin 2B

Sin 2*70= sin2*B

0.6427=sin2B

B=sin inverse(0.6427)=40/2=20°

7 0

3 years ago

A solenoid of radius r 5 1.25 cm and length , 5 30.0 cm has 300 turns and carries 12.0 A. (a) Calculate the flux through the sur

Elina [12.6K]

Answer:

$\phi = 7.4 \times 10^{-6} Wb$

Explanation:

Magnetic field due to a long solenoid at the center is given by

$B = \frac{\mu_o N i}{L}$

here we know that

N = 300

L = 30 cm

i = 12 A

now magnetic field is given as

$B = \frac{(4\pi \times 10^{-7})(300)(12)}{0.30}$

$B = 0.015 T$

Now magnetic flux through the disc is given as

$\phi = B.A$

$\phi = (0.015)(\pi r^2)$

$\phi = (0.015)(\pi)(0.0125)^2$

$\phi = 7.4 \times 10^{-6} Wb$

8 0

3 years ago

Which is NOT a property of an acid

nikdorinn [45]

What are the options?

5 0

3 years ago