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3 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.

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PLEASE HELP!!!! OFFERING 99 POINTS!!

Nitella [24]

The Gravitational PE (U) depends on three things: the object’s mass (m), its height (h), and gravitational acceleration (g), which is 9.81 m/s^2 on Earth’s surface.

so U = mgh = 9.81mh on earth

mass of the car = 50.0 grams = 0.05kg

height, h:

Hill 1 = 90.0 cm = 0.9m,

Hill 2 = 65.0 cm = 0.65m,

Hill 3 = 20.0 cm = 0.2m

substitute into eqn U = mgh

U @ top of Hill 1 = 0.05*9.81*0.9 = 0.4415J

U @ top of Hill 2 = 0.05*9.81*0.65 = 0.3188J

U @ top of Hill 3 = 0.05*9.81*0.2 = 0.0981J

difference in Gravitational Potential Energy from the top of Hill 1 to the top of Hill 3 = 0.4415 - 0.0981

= 0.3434J where J is the unit for energy, Joules

7 0

3 years ago

Read 2 more answers

A dragster completed a 402.3-(0.2500-mi) run in 5.023s. If the car had a constant acceleration, what was its acceleration and fi

Ray Of Light [21]

Answer:

v = 80.092 m/s

a = 15.945 m/s²

Explanation:

Given,

The distance completed by the dragster, d = 402.3 m

The time taken by the dragster to complete that distance is, t = 5.023 s

The initial velocity of the dragster, u = 0

The final velocity of the dragster, v = ?

The acceleration of the dragster, a = ?

The velocity of the body is given by the formula

v = d/t m/s

Substituting the above values inn the equation

v = 402.3 m / 5.023 s

= 80.092 m/s

So, the velocity of the dragster is, v = 80.092 m/s

The acceleration of the body is given by the formula

a = (v - u)/t m/s²

Substituting the above values in the equation

= (80.092 - 0) / 5.023

= 15.945 m/s²

Hence, the acceleration of the dragster is, a = 15.945 m/s²

6 0

3 years ago

Babe Ruth steps to the plate and casually points to left centerfield to indicate the location of his next home run. Themighty Ba

Elden [556K]

Answer:

Fs = 32N

Fh = 21 N

Explanation:

Statics

Let Fs be the force of the shoulder, assumed upward

Let Fh be the force of the hand, assumed downward

Sum moments about the hand to zero

1.1(9.8)[67] - Fs[22.5] = 0

Fs = 32N

Sum vertical forces to zero

32 - Fh - 1.1(9.8) = 0

Fh = 21 N

Or we could sum moments about the shoulder point

1.1(9.8)[67 - 22.5] - Fh[22.5] = 0

Fh = 21 N

6 0

3 years ago

What kind of organism always comes last in every food chain

Elena-2011 [213]

Tertiary consumer is always at the top.
The producer is always at the bottom.

4 0

3 years ago

A missile is moving 1350 m/s at 25.0° angle. It needs to hit in a 55.0° direction in 10.20 s. What is the direction of its final

77julia77 [94]

Answer:

final velocity = 3504 m/s

Explanation:

Given data:

velocity of missile = Vi = 1350m/s

angle at which missile is moving = 25degree

distance between missile and targets = 23500m

angle between target and missile=55degree

time=10.2s

To find:

Final velocity: ?

Formula:

x = Vx*t + ½*ax*t²

Let x be the horizontal component of distance

x = ertical component of distance

t-time

ax = horizontal component of acceleration

ay = Vertical component of acceleration

Vx = horizontal component of velocity

Vy = Vertical component of velocity

Solution:

x = Vx*t + ½*ax*t²

23500m * cos55.0º = 1350m/s * cos25.0º * 10.20s + ½ * ax * (10.20s)²

ax = 19.2 m/s²

V'x = Vx + ax*t = 1350m/s * cos25.0º + 19.2m/s² * 10.20s = 1419 m/s

similarly vertically:

y = Vy*t + ½*ay*t²

23500m * sin55.0º = 1350m/s * sin25.0º * 10.20s + ½ * ay * (10.20s)²

ay = 258 m/s²

V'y = Vy + ay*t

= 1350m/s * sin25.0º + 258m/s² * 10.20s = 3204 m/s

V = √(V'x² + V'y²)

= 3504 m/s

8 0

3 years ago