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Softa [21]
3 years ago
15

A differential amplifier is very useful for removing common mode noise voltage that might be fed or induced in the signal cables

. Suppose the input signal coming from a signal wire is Vs. Because of potential corruption in the measurement circuit, a common mode noise Vn is induced. In this case, the input voltages at both input wires become Vs Vn at the inverting input and Vn at the non-inverting input. Derive the circuit transfer equation and give the condition for removal of common mode noise.
Engineering
1 answer:
Reptile [31]3 years ago
7 0
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What do you mean by searching?​
nikitadnepr [17]

Answer:

thoroughly scrutinizing, especially in a disconcerting way.

Explanation:

8 0
3 years ago
Which of the following best describes the relationship between the World Wide Web and the Internet? А The World Wide Web is a pr
Gwar [14]

Answer:

C

Explanation:

5 0
2 years ago
A cylindrical drill with radius 4 is used to bore a hole through the center of a sphere of radius 5. Find the volume of the ring
ANTONII [103]

Answer:

The volume of the ring shaped solid that remains is 21 unit^3.

Explanation:

The total volume of the sphere is given as:

Volume of Sphere = (4/3)πr^3

where, r = radius of sphere

Volume of Sphere = (4/3)(π)(5)^3

Volume of Sphere = 523.6 unit^3

Now, we find the volume of sphere removed by the drill:

Volume removed = (Cross-sectional Area of drill)(Diameter of Sphere)

Volume removed = (πr²)(D)

where, r = radius of drill = 4

D = diameter of sphere = 2*5 = 10

Therefore,

Volume removed = (π)(4)²(10)

Volume removed = 502.6 unit^3

Therefore, the volume of ring shaped solid that remains will be the difference between the total volume of sphere, and the volume removed.

Volume of Ring = Volume of Sphere - Volume removed

Volume of Ring = 523.6 - 502.6

<u>Volume of Ring = 21 unit^3</u>

5 0
3 years ago
Two thousand pieces will flow through from the first machine A to the final machine F based on the given sequence of operations.
Vlad1618 [11]

The total number of trips that the vehicle has to make based on the given sequence of operation is 120 trips.

<em>"Your</em><em> </em><em>question is not complete, it seems to be missing the following information;"</em>

The sequence of operation is A - E - D - C - B - A - F

The given parameters;

  • <em>number of pieces that will flow from the first machine A to machine F, = 2,000 pieces</em>
  • <em>initial unit load specified in the first machine, L₁ = 50</em>
  • <em>final unit load, L₂ = 100 </em>
  • <em>the capacity of the vehicle = 1 unit load</em>

<em />

The given sequence of operation of the vehicle;

A - E - D - C - B - A - F

<em>the vehicle makes </em><em>6 trips</em><em> for </em><em>100</em><em> unit </em><em>loads</em>

The total number of trips that the vehicle has to make, in order to transport the 2000 pieces of the load given, is calculated as follows.

100 unit loads ----------------- 6 trips

2000 unit loads --------------- ?

= \frac{2000}{100} \times 6\\\\= 120 \ trips

Thus, the total number of trips that the vehicle has to make based on the given sequence of operation is 120 trips.

Learn more here:brainly.com/question/21468592

6 0
2 years ago
A 10 hp motor is used to raise a 1000 Newton weight at a vertical distance of 5 meters. What is the work the motor performs?
Aleksandr [31]

The work done by a 10 HP motor when it raises a 1000 Newton weight at a vertical distance of 5 meters is <u>5kJ</u>.

Define work. Explain the rate of doing work.

Work is <u>the energy that is moved to or from an item by applying force along a displacement</u> in physics. For a constant force acting in the same direction as the motion, work is <u>easiest expressed as the product of </u><u>force </u><u>magnitude and distance traveled</u>.

Since the <u>force </u><u>transfers one unit of energy for every unit of </u><u>work </u><u>it performs</u>, the rate at which work is done and energy is used are equal.

Solution Explained:

Given,

Weight = 1000N and distance = 5m

A/Q, the work here is done in lifting then

Work = (weight) × (distance moved)

         = 1000 X 5

         = 5000Nm or 5000J = 5kJ

Therefore, the work done in lifting a 1000 Newton weight at a vertical distance of 5 meters is 5kJ.

To learn more about work, use the link given
brainly.com/question/25573309
#SPJ9

<u />

4 0
1 year ago
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