1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Annette [7]
3 years ago
10

The lab you are working in has asked you to select a data acquisition board as part of a new instrument designed to record a vol

tage waveform. The waveform to be measured is a triangle shape with a duration of 100 ms and an amplitude of 50 mV. Your budget is $1500 and you must purchase a data acquisition board from National Instruments with a USB interface.
Required:
Draw a model to purchase and include an explanation of why you selected this model.
Physics
1 answer:
Thepotemich [5.8K]3 years ago
5 0

Answer:

USB-6000

Explanation:

For this strange exercise, you must take into account two fundamental aspects of the data acquisition cards;

* The resolution of the cards, which for a 12-bit card is 2.24mV =2.24 10⁻³V , for which a card of this resolution is sufficient

    In your case, the amplitude is 50 mv, which is why it is much higher than the resolution of the card.

* The sampling rate or the frequency in which the card can acquire the data, the typical frequencies go from 200 Khz to the MHz, in general a card that has more than double the frequency to be measured should be selected.

In this case the frequency of your wave is

           f = 1 / T

           f = 1/100 10⁻³

            f = 10 Hz

therefore a low frequency card is suitable.

- The third point is the requirement that it has a USB interface, most of the card do not have it

If you check the National Instruments catalog, there are low-cost cards for example: USB-6000 of 12bits and 10KS / s the latter is the samples taken per second

You might be interested in
NEED THE ANSWER PLZ<br><br>What are lanthanides?
geniusboy [140]
It is the the ecrins of light
5 0
3 years ago
Please Help. I would Really Appreciate it
Readme [11.4K]

Answer:

a) 0.60 kg cart has final velocity 3.0 m/s [E]

0.80 kg cart has final velocity 4.0 m/s [W]

b) 0.12 m

Explanation:

Take east to be positive.

a) Momentum is conserved.

m₁u₁ + m₂u₂ = m₁v₁ + m₂v₂

(0.60) (-5.0) + (0.80) (2.0) = (0.60) v₁ + (0.80) v₂

-1.4 = 0.6 v₁ + 0.8 v₂

Kinetic energy is conserved in elastic collisions.

½ m₁u₁² + ½ m₂u₂² = ½ m₁v₁² + ½ m₂v₂²

m₁u₁² + m₂u₂² = m₁v₁² + m₂v₂²

(0.60) (-5.0)² + (0.80) (2.0)² = (0.60) v₁² + (0.80) v₂²

18.2 = 0.6 v₁² + 0.8 v₂²

Solve the system of equations.

-1.4 = 0.6 v₁ + 0.8 v₂

-1.4 − 0.6 v₁ = 0.8 v₂

-1.75 − 0.75 v₁ = v₂

18.2 = 0.6 v₁² + 0.8 (-1.75 − 0.75 v₁)²

18.2 = 0.6 v₁² + 0.8 (3.0625 +2.625 v₁ + 0.5625 v₁²)

182 = 6 v₁² + 8 (3.0625 + 2.625 v₁ + 0.5625 v₁²)

182 = 6 v₁² + 24.5 + 21 v₁ + 4.5 v₁²

0 = 10.5 v₁² + 21 v₁ − 157.5

0 = v₁² + 2 v₁ − 15

0 = (v₁ − 3) (v₁ + 5)

v₁ = 3 or -5

Since u₁ = -5.0 m/s, v₁ must be 3.0 m/s.

Solving for v₂:

v₂ = -0.75 v₁ − 1.75

v₂ = -4.0 m/s

b) The compression of the spring is a maximum when the carts have the same velocity.

Momentum is conserved.

m₁u₁ + m₂u₂ = (m₁ + m₂) v

(0.60) (-5.0) + (0.80) (2.0) = (0.60 + 0.80) v

-1.4 = 1.4 v

v = -1.0

Energy is conserved.

½ m₁u₁² + ½ m₂u₂² = ½ (m₁ + m₂) v² + ½ kx²

m₁u₁² + m₂u₂² = (m₁ + m₂) v² + kx²

(0.60) (-5.0)² + (0.80) (2.0)² = (0.60 + 0.80) (-1.0)² + (1200) x²

18.2 = 1.4 + 1200 x²

16.8 = 1200 x²

x² = 0.014

x = 0.12

5 0
4 years ago
When blueshift occurs,the preceived frequency of the wave would be?​
LiRa [457]

Answer:

When blueshift happens, the perceived frequency of the wave would be higher than the actual frequency.

Explanation:

As the name suggests, when blueshift happens to electromagnetic waves, the frequency of the observed wave would shift towards the blue (high-frequency) end of the visible spectrum. Hence, there would be an increase to the apparent frequency of the wave.

Blueshifts happens when the source of the wave and the observer are moving closer towards one another.

Assume that the wave is of frequency f\; {\rm Hz} at the source. In other words, the source of the wave sends out a peak after every (1/f)\; {\text{seconds}}.

Assume that the distance between the observer and the source of the wave is fixed. It would then take a fixed amount of time for each peak from the source to reach the observer.

The source of this wave sends out a peak after each period of (1/f)\; {\text{seconds}}. It would appear to the observer that consecutive peaks arrive every (1/f)\; {\text{seconds}}\!. That would correspond to a frequency of f\; {\rm Hz}.

On the other hand, for a blueshift to be observed, the source of the wave needs to move towards the observer. Assume that the two are moving towards one another at a constant speed of v \; {\rm m \cdot s^{-1}}.

Again, the source of this wave would send out a peak after each period of (1/f)\; {\text{seconds}}. However, by the time the source sends out the second peak, the source would have been v \cdot (1 / f) \; { \rm m}= (v / f)\; {\rm m} closer to the observer then when the source sent out the first peak.

When compared to the first peak, the second peak would need to travel a slightly shorter distance before it reach the observer. Hence, from the perspective of the observer, the time difference between the first and the second peak would be shorter than (1/f)\; {\text{seconds}}. The observed frequency of this wave would be larger than the original f\; {\rm Hz}.

6 0
3 years ago
BWhat Do You Get When You Multiply An Object's mass times the acceleration?
STatiana [176]

You get the net force acting on the object.

6 0
4 years ago
A gray kangaroo can bound across a flat stretch of ground with each jump carrying it 10 m from the takeoff point. you may want t
Stells [14]

<u>Answer:</u>

Takeoff speed of Kangaroo = 12.91 m/s

<u>Explanation:</u>

  The distance reached by Kangaroo =  10 meter

  Angle at which it jumps = 18°

  The motion of Kangaroo is like a projectile, the distance traveled is the range of projectile.

   Range of projectile = Time taken for the projectile to reach ground* Horizontal velocity

   Time taken for the projectile to reach ground:

       Time taken = Two times of the time taken for the projectile to reach maximum height

       Time taken for the projectile to reach maximum height = Vertical speed / Acceleration = u sin θ/g

      Time taken for the projectile to reach ground = 2 u sin θ/g

  So Range of projectile = ucos\theta*\frac{2usin\theta}{g} =\frac{u^2sin2\theta}{g}

 We have Range = 10 meter, θ = 18⁰

    Substituting

        10=\frac{u^2sin(2*18)}{9.8}\\ \\ u^2= 166.73\\ \\ u=12.91 m/s

 Takeoff speed of Kangaroo = 12.91 m/s

3 0
3 years ago
Read 2 more answers
Other questions:
  • How to treat thyroid at home​
    9·2 answers
  • The change in color when acetone is placed on the wing is due to the difference between the indices of refraction of acetone and
    6·1 answer
  • A small object at rest is let to slide straight down a 35° incline. After 0.45 m its speed
    5·1 answer
  • Calculate the mass of the air contained in a room that measures 2.50 m x 5.50 m x 3.00 m if the density of air is 1.29 g/dm3.53.
    14·1 answer
  • What's the relationship between potential energy and a position of a rock
    14·1 answer
  • QUICK!!!!!!!!!!!! QUICKKKK!!! HELP!!!! What is the slope of the line????
    14·1 answer
  • You can identify a pure substance if you know it's ?
    8·2 answers
  • The train with a mass of 49 kg × 10⁴ kg departs from the place and in 25 seconds reaches a speed of 5m / s, determine the attrac
    12·1 answer
  • The eiffel tower is 300 meters tall. Disregarding air friction, at what velocity would an object be traveling when it reaches th
    9·1 answer
  • According to max weber, flaunting of one’s wealth to show one’s status is called?
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!