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Suppose you want to determine the resistance of a resistor that is nominally 100 . You should be able to apply 10 V across the r

Butoxors [25]

Answer:

a) For y = 102 mA, R = 98.039 ohms

For y = 97 mA, R = 103.09 ohms

b) Check explanatios for b

Explanation:

Applied voltage, V = 10 V

For the first measurement, current $y_{1} = 102 mA = 0.102 A$

According to ohm's law, V = IR

R = V/I

Here, $I = y_{1}$

$R = \frac{V}{y_{1} } \\R = \frac{10}{0.102} \\R = 98.039 ohms$

For the second measurement, current $y_{2} = 97 mA = 0.097 A$

$R = \frac{V}{y_{2} }$

$R = \frac{10}{0.097} \\R = 103 .09 ohms$

b) $y = \left[\begin{array}{ccc}y_{1} &y_{2} \end{array}\right] ^{T}$

$y = \left[\begin{array}{ccc}y_{1} \\y_{2} \end{array}\right]$

$y = \left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right]$

A linear equation is of the form y = Gx

The nominal value of the resistance = 100 ohms

$x = \left[\begin{array}{ccc}100\end{array}\right]$

$\left[\begin{array}{ccc}102*10^{-3} \\97*10^{-3} \end{array}\right] = \left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] \left[\begin{array}{ccc}100\end{array}\right]\\\left[\begin{array}{ccc}G_{1} \\G_{2} \end{array}\right] = \left[\begin{array}{ccc}102*10^{-5} \\97*10^{-5} \end{array}\right]$

3 0

2 years ago

A gold bar has a density of 19.3 g/mL. If the gold bar has a mass of 6.3 grams, what is the volume?

Natali [406]

Answer:

The correct answer is "0.32 mL".

Explanation:

The given values are:

Density of gold bar,

d = 19.3 g/mL

Mass of gold bar,

m = 6.3 grams

Now,

The volume will be:

⇒ $Density = \frac{Mass}{Volume}$

or,

⇒ $Volume=\frac{Mass}{Density}$

On substituting the values, we get

⇒ $=\frac{6.3 \ g}{19.3 \ g/mL}$

⇒ $=0.32 \ mL$

7 0

2 years ago

Find the lengths of each of the following vectors

Irina18 [472]

Answer:

Explanation:

Generally, length of vector means the magnitude of the vector.

So, given a vector

R = a•i + b•j + c•k

Then, it magnitude can be caused using

|R|= √(a²+b²+c²)

So, applying this to each of the vector given.

(a) 2i + 4j + 3k

The length is

L = √(2²+4²+3²)

L = √(4+16+9)

L = √29

L = 5.385 unit

(b) 5i − 2j + k

Note that k means 1k

The length is

L = √(5²+(-2)²+1²)

Note that, -×- = +

L = √(25+4+1)

L = √30

L = 5.477 unit

(c) 2i − k

Note that, since there is no component j implies that j component is 0

L = 2i + 0j - 1k

The length is

L = √(2²+0²+(-1)²)

L = √(4+0+1)

L = √5

L = 2.236 unit

(d) 5i

Same as above no is j-component and k-component

L = 5i + 0j + 0k

The length is

L = √(5²+0²+0²)

L = √(25+0+0)

L = √25

L = 5 unit

(e) 3i − 2j − k

The length is

L = √(3²+(-2)²+(-1)²)

L = √(9+4+1)

L = √14

L = 3.742 unit

(f) i + j + k

The length is

L = √(1²+1²+1²)

L = √(1+1+1)

L = √3

L = 1.7321 unit

3 0

2 years ago

Is the matching correct?

iragen [17]

Yes thats right :)

have a great day!!!

7 0

3 years ago

If you pull horizontally on a desk with a force of 150 N and the desk doesn't move, the friction force must be 150 N. Now if you

s344n2d4d5 [400]

Answer:

The friction force is 250 N

Explanation:

The desk is moving at constant velocity. This means that its acceleration is zero: a = 0. Newton's second law states that the resultant of the forces acting on the desk is equal to the product between mass (m) and acceleration (a):

$\sum F=ma$

In this case, we know that the acceleration is zero: a = 0, so also the resultant of the forces must be zero:

$\sum F = 0$ (1)

We are only interested in the forces acting along the horizontal direction, since it is the direction of motion. There are two forces acting in this direction:

- the pull, forward, F = 250 N

- the friction force, backward, $F_f$

Given (1), we have

$F-F_f = 0$

So the force of friction must be equal to the pull:

$F=F_f = 250 N$

8 0

2 years ago