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My name is Ann [436]

4 years ago

6

Does time machine really exist?

2 answers:

Genrish500 [490]4 years ago

7 0

No...and even if it did the present that you are in would not be the same...theres no need for time machine because whatever happens , happens for good because its from GOD.

Valentin [98]4 years ago

7 0

Not currently,but it's unpredictable.

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A 12-bit analog to digital data acquisition module has an input range of 0 vdc to + 10 vdc. calculate the least significant bit

trasher [3.6K]

Correct answer is 2.441mV

What is Digital data?

Digital data is information represented as a string of discrete symbols, each of which can take on one of only a finite number of values from some alphabet, such as letters or digits, in information theory and information systems. s. Binary data is the most common type of digital data in modern information systems. It is represented by a string of binary digits (bits), each of which can have one of two values, 0 or 1.

In contrast to digital data, analog data is represented by a value from a continuous range of real numbers.

To learn more about Digital output from the given link:

https://brainly.in/question/2989269

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7 0

2 years ago

Four electrons are located at the corners of a square 10.0 nm on a side, with an alpha particle at its midpoint. How much work i

Elis [28]

Four electrons are placed at the corner of a square

So we will first find the electrostatic potential at the center of the square

So here it is given as

$V = 4\frac{kQ}{r}$

here

r = distance of corner of the square from it center

$r = \frac{a}{\sqrt2}$

$r = \frac{10nm}{\sqrt2} = 7.07 nm$

$Q = e = -1.6 * 10^{-19} C$

now the net potential is given as

$V = \frac{4 * 9*10^9 * (-1.6 * 10^{-19})}{7.07 * 10^{-9}}$

$V = 0.815 V$

now potential energy of alpha particle at this position

$U_i = qV = 2*1.6 * 10^{-19} * (-0.815) = -2.6 * 10^{-19} J$

Now at the mid point of one of the side

Electrostatic potential is given as

$V = 2\frac{kQ}{r_1} + 2\frac{kQ}{r_2}$

here we know that

$r_1 = \frac{a}{2} = 5 nm$

$r_2 = \sqrt{(a/2)^2 + a^2} = \frac{\sqrt5 a}{2}$

$r_2 = 11.2 nm$

now potential is given as

$V = 2\frac{9 * 10^9 * (-1.6 * 10^{-19})}{5 * 10^{-9}} + 2\frac{9*10^9 * (-1.6 * 10^{-19})}{11.2 * 10^{-9}}$

$V = -0.576 - 0.257 = -0.833 V$

now final potential energy is given as

$U_f = q*V = 2*1.6 * 10^{-19}* (-0.833) = -2.67 * 10^{-19} J$

Now work done in this process is given as

$W = U_f - U_i$

$W = (-0.267 * 10^{-19}) - (-0.26 * 10^{-19}}$

$W = -7 * 10^{-22} J$

8 0

3 years ago

The strength of an electromagnet CANNOT be increased by?

zmey [24]

Answer:

reversing the current

Explanation:

4 0

3 years ago

The mass of a string is 5.9 × 10-3 kg, and it is stretched so that the tension in it is 200 n. a transverse wave traveling on th

bagirrra123 [75]

The velocity of the wave on the string is given by

$v=\sqrt{\frac{T}{\frac{m}{L}}} \\ v=\sqrt{\frac{TL}{m}}$

Solving the above equation,

$v^2=\frac{TL}{m} \\ L=\frac{v^2m}{T}$

The frequency of the wave $f=300$ and wave length is $0.76$

The velocity is $v=(300)(0.76)=228$

Substituting numerical values,

$L=\frac{228^2(0.0059)}{200}\\ T=1.534$

The length of the string is $1.534 m$

4 0

3 years ago

If systems interacts with objects outside itself . why would the momentum of the system change ?

Masteriza [31]

Vibration
speed
heart rate

4 0

4 years ago