All categories

2 years ago

Why is the more cumbersome Two's complement representation preferred instead of the more intuitive sign bit magnitude approach?

Physics

1 answer:

Troyanec [42]2 years ago

7 0

Explanation:

The two's-complement mechanism has the benefit that it does not require the addition and subtraction circuitry to investigate the operands ' signs to evaluate either to add or subtract. This property makes the whole thing both easier to accomplish and able to handle arithmetic of higher accuracy with ease. Also Zero has only one interpretation, bypassing the subtle nuances associated with negative one that arise in the complement-systems of ones.

You might be interested in

A 4kg block is sliding on a horizontal frictionless floor at a speed of 2.5ms and runs into a horizontal spring. The spring has

alexgriva [62]

Answer:

Explanation:

Given

mas of block $m=4\ kg$

speed of block $v=2.5\ m/s$

spring constant $k=30\ N-m$

As the mass collides with the spring its kinetic energy is converted to the Elastic Potential energy of the spring

$\frac{1}{2}mv^2=\frac{1}{2}kx^2$

$x=v\sqrt{\frac{m}{k}}$

$x=2.5\times \sqrt{\frac{4}{30}}$

$x=0.912\ m$

8 0

2 years ago

Read 2 more answers

Consider an alcohol and a mercury thermometer that read exactly 0 oC at the ice point and 100 oC at the steam point. The distanc

Alexeev081 [22]

Answer:

No, both the thermometers will give the different reading.

Explanation:

Given,

Both thermometer has same ice point = $T_i\ =\ 0^o C$
Both thermometer has same steam point = $T_s\ =\ 100^o C$

Distance between the ice point and steam point in both the thermometer is same of 100 division,

All the data given in both the thermometers are same, but the material in the thermometer is different due to this the reading at 60^o C will differ in both the thermometer. Because the reading on both the thermometer is depended upon the thermal expansion of the material inside it, but both the materials are different. Due to this the rise of fluid in the thermometer, i,e,. the volume of the fluid material in the thermometer will depend upon the thermal expansion. Hence both the material alcohol and mercury have the different thermal expansion, therefore the rise of the fluid in the thermometer also differ in both the thermometer.

7 0

2 years ago

A thin stream of water flows smoothly from a faucet and falls straight down. at one point the water is flowing at a speed of v1

kati45 [8]

<span>The formulas are, v1d1Â˛ = v2d2Â˛ ........ (1) h = (v2Â˛-v1Â˛)/2g ...... (2) Given that, v1 = 1.71 m/s we assume that the stream has decreased by a factor d2 =0.805d1 then, v1d1Â˛ = v2 (0.805d1)Â˛ cancelled both side d1Â˛ then we get, v1 = v2 (0.805)Â˛ v1 = v2 (0.648025) Sub v1 = 1.71, 1.71 = v2 (0.648025) v2 = 1.71/0.648025 v2 = 2.638787083831642 v2 = 2.64 m/s The vertical distance formula, h = (v2Â˛-v1Â˛)/2g We know that value of gravity constant is 9.8 m/sÂ˛ h = {(2.64)Â˛ - (1.71)Â˛)/2(9.8) h = {(6.9696) - (2.9241)}/19.6 h = (4.0455)/19.6 h = 0.2064030612244898 h = 0.21 cm Therefore, the vertical distance h = 0.21 cm.</span>

8 0

2 years ago

A circular loop of wire with a radius of 12.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne

Ulleksa [173]

(a) 34 V

The average emf induced in the loop is given by Faraday-Newmann-Lenz law:

$\epsilon = -\frac{\Delta \Phi_B}{\Delta t}$ (1)

where

$\Delta \Phi_B$ is the variation of magnetic flux through the coil

$\Delta t = 2.0 ms = 0.002 s$ is the time interval

We need to find the magnetic flux before and after. The magnetic flux is given by:

$\Phi_B = BA$

where

B is the magnetic field intensity

A is the area of the coil

The radius of the coil is r = 12.0 cm = 0.12 m, so its area is

$A=\pi r^2 = \pi (0.12 m)^2 = 0.045 m^2$

At the beginning, the magnetic field is

$B_i = 1.5 T$

so the flux is

$\Phi_i = B_i A = (1.5 T)(0.045 m^2)=0.068 Wb$

while after the removal of the coil, the magnetic field is zero, so the flux is also zero:

$\Phi_f = 0$

so the variation of magnetic flux is

$\Delta \Phi = 0-0.068 Wb=-0.068 Wb$

And substituting into (1) we find the average emf in the coil

$\epsilon=-\frac{-0.068 Wb}{0.002 s}=34 V$

(b) Counterclockwise

In order to understand the direction of the induced current, we have to keep in mind the negative sign in Lenz's law (1), which tells that the direction of the induced current must be such that the magnetic field produced by this current opposes the variation of magnetic flux in the coil.

In this situation, the magnetic flux through the coil is decreasing, since the coil is removed from the field. So, the induced current must be such that it produces a magnetic field whose direction is the same as the direction of the external magnetic field, which is upward along the positive z-direction.

Looking down from above and using the right-hand rule on the loop (thumb: direction of the current, other fingers wrapped: direction of magnetic field), we see that in order to produce at the center of the coil a magnetic field which is along positive z-direction, the induced current must be counterclockwise.

4 0

2 years ago

In a coordinate system in which the x-axis is east, for what range of angles is the x- component positive? For what range is it

trasher [3.6K]

If i remeber correctly when dealing with real world cordinate systems as you rotate around clockwise you move in a positive direction. but all the examples i have done said north was 0 degrees, so i may be wrong

6 0

2 years ago