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1 year 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]1 year 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.

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When explaining chemical reactions to a friend, Brianna models a reaction by taking several colors of modeling clay and making a

Drupady [299]

Answer: synthesis

Explanation:

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2 years ago

Read 2 more answers

A certain amusement park ride consists of a large rotating cylinder of radius R=3.05 m.R=3.05 m. As the cylinder spins, riders i

aniked [119]

Answer:

a. N = 2.49W b. 0.40

Explanation:

a. What is the magnitude of the normal force FNFN between a rider and the wall, expressed in terms of the rider's weight W?

Since the normal force equals the centripetal force on the rider, N = mrω² where r = radius of cylinder = 3.05 m and ω = angular speed of cylinder = 0.450 rotations/s = 0.450 × 2π rad/s = 2.83 rad/s

Now N = mrω² = m(3.05 m) × (2.83 rad/s)² = 24.43m

The rider's weight W = mg = 9.8m

The ratio of the normal force to the rider's weight is

N/W = 24.43m/9.8m = 2.49

So the normal force expressed in term's of the rider's weight is

N = 2.49W

b. What is the minimum coefficient of static friction µsμs required between the rider and the wall in order for the rider to be held in place without sliding down?

The frictional force, F on the rider by the wall of the cylinder equals the weight, W of the rider. F = W.

Since the frictional force F = μN, where μ = coefficient of static friction between rider and wall of cylinder and N = normal force between rider and wall of cylinder.

So, the normal force equals

N = F/μ = W/μ = mg/μ = mrω²

μ = mg/mrω²

= W/N

= 9.8m/24.43m

= 0.40

6 0

1 year ago

A heavy stone of mass m is hung from the ceiling by a thin 8.25-g wire that is 65.0 cm long. When you gently pluck the upper end

Triss [41]

Answer: m= 35.6 kg

Explanation:

For finding the mass of the stone we have the formula

v= $\sqrt{\frac{Tension}{Linear. Mass. density} }$

Here, Tension= m*g = m*9.81

and linear mass density= $\frac{8.25 g}{65 cm}$

Linear mass density= $\frac{8.25*10^-3}{65*10^-2}$

Linear mass density= 0.0127 kg/m

Velocity= $2*\frac{l}{t}$

Velocity= 2 * $\frac{65*10^-2}{7.84}$

Velocity= 165.8 m/s

So putting all these values in equation we get

v= $\sqrt{\frac{Tension}{Linear. Mass. density} }$

165.8= $\sqrt{\frac{m*9.81}{0.0127} }$

Solving we get

m= 35.58 kg

or m= 35.6 kg

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2 years ago

Before you start taking measurements though, we’ll first make sure you understand the underlying concepts involved. By what meth

Svetradugi [14.3K]

Answer:

If they are metallic spheres they are connected to earth and a charged body approaches

non- metallic (insulating) spheres in this case are charged by rubbing

Explanation:

For fillers, there are two fundamental methods, depending on the type of material.

If they are metallic spheres, they are connected to earth and a charged body approaches, this induces a charge of opposite sign and of equal magnitude, then it removes the contact to earth and the sphere is charged.

If the non- metallic (insulating) spheres in this case are charged by rubbing with some material or touching with another charged material, in this case the sphere takes half the charge and when separated each sphere has half the charge and with equal sign.

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2 years ago

Suppose you wanted to hold up an electron against the force of gravity by the attraction of a fixed proton some distance above i

SCORPION-xisa [38]

Answer:

The value is $r = 5.077 \ m$