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.
Answer:
Acceleration=24.9ft^2/s^2
Angular acceleration=1.47rads/s
Explanation:
Note before the ladder is inclined at 30° to the horizontal with a length of 16ft
Hence angular velocity = 6/8=0.75rad/s
acceleration Ab=Aa +(Ab/a)+(Ab/a)t
4+0.75^2*16+a*16
0=0.75^2*16cos30°-a*16sin30°---1
Ab=0+0.75^2sin30°+a*16cos30°----2
Solving equation 1
(0.75^2*16cos30/16sin30)=angular acceleration=a=1.47rad/s
Also from equation 2
Ab=0.75^2*16sin30+1.47*16cos30=24.9ft^2/s^2
Answer:
The classification of that same issue in question is characterized below.
Explanation:
The given values are:
Current, I = 50.0 A
Diameter, d = 0.10 cm
(a)...
As we know,
⇒ Magnetic force = Copper wire's weight
So,
⇒ 
On putting the estimated values, we get
⇒ 
⇒ 
⇒ 
(b)...
As we know,
⇒ 
⇒ 
⇒ 
⇒ 
Answer:
E.true only when no charge is enclosed within the Gaussian surface.
Explanation:
Because Gauss’s law states that the net flux of an electric field in a closed surface is directly proportional to the enclosed electric charge.
Answer:
The distance the planet Neptune travels in a single orbit around the Sun is <em>60.2π </em><em>AU.</em>
Explanation:
As it is given that the Neptune's orbit is circular, the formula that we have to use is the circumference of a circle in order to find the distance it travels in a single orbit around the Sun. In other words, you can say that the circumference of the circle is <em>equivalent</em> to the distance it travels around the Sun in a single orbit.
<em>The circumference of the circle = Distance Travelled (in a single orbit) = 2*π*R ---- (A)</em>
Where,
<em>R = Orbital radius (in this case) = 30.1 AU</em>
<em />
Plug the value of R in the equation (A):
<em>(A) => The circumference of the circle = 2*π*(30.1)</em>
<em> The circumference of the circle = </em><em>60.2π</em>
Therefore, the distance the planet Neptune travels in a single orbit around the Sun is <em>60.2π </em><em>AU.</em>
