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.
The charge density of the sheet is 1.384×10⁻⁷C/m².
Charge density is defined as the charge per unit area.
The sheet is a square of length l=17 cm.
Calculate the area A of the sheet .
The charge Q on the sheet is
The charge density σ is given by,
Substitute 4×10⁻⁹C for Q and 0.0289 m² for A.
Thus, the charge density of the sheet is <u>1.384×10⁻⁷C/m².</u>
Answer:
The current is changing at the rate of 0.20 A/s
Explanation:
Given;
inductance of the inductor, L = 5.0-H
current in the inductor, I = 3.0 A
Energy stored in the inductor at the given instant, E = 3.0 J/s
The energy stored in inductor is given as;
E = ¹/₂LI²
E = ¹/₂(5)(3)²
E = 22.5 J/s
This energy is increased by 3.0 J/s
E = 22.5 J/s + 3.0 J/s = 25.5 J/s
Determine the new current at this given energy;
25.5 = ¹/₂LI²
25.5 = ¹/₂(5)(I²)
25.5 = 2.5I²
I² = 25.5 / 2.5
I² = 10.2
I = √10.2
I = 3.194 A/s
The rate at which the current is changing is the difference between the final current and the initial current in the inductor.
= 3.194 A/s - 3.0 A/s
= 0.194 A/s
≅0.20 A/s
Therefore, the current is changing at the rate of 0.20 A/s.
Answer: the correct answer is 7.8026035971 x 10^(-13) joule
Explanation:
Use Energy Conservation. By ``alpha decay converts'', we mean that the parent particle turns into an alpha particle and daughter particles. Adding the mass of the alpha and daughter radon, we get
m = 4.00260 u + 222.01757 u = 226.02017 u .
The parent had a mass of 226.02540 u, so clearly some mass has gone somewhere. The amount of the missing mass is
Delta m = 226.02540 u - 226.02017 u = 0.00523 u ,
which is equivalent to an energy change of
Delta E = (0.00523 u)*(931.5MeV/1u)
Delta E = 4.87 MeV
Converting 4.87 MeV to Joules
1 joule [J] = 6241506363094 mega-electrón voltio [MeV]
4 mega-electrón voltio = 6.40870932 x 10^(-13) joule
4.87 mega-electrón voltio = 7.8026035971 x 10^(-13) joule
and velocity of Abigail =
Explanation: Consider Vs as sam velocity and Va as bigail velocity. Now consider the north as positive x and the east as positive y for our reference. Before collision, their velocity vectors can be represented as
= Vs i
=Va j Now after collision their velocity vectors are given as
= 6 cos 37 i + 6 sin 37 j = 4.79 i + 3.61 j
= 9 cos 23 i + 9 sin 23 j = 8.28 i - 3.52 j As we know in the absence of external force the momentum before and after collision will remain the same therefore P1 = P2 ∴ P = mv Momentum = mass x velocity
As sliding is on frictionless surface therefore before and after the collision, momentum remains conserved in the absence of external force. we can write it as
Now putting values we get 80 x Vs + 50 x 0 = 80 x 4.79 + 50 x 8.28 (First Consider horizontal vector components) Velocity of sam before collision= Vs = 9.97
Now consider vertical vector components 80 x 0 + 50 x Va = 80 x 3.61 + 50 x (-3.52) (Before Collision direction opposite so put negative sign) velocity of abigail before collision = Va = 2.26