Answer:
The magnitude of the magnetic force acting on the wire is zero, because the magnetic field is parallel to the wire.
In fact, the magnetic force exerted by the magnetic field on the wire is
where I is the current in the wire, L the length of the wire, B the magnetic field intensity and the angle between the direction of B and the wire. In our problem, B and the wire are parallel, so the angle is and so , therefore the magnetic force is zero: F=0.
Firstly they have a acceleration downwards due the force downwards due they gravitational field acting on it's mass.
as it falls it gains speed, and as it gains speed the air Resistance which is a upward force actin on the drop increases, eventually the rain drop's upward and downward forces are balanced and hence there is no RESULTANT force therefore no acceleration, so the drops falls in constant speed (terminal verlocity is a better term)
Are you wondering that why is the raindrop still moving given that the forces are balanced? If so according to Newton's 1st law an object will keep moving or Remain at rest until a RESULTANT force acts on it.
Answer:
Answer is: c. It must lose two electrons and become an ion.
Magnesium (Mg) is metal from 2. group of Periodic table of elements and has low ionisation energy and electronegativity, which means it easily lose valence electons (two valence electrons).
Magnesium has atomic number 12, which means it has 12 protons and 12 electrons. It lost two electrons to form magnesium cation (Mg²⁺) with stable electron configuration like closest noble gas neon (Ne) with 10 electrons.
Electron configuration of magnesium ion: ₁₂Mg²⁺ 1s² 2s² 2p⁶.
Explanation:
Answer:
The angular acceleration α = 14.7 rad/s²
Explanation:
The torque on the rod τ = Iα where I = moment of inertia of rod = mL²/12 where m =mass of rod and L = length of rod = 4.00 m. α = angular acceleration of rod
Also, τ = Wr where W = weight of rod = mg and r = center of mass of rod = L/2.
So Iα = Wr
Substituting the value of the variables, we have
mL²α/12 = mgL/2
Simplifying by dividing through by mL, we have
mL²α/12mL = mgL/2mL
Lα/12 = g/2
multiplying both sides by 12, we have
Lα/12 × 12 = g/2 × 12
αL = 6g
α = 6g/L
α = 6 × 9.8 m/s² ÷ 4.00 m
α = 58.8 m/s² ÷ 4.00 m
α = 14.7 rad/s²
So, the angular acceleration α = 14.7 rad/s²
