Answer:
Explanation:
Givens
delta mv = ?
m = 2 kg
vi = 6 m/s right is positive
vf = - 4 m/s left
Formula
delta mv = m (vf - vi)
Solution
delta mv = 2 * (-4 - 6)
delta mv = 2 * - 10
delta mv = - 20 kg m/s
Answer:
Mass of B is 10 kg
Explanation:
M1v1a + m2v1b = m1v2a + m2v2b
Let m2 = x
5*20 + x10 = 5*10 + x15
100 + x10 = 50 + x15
50 = x5
X= 10
Mass of B is 10 kg
M1 = mass of A
V1a = initial velocity of A
V2a = final velocity of A
M2 = mass of B
V1b = initial velocity of B
V2b = final velocity of B
Answer:
<em>C. Zero </em><em>acceleration</em><em>.</em><em>.</em><em>.</em><em>.</em>
Answer:
continental drift
Explanation:
Wegener proposed that the continents plowed through crust of ocean basins, which would explain why the outlines of many coastlines look like they fit together like a puzzle
Answer:
The average emf induced by rotating a turn of coil in the earth's magnetic field is given by Faraday's law of induction. This law states that the induced emf in a closed loop equals the negative of the time rate of change of magnetic flux through the loop.
Explanation:
When a coil is placed in a magnetic field, it experiences a flux equal to magnetic field strength multiplied by the area enclosed by the loop times the cosine of the angle between the normal to the area and the magnetic field. Mathematically
The flux Ф = BACosθ
When the loop is placed in the earth's magnetic field without any motion, this flux value is constant and no emf is induced. The moment rotation begins in the magnetic field, the angle between normal to the plane of the area and the magnetic field lines starts to change/vary though the area and the strength of the magnetic field remain constant. This causes the net magnetic flux through the loop to change as the magnetic flux is proportional to the angle θ depending on how fast the rotation is made that is the time interval dt an average emf is induced in the coil.
E = –dФ/dt = – d(BACosθ)/dt
Since B and A are constant
E = –BAd(Cosθ)/dt
Where θ is a function of time.
θ = ωt
Where ω = angular frequency measured in rad/s
E = –BAd(Cosωt)/dt
E = –(–ω)BASinωt = ωBASinωt
E = ωBASinωt= –d(BACosθ)/dt
Depending on the case presented both formulas work out well.
Now given that the plane becomes parallel with the field, the normal now becomes perpendicular to the field and θ = 90° so
E = ωBASin90° = ωBA(1) = ωBA