To solve this problem, we will apply the concepts related to Faraday's law that describes the behavior of the emf induced in the loop. Remember that this can be expressed as the product between the number of loops and the variation of the magnetic flux per unit of time. At the same time the magnetic flux through a loop of cross sectional area is,

Here,
= Angle between areal vector and magnetic field direction.
According to Faraday's law, induced emf in the loop is,





At time
, Induced emf is,


Therefore the magnitude of the induced emf is 10.9V
A gyre is a set of currents that form b. a loop. The circulation of gyres are affected by global wind patterns, landmasses, and the planet's rotation. The circulation is also affected by temperature, as warm water goes up and cold water sinks. There are five major gyres in the world: <span>North Atlantic, South Atlantic, Indian, North Pacific, and South Pacific.</span>
<span>The angular momentum of a particle in orbit is
l = m v r
Assuming that no torques act and that angular momentum is conserved then if we compare two epochs "1" and "2"
m_1 v_1 r_1 = m_2 v_2 r_2
Assuming that the mass did not change, conservation of angular momentum demands that
v_1 r_1 = v_2 r_2
or
v1 = v_2 (r_2/r_1)
Setting r_1 = 40,000 AU and v_2 = 5 km/s and r_2 = 39 AU (appropriate for Pluto's orbit) we have
v_2 = 5 km/s (39 AU /40,000 AU) = 4.875E-3 km/s
Therefore, </span> the orbital speed of this material when it was 40,000 AU from the sun is <span>4.875E-3 km/s.
I hope my answer has come to your help. Thank you for posting your question here in Brainly.
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Given values:
Mass of the steel ball, m = 100 g = 0.1 kg
Height of the steel ball, h1 = 1.8 m
Rebound height, h2 = 1.25 m
a. PE= mgh
0.1 x 9.8 x 1.8 =
1.764 Joules
b. KE = PE ->
1.764 Joules
c. KE= 1/2 mv square
so v = square root 2ke/m
square root 2 x 1.764/ 0.1
= 5.93 m/s
d. KE=PE=mgh square
0.1 x 9.8 x 1.21 =
1.186 joules
velocity of rebond is square root 2x 1.186/ 0.1 = 4.87 m/s