Answer:
it decreases because of the differences in hormones
Answer:
The induced current is 0.084 A
Explanation:
the area given by the exercise is
A = 200 cm^2 = 200x10^-4 m^2
R = 5 Ω
N = 7 turns
The formula of the emf induced according to Faraday's law is equal to:
ε = (-N * dφ)/dt = (N*(b2-b1)*A)/dt
Replacing values:
ε = (7*(38 - 14) * (200x10^-4))/8x10^-3 = 0.42 V
the induced current is equal to:
I = ε /R = 0.42/5 = 0.084 A
3. The sum of the players' momenta is equal to the momentum of the players when they're stuck together:
(75 kg) (6 m/s) + (80 kg) (-4 m/s) = (75 kg + 80 kg) v
where v is the velocity of the combined players. Solve for v :
450 kg•m/s - 320 kg•m/s = (155 kg) v
v = (130 kg•m/s) / (155 kg)
v ≈ 0.84 m/s
4. The total momentum of the bowling balls prior to collision is conserved and is the same after their collision, so that
(6 kg) (5.1 m/s) + (4 kg) (-1.3 m/s) = (6 kg) (1.5 m/s) + (4 kg) v
where v is the new velocity of the 4-kg ball. Solve for v :
30.6 kg•m/s - 5.2 kg•m/s = 9 kg•m/s + (4 kg) v
v = (16.4 kg•m/s) / (4 kg)
v = 4.1 m/s
Let's be clear: The plane's "395 km/hr" is speed relative to the
air, and the wind's "55 km/hr" is speed relative to the ground.
Before the wind hits, the plane moves east at 395 km/hr relative
to both the air AND the ground.
After the wind hits, the plane still maintains the same air-speed.
That is, its velocity relative to the air is still 395 km/hr east.
But the wind vector is added to the air-speed vector, and the
plane's velocity <span>relative to the ground drops to 340 km/hr east</span>.
Answer:
D. move up to another shell that would form
Explanation:
An atom has protons, neutrons and electrons. Protons and neutrons are present in the nucleus and electrons orbit the nucleus in fixed shells. An electron can jump to higher shell when it gains energy and lower one when it loses energy. Thus, when single electron in hydrogen atom is given a small amount of energy, it would jump to another higher shell.
