Answer:
Spin a coil of wire inside a magnetic field or move a magnet inside a coil of wire.
Explanation:
This is according to Faraday's law of electromagnetic induction.
Answer:
1. A burning streak of light, called a Meteor.
2.The rocky bodies that survives the fall to the planet’s surface are Meteorites.
Explanation:
Let us see the difference between them.
- Asteroids: They are the remnants of a broken planet orbiting the sun in between Mars and Saturn. They are small rocky bodies from a tennis ball size to larger ones weighing a few tones.
- Meteoroids: They are the small rocky objects of pebble size formed due to the collision of asteroids.
-
Meteors: These are pebble size objects that enter Earth's atmosphere and burns off by forming streaks of light.
- Meteorites: These are rocky objects bigger than meteors than don't burn out completely and fall on the earth's surface.
-
Comets: These are objects mainly made of dust and ice that orbit the sun in a highly elliptical orbit. This composition makes a long tail when approaching the sun.
Answer:
a) v = 0.4799 m / s, b) K₀ = 1600.92 J, K_f = 5.46 J
Explanation:
a) How the two players collide this is a momentum conservation exercise. Let's define a system formed by the two players, so that the forces during the collision are internal and also the system is isolated, so the moment is conserved.
Initial instant. Before the crash
p₀ = m v₁ + M v₂
where m = 95 kg and his velocity is v₁ = -3.75 m / s, the other player's data is M = 111 kg with velocity v₂ = 4.10 m / s, we have selected the direction of this player as positive
Final moment. After the crash
p_f = (m + M) v
as the system is isolated, the moment is preserved
p₀ = p_f
m v₁ + M v₂ = (m + M) v
v =
let's calculate
v = 
v = 0.4799 m / s
b) let's find the initial kinetic energy of the system
K₀ = ½ m v1 ^ 2 + ½ M v2 ^ 2
K₀ = ½ 95 3.75 ^ 2 + ½ 111 4.10 ^ 2
K₀ = 1600.92 J
the final kinetic energy
K_f = ½ (m + M) v ^ 2
k_f = ½ (95 + 111) 0.4799 ^ 2
K_f = 5.46 J
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<span>View the entire bridge as the system, and find the torque about a pivot chosen to eliminate the torque from one of the unknowns
</span>Find the sum of the torques ∑ τ Q about the right pier Q of the bridge. Remember that counterclockwise torque is positive
∑τQ - 2MgL− 3FpL
<span>The sum of the torques is zero
</span>
FP = 2/3 Mg
