Answer:
Wrecking yards employ extremely powerful electromagnets to move heavy pieces of scrap metal or even entire cars from one place to another. ... This creates a magnetic field around the coiled wire, magnetizing the metal as if it were a permanent magnet
Explanation:
I believe the answer is Natural Laws
Bohr's equation for the change in energy is

where
h = Planck's constant
c == the velocity of light
λ = wavelength.
The velocity is related to wavelength and frequency, f, by
c = fλ
Let us examine the given answers on the basis of the given equations.
a. As λ increases, f decreases and ΔE decreases.
TRUE
b. As λ increases, f increases and ΔE increases.
FALSE
c. As λ increases, f increases and ΔE decreases.
FALSE
Answer:
As the wavelength increases, the frequency decreases and energy decreases.
<u>Momentum</u>
- a vector quantity; has both magnitude and direction
- has the same direction as object's velocity
- can be represented by components x & y.
Find linebacker momentum given m₁ = 120kg, v₁ = 8.6 m/s north
P₁ = m₁v₁
P₁ = (120)(8.6)
[ P₁ = 1032 kg·m/s ] = y-component, linebacker momentum
Find halfback momentum given m₂ = 75kg, v₂ = 7.4 m/s east
P₂ = m₂v₂
P₂ = (75)(7.4)
[ P₂ = 555 kg·m/s ] = x-component, halfback momentum
Find total momentum using x and y components.
P = √(P₁)² + (P₂)²
P = √(1032)² + (555)²
[[ P = 1171.77 kg·m/s ]] = magnitude
!! Finally, to find the magnitude of velocity, take the divide magnitude of momentum by the total mass of the players.
P = mv
P = (m₁ + m₂)v
1171.77 = (120 + 75)v <em>[solve for v]</em>
<em />v = 1171.77/195
v = 6.0091 ≈ 6.0 m/s
If asked to find direction, take inverse tan of x and y components.
tanθ = (y/x)
θ = tan⁻¹(1032/555)
[ θ = 61.73° north of east. ]
The magnitude of the velocity at which the two players move together immediately after the collision is approximately 6.0 m/s.
Hooke's Law states that the extension is directly proportional to the force applied so:
F/x = constant
F₁/x₁ = F₂/x₂
2 / 0.02 = 1600 / x₂
x₂ = 16 m
Elastic work = 1/2 Fx
= 1/2 * 1600 * 16
= 12.8 kJ