Kinetic energy = (1/2) (mass) (speed squared)
Kinetic energy = (1/2) (400 kg) (17 m/s)²
Kinetic energy = (1/2) (400 kg) (289 m²/s²)
<em>Kinetic energy = 57,800 Joules</em>
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(That's some amazing house. I'd like to be there to see it.)
Answer:
244mm
Explanation:
I₁ = 3.35A
I₂ = 6.99A
μ₀ = 4π*10^-7
force per unit length (F/L) = 6.03*10⁻⁵N/m
B = (μ₀ I₁ I₂ )/ 2πr .........equation i
B = F / L ..........equation ii
equating equation i & ii,
F / L = (μ₀ I₁ I₂ )/ 2πr
Note F/L = B = F
F = (μ₀ I₁ I₂ ) / 2πr
2πr*F = (μ₀ I₁ I₂ )
r = (μ₀ I₁ I₂ ) / 2πF
r = (4π*10⁻⁷ * 3.35 * 6.99) / 2π * 6.03*10⁻⁵
r = 1.4713*10⁻⁵ / 6.03*10⁻⁵
r = 0.244m = 244mm
The distance between the wires is 244m
Because gravity is constant
<span>the only force acting in free-fall is gravity which points downward at 9.8 m/s</span>
The Bohr's proposal for the angular momentum of an electron in Bohr's model of the hydrogen atom is:
L=(n*h)/(2π), where n is the number of the energy level and h is the Planck's constant. This equation shows us the quantization of angular momentum of the electron. So the correct answer is the second one: Planck's constant.
The solution you should use is Hooke's law: F=-kx
It should have the same signs because they repel due to the stretch of the spring.
a. Since there is a constant energy within the spring, then Hooke's law will determine the possible algebraic signs. The solution should be
<span>F = kx
270 N/m x 0.38 m = 102.6 N
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b. Then use Coulomb's law; F=kq1q2/r^2 to find the charges produced in the force.
