F = G mM / r^2, where
<span>F = gravitational force between the earth and the moon, </span>
<span>G = Universal gravitational constant = 6.67 x 10^(-11) Nm^2/(kg)^2, </span>
<span>m = mass of the moon = 7.36 × 10^(22) kg </span>
<span>M = mass of the earth = 5.9742 × 10^(24) and </span>
<span>r = distance between the earth and the moon = 384,402 km </span>
<span>F </span>
<span>= 6.67 x 10^(-11) * (7.36 × 10^(22) * 5.9742 × 10^(24) / (384,402 )^2 </span>
<span>= 1.985 x 10^(26) N</span>
The answer is b 12N because
We know that<span>
W = F × d × c o s(θ)</span>
assuming theta=0 we then solve and have<span>
F=<span>W/d</span></span>
substitute known values to get:<span><span>
F=<span><span>60J/</span><span>5m</span></span>=12N</span></span>
Answer:
This relationship is explained by Ohm's law
Explanation:
Ohm's law states that the current flowing through a circuit or a resistor is directly proportional to the voltage across the resistor and inversely proportional to the resistance. Where current is i, voltage is v and resistance is r, Ohm's law can be represented mathematically as
V= IR
In short, and in general:
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<span>Each of these systems has exactly one degree of freedom and hence only one natural frequency obtained by solving the differential equation describing the respective motions. For the case of the simple pendulum of length L the governing differential equation is d^2x/dt^2 = - gx/L with the natural frequency f = 1/(2π) √(g/L). For the mass-spring system the governing differential equation is m d^2x/dt^2 = - kx (k is the spring constant) with the natural frequency ω = √(k/m). Note that the normal modes are also called resonant modes; the Wikipedia article below solves the problem for a system of two masses and two springs to obtain two normal modes of oscillation.</span>
