The gravitational field gx at the surface of planet x (M/m)(r/R)² times the gravitational field gy at the surface of planet y.
<h3>What is gravitational field?</h3>
A gravitational field, as defined by physics, is the influence a massive body has on the area surrounding it, exerting a force on another huge body. In other words, a gravitational field, which is measured in newtons per kilograms (N/kg), aids in the explanation of gravitational field.
Now, Let planet x has mass M and radius R: then gravitational field at the surface of planet: gₓ = GM/R².
Let planet y has mass m and radius r: then gravitational field at the surface of planet:
= Gm/r².
Hence, gₓ/
= (M/m)(r/R)².
So, the gravitational field gx at the surface of planet x (M/m)(r/R)² times the gravitational field gy at the surface of planet y.
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Answer:
(a) τ=2.88 s, (b) q₀= 30.6μC and (c) t=1.434s
Explanation:
A RC circuit is an resistor(R)-capacitor(C) electric circuit.
(a) In a resistor-capacitor circuit, the time constant (τ) can be calculated by:
<em>where R: is the resistence and C: the capacitance of the capacitor</em>
(b) The maximum charge (q₀) is giving by:
<em>where ε: is the voltage across the capacitor</em>
(c) The time (t) that take the charge (q) to build up to 12 μC can be calculated from the next equation:
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Copper conducts electricity
To solve this problem we will apply the concepts related to the electric field. This is defined as the product between the angular frequency, the number of turns of the body (solenoid in this case) the magnetic field and the sine of the angular frequency and time. Mathematically this can be described as

Here,
= Angular frequency
N = Number of turns
B = Magnetic field
The emf has its maximum value when 
Thus the amplitude of the emf is

When number of turns of armature, area and applied magnetic field remains constant, induced emf is proportional to angular speed.

Further it can be written as follows,




Therefore the maximum amplitude of induced emf when armature rotates at 10.0rad/s is 18V