The distance between Dustin and the planet is larger than the distance between Barb and the planet
Explanation:
The magnitude of the gravitational force between each astronaut and the planet is given by
where
:
is the gravitational constant
M is the mass of the planet
m is the mass of the astronaut
r is the separation between the astronaut and the planet
In this problem, we have:
- The force of gravity between Dustin and the planet is 120,265 N
- The force of gravity between Barb and the planet is 354,999 N
We see that the force exerted by the Planet on Barb is much greater than the force exerted by the planet on Dustin. Assuming that the mass of Dustin and Barb is similar, then we can say that the magnitude of the force of gravity depends mainly on the distance:
And since the force is inversely proportional to the square of the distance, this means that the distance between Dustin and the planet is larger than the distance between Barb and the planet.
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Answer:
This is a conceptual problem so I will try my best to explain the impossible scenario. First of all the two dust particles ara virtually exempt from any external forces and at rest with respect to each other. This could theoretically happen even if it's difficult for that to happen. The problem is that each of the particles have an electric charge which are equal in magnitude and sign. Thus each particle should feel the presence of the other via a force. The forces felt by the particles are equal and opposite facing away from each other so both charges have a net acceleration according to Newton's second law because of the presence of a force in each particle:
Having seen Newton's second law it should be clear that the particles are actually moving away from each other and will not remain at rest with respect to each other. This is in contradiction with the last statement in the problem.
the common attributes are positive and negatively charged
Answer:
a) T = 1,467 s
, b) A = 0.495 m
, c) v = 4.97 10⁻² m / s
Explanation:
The simple harmonic movement is described by the expression
x = A cos (wt + Ф)
Where the angular velocity is
w = √ k / m
a) Ask the period
Angular velocity, frequency and period are related
w = 2π f = 2π / T
T = 2π / w
T = 2pi √ m / k
T = 2π √ (1.2 / 22)
T = 1,467 s
f = 1 / T
f = 0.68 Hz
b) ask the amplitude
The mechanical energy of a harmonic oscillator
E = ½ k A²
A = √2 E / k
A = √ (2 2.7 / 22)
A = 0.495 m
c) the mass changes to 8.0 kg
As released from rest Ф = 0, the equation remains
x = A cos wt
w = √ (22/8)
w = 1,658
x = 3.0 cos (1,658 t)
Speed is
v = dx / dt
v = -A w sin wt
The speed is maximum when without wt = ±1
v = Aw
v = 0.03 1,658
v = 4.97 10⁻² m / s
The answer is a, series circuit.
