If gravity between the sun and Earth suddenly vanished; describe the expected Earth motion.
1 answer:
Answer:
When the gravity between the Sun and the Earth suddenly vanishes, then Earth will keep moving in a straight line in a direction where it was moving at the moment when gravity vanished.
Explanation:
The gravitational force between Sun and Earth is given by
Where F is the gravitational force between the Sun and the Earth. M₁ and M₂ are the masses of Sun and Earth and R is the distance between them.
If we assume that the gravity between the sun and Earth suddenly vanishes, then there would not be any force between the Sun and the Earth and the Earth will keep on moving in a straight line. This is endorsed by the Newton's first law that a body in motion remains in motion if no external force is acting on it.
The direction of Earth's motion will be determined by the previous direction of the motion that is the moment when the gravity was vanished.
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Answer:
350x
Explanation:
In a microscope the objective has higher magnification than the eyepiece so, this is a microscope
The magnification of a microscope is given by the product of the magnifications of the eyepiece and and the objective.
Objective lens magnification = 35x =
Eyepiece magnification = 10x =
Total magnification
Total magnification is 350x
Answer:
a) K = 0.63 J, b) h = 0.153 m
Explanation:
a) In this exercise we have a physical pendulum since the rod is a material object, the angular velocity is
w² =
where d is the distance from the pivot point to the center of mass and I is the moment of inertia.
The rod is a homogeneous body so its center of mass is at the geometric center of the rod.
d = L / 2
the moment of inertia of the rod is the moment of a rod supported at one end
I = ⅓ m L²
we substitute
w =
w =
w =
w = 4.427 rad / s
an oscillatory system is described by the expression
θ = θ₀ cos (wt + Φ)
the angular velocity is
w = dθ /dt
w = - θ₀ w sin (wt + Ф)
In this exercise, the kinetic energy is requested in the lowest position, in this position the energy is maximum. For this expression to be maximum, the sine function must be equal to ±1
In the exercise it is indicated that at the lowest point the angular velocity is
w = 4.0 rad / s
the kinetic energy is
K = ½ I w²
K = ½ (⅓ m L²) w²
K = 1/6 m L² w²
K = 1/6 0.42 0.75² 4.0²
K = 0.63 J
b) for this part let's use conservation of energy
starting point. Lowest point
Em₀ = K = ½ I w²
final point. Highest point
Em_f = U = m g h
energy is conserved
Em₀ = Em_f
½ I w² = m g h
½ (⅓ m L²) w² = m g h
h = 1/6 L² w² / g
h = 1/6 0.75² 4.0² / 9.8
h = 0.153 m