<em>The gravitational force between two objects is inversely proportional to the square of the distance between the two objects.</em>
The gravitational force between two objects is proportional to the product of the masses of the two objects.
The gravitational force between two objects is proportional to the square of the distance between the two objects. <em> no</em>
The gravitational force between two objects is inversely proportional to the distance between the two objects. <em> no</em>
The gravitational force between two objects is proportional to the distance between the two objects. <em> no</em>
The gravitational force between two objects is inversely proportional to the product of the masses of the two objects. <em> no</em>
Let the vector position of the object in the (x-y) plane be

The applied force is

By definition, the applied torque is

Answer:
This question is in complete.The question is
A coin with a diameter 3.00 cm rolls up a 30.0° inclined plane. The coin starts out with an initial angular speed of 60.0 rad/s and rolls in a straight line without slipping. If the moment of inertia of the coin is(1/2) MR² , how far will the coin roll up the inclined plane (length along the ramp)? Hint: Conservation of mechanical energy.
Answer:
distance=0.124 m
Explanation:

Because,
In left image pin is not touch to the wire.
In right image pin is touch to the wire.
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Answer:
Δy= 5,075 10⁻⁶ m
Explanation:
The expression that describes the interference phenomenon is
d sin θ = (m + ½) λ
As the observation is on a distant screen
tan θ = y / x
tan θ= sin θ/cos θ
As in ethanes I will experience the separation of the vines is small and the distance to the big screen
tan θ = sin θ
Let's replace
d y / x = (m + ½) λ
The width of a bright stripe at the difference in distance
y₁ = (m + ½) λ x / d
m = 1
y₁ = 3/2 λ x / d
Let's use m = 1, we look for the following interference,
m = 2
y₂ = (2+ ½) λ x / d
The distance to the screen is constant x₁ = x₂ = x₀
The width of the bright stripe is
Δy = λ x / d (5/2 -3/2)
Δy = 630 10⁻⁹ 2.90 /0.360 10⁻³ (1)
Δy= 5,075 10⁻⁶ m