What is the border between two tectonic plates called?
1 answer:
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1) The gravitational force between Ellen and the moon is
2) The two forces are equal, while the acceleration of the bus is smaller than the acceleration of the bicycle.
Explanation:
1)
The magnitude of the gravitational force between two objects is given by
where
is the gravitational constant
are the masses of the two objects
r is the separation between them
In this problem, we have:
is the mass of Ellen
is the mass of the moon
is the distance between Ellen and the moon
Substituting, we find the gravitational force between Ellen and the moon:
2)
We can analyze the forces acting in the collision between the bus and the bicycle by using Newton's third law of motion, which states that:
"When an object A exerts a force (called action) on an object B, then object B exerts an equal and opposite force (called reaction) on object A"
Applied to our problem, this means that the force exerted by the bus on the bicycle during the collision (action force) is equal (and opposite) to the force exerted by the bicycle on the bus (reaction force).
Now let's analyze the accelerations of the two vehicles. We can find the acceleration of each vehicle by using Newton's second law:
where
a is the acceleration
F is the force exerted on the vehicle
m is the mass of the vehicle
As we said previously, the force F exerted on each of the two vehicles: so, the acceleration only depends on the mass. In particular, the acceleration is inversely proportional to the mass: therefore, the larger the mass of the vehicle, the smaller the acceleration. This means that the acceleration of the bus is smaller than the acceleration of the bicycle.
Learn more about gravitational force:
And about Newton's third law:
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Answer:
= 1.75 × 10⁻⁴ m/s
Explanation:
Given:
Density of copper, ρ = 8.93 g/cm³
mass, M = 63.5 g/mol
Radius of wire = 0.625 mm
Current, I = 3A
Area of the wire, =
Now,
The current density, J is given as
= 2444619.925 A/mm²
now, the electron density,
where,
=Avogadro's Number
Now,
the drift velocity,
where,
e = charge on electron = 1.6 × 10⁻¹⁹ C
thus,
= 1.75 × 10⁻⁴ m/s