Which data set has the largest range?
2 answers:
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Answer:
It is called force of friction
Explanation:
The force of friction is a force that acts between two objects whose surfaces are in contact with each other.
Consider the typical case of an object sliding along a certain surface. There are two types of frictions:
- Static friction: this is the force of friction that acts when the object is not in motion yet. If you push the object forward with a force F, the object will not move immediately, but it will "oppose" to this motion with a force of static friction exactly equal to the push applied:
However, this force of static friction has a maximum value, which is given by
where
is the coefficient of static friction
N is the normal reaction exerted by the surface on the object
So, when becomes greater than
, the static friction is no longer able to balance the push applied, and the object will start sliding forward.
- Kinetic friction: this is the force of friction that acts when the object is already in motion. Its magnitude is given by
where
is the coefficient of kinetic friction, and its value is generally smaller than
. The direction of this force is also opposite to the direction of motion of the object.
Complete Question
The complete question is shown on the first uploaded image
Answer:
The electric field at that point is
Explanation:
From the question we are told that
The radius of the inner circle is
The radius of the outer circle is
The charge on the spherical shell
The magnitude of the point charge at the center is
The position we are considering is x = 0.60 m from the center
Generally the electric field at the distance x = 0.60 m from the center is mathematically represented as
substituting values
where k is the coulomb constant with value
substituting values
The strength of the gravitational field is given by:
where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.
In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
And now we can use the previous equation to calculate the field strength at that altitude:
And we can see this value is a bit less than the gravitational strength at the surface, which is
.
where
G is the gravitational constant
M is the Earth's mass
r is the distance measured from the centre of the planet.
In our problem, we are located at 300 km above the surface. Since the Earth radius is R=6370 km, the distance from the Earth's center is:
And now we can use the previous equation to calculate the field strength at that altitude:
And we can see this value is a bit less than the gravitational strength at the surface, which is
<h2>Answer:</h2><h2>The acceleration of the meteoroid due to the gravitational force exerted by the planet = 12.12 m/
</h2>
Explanation:
A meteoroid is in a circular orbit 600 km above the surface of a distant planet.
Mass of the planet = mass of earth = 5.972 x Kg
Radius of the earth = 90% of earth radius = 90% 6370 = 5733 km
The acceleration of the meteoroid due to the gravitational force exerted by the planet = ?
By formula, g =
where g is the acceleration due to the gravity
G is the universal gravitational constant = 6.67 x
M is the mass of the planet
r is the radius of the planet
Substituting the values, we get
g =
g = 12.12 m/
The acceleration of the meteoroid due to the gravitational force exerted by the planet = 12.12 m/