Now suppose that the Earth had the same mass and radius as it currently does, but all of the mass was concentrated into a thin (
1 answer:
Answer:
The force of gravity at the shell will be extremely great on me due to the huge mass collapsed into the small radius.
<em>At the center of the shell, the gravitational forces all around should cancel out, giving me a feeling of weightlessness; which will be a lesser force compared to that felt while standing on the shell.</em>
<em></em>
Explanation:
For the collapsed earth:
mass = 5.972 × 10^24 kg
radius = 1 ft
according to Newton's gravitation law, the force of gravity due to two body with mass is given as
Fg = GMm/
Where Fg is the gravitational force between the two bodies.
G is the gravitational constant
M is the mass of the earth
m is my own mass
R is the distance between me and the center of the earths in each case
For the case where I stand on the shell:
radius R will be 1 ft
Fg = GMm/
Fg = GMm
For the case where I stand stand inside the shell, lets say I'm positioned at the center of the shell. The force of gravity due to my mass will be balanced out by all other masses around due to the shell of the hollow earth. This cancelling will produce a weightless feeling on me.
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Answer:
a. The component of the net force which make up the apparent weight are added to each other at the bottom and subtracted (the centripetal force from the weight) at the top)
b. Apparent weight at the top is approximately 519.06 N
Apparent weight at the bottom is approximately 558.94 N
Explanation:
a. The apparent weight at the top is different from the apparent weight at the bottom of a moving Ferris wheel because of the opposite direction in which the centripetal force acts at the top and the bottom, which are upwards and downwards respectively, while the weight acts downwards constantly
b. The given parameters are
The radius of the Ferris wheel, r = 7.2 m
The period for one complete revolution, t = 28 seconds
The angle covered in one revolution, θ = 2·π radian
The mass of the person riding on the Ferris wheel, the passenger = 55 kg
Therefore, we have;
The angular speed, ω = Δθ/Δt = 2·π/(28)
From which we have;
Centripetal force, = m × ω² × r
Substituting the known values, we have = 55 kg × (2·π/(28 s))² × 7.2 m ≈ 19.94 N
The centripetal force, = 19.94 N always acting outward from the center
Weight = Mass × Acceleration due to gravity
The weight of the passenger = 55 kg × 9.8 m/s² = 539 N
The weight of the passenger = 539 N always acting downwards
At the top of the Ferris wheel the the centripetal force is acting upwards and the weight is acting downwards
Therefore;
The net force, which is the apparent weight of the passenger at the top = 539 N - 19.94 N ≈ 519.06 N
Apparent weight at the top ≈ 519.06 N
At the bottom of the Ferris wheel the weight is acting downwards and the centripetal force is also acting downwards
Therefore;
The net force at the bottom, which is the apparent weight of the passenger at the bottom = 539 N + 19.94 N ≈ 558.94 N
Apparent weight at the bottom ≈ 558.94 N.
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Answer:
<h2>206.67N</h2>Explanation:
The sum of force along both components x and y is expressed as;
The magnitude of the net force which is also known as the resultant will be expressed as
To get the resultant, we need to get the sum of the forces along each components. But first lets get the acceleration along the components first.
Given the position of the object along the x-component to be x = 6t² − 4;
Similarly,
Hence, the magnitude of the net force acting on this object at t = 2.15 s is approximately 206.67N
The hawk’s centripetal acceleration is 2.23 m/s²
The magnitude of the acceleration under new conditions is 2.316 m/s²
radius of the horizontal arc = 10.3 m
the initial constant speed = 4.8 m/s
we know that the centripetal acceleration is given by
=
= 23.04/10.3
= 2.23 m/s²
It continues to fly but now with some tangential acceleration
= 0.63 m/s²
therefore the net value of acceleration is given by the resultant of the centripetal acceleration and the tangential acceleration
so
=
=
= 2.316 m/s²
So the magnitude of net acceleration will become 2.316 m/s².
learn more about acceleration here :
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