If gravity between the Sun and the Earth suddenly vanished, Earth would continue moving
1 answer:
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We can use the law of conservation of energy to solve the problem.
The total mechanical energy of the system at any moment of the motion is:
where U is the potential energy and K the kinetic energy.
At the beginning of the motion, the ball starts from the ground so its altitude is h=0 and therefore its potential energy U is zero. So, the mechanical energy is just kinetic energy:
When the ball reaches the maximum altitude of its flight, it starts to go down again, so its speed at that moment is zero: v=0. So, its kinetic energy at the top is zero. So the total mechanical energy is just potential energy:
But the mechanical energy must be conserved, Ef=Ei, so we have
and so, the potential energy at the top of the flight is
The total mechanical energy of the system at any moment of the motion is:
where U is the potential energy and K the kinetic energy.
At the beginning of the motion, the ball starts from the ground so its altitude is h=0 and therefore its potential energy U is zero. So, the mechanical energy is just kinetic energy:
When the ball reaches the maximum altitude of its flight, it starts to go down again, so its speed at that moment is zero: v=0. So, its kinetic energy at the top is zero. So the total mechanical energy is just potential energy:
But the mechanical energy must be conserved, Ef=Ei, so we have
and so, the potential energy at the top of the flight is
Answer:
1.Theimage will be located at -0.13m or -13 cm
2.The height of the image will be 0.052m or 5.2cm
Explanation:
Given that;
Height of object, h=20 cm = 0.2m
Object distance in front of convex mirror, o,= 50 cm =0.5m
Radius of curvature, r, =34 cm =0.34m
Let;
Image distance, i,=?
Image height, h'=?
You know that focal length,f, is half the radius of curvature,hence
f=r/2 = 0.34/2 = 0.17m ( this length is inside the mirror, in a virtual side, thus its is negative)
f= -0.17m
Apply the relationship that involves the focal length;
Re-arrange to get i
This is a virtual image formed at a negative distance produced through extension of drawing rays behind the mirror if you use rays to locate the image behind the mirror
Apply the magnification formula
magnification, m=height of image÷height of object
substitute the values to get the height of image h'
The correct answer is
<span>c. one person exerts more force than the other so that the forces are unbalanced.
In fact, the door is initially at rest. In order to move the door, a net force different from zero should be applied, according to Newton's second law:
</span>
<span>where the term on the left is the resultant of the forces acting on the door, m is the door mass and a its acceleration.
In order to move the door, the acceleration must be different from zero. But this means that the resultant of the forces acting on it must be different from zero: this is possible only if the forces applied by the two persons are unbalanced, i.e. one person exerts more force than the other.</span>
<span>c. one person exerts more force than the other so that the forces are unbalanced.
In fact, the door is initially at rest. In order to move the door, a net force different from zero should be applied, according to Newton's second law:
</span>
<span>where the term on the left is the resultant of the forces acting on the door, m is the door mass and a its acceleration.
In order to move the door, the acceleration must be different from zero. But this means that the resultant of the forces acting on it must be different from zero: this is possible only if the forces applied by the two persons are unbalanced, i.e. one person exerts more force than the other.</span>