If you're holding the apple at your waist, lift it to your mouth.
Potential energy relative to any level is proportional to its height
above that level. Increase that height, and you've increased the
potential energy.
Since energy is conserved ... it never magically appears or
disappears ... you need to tell where that extra energy for the
apple came from.
It's exactly the work you did ... the force of your muscles acting
through the distance you raised the apple ... that became the
additional potential energy that the apple gained.
Answer:
16.03m(2dp)
Explanation:
Ep=m x g x h
1100=7.0x 9.8( gravitational field strength) x h
Height= 1100/7.0 x 9.8
=16.03498542
= 16.03m (2dp)
Answer:
Explanation:
a )
The stored elastic energy of compressed spring
= 1 / 2 k X²
= .5 x 19.6 x (.20)²
= .392 J
b ) The stored potential energy will be converted into gravitational potential energy of the block earth system when the block will ascend along the incline . So change in the gravitational potential energy will be same as stored elastic potential energy of the spring that is .392 J .
c ) Let h be the distance along the incline which the block ascends.
vertical height attained ( H ) =h sin30
= .5 h
elastic potential energy = gravitational energy
.392 = mg H
.392 = 2 x 9.8 x .5 h
h = .04 m
4 cm .
=
A) See ray diagram in attachment (-6.0 cm)
By looking at the ray diagram, we see that the image is located approximately at a distance of 6-7 cm from the lens. This can be confirmed by using the lens equation:
where
q is the distance of the image from the lens
f = -10 cm is the focal length (negative for a diverging lens)
p = 15 cm is the distance of the object from the lens
Solving for q,
B) The image is upright
As we see from the ray diagram, the image is upright. This is also confirmed by the magnification equation:
where 
are the size of the image and of the object, respectively.
Since q < 0 and p > o, we have that 
, which means that the image is upright.
C) The image is virtual
As we see from the ray diagram, the image is on the same side of the object with respect to the lens: so, it is virtual.
This is also confirmed by the sign of q in the lens equation: since q < 0, it means that the image is virtual
