A 1 kg box is sitting on a flat horizontal surface that has a coefficient of static friction value of 0.8. What answer best stat
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This question involves the concept of the conservation of energy.
(a) The child's speed halfway will be "12.13 m/s".
(b) The child's speed three-fourth way will be "14.86 m/s".
(a)
Using the <em>law of conservation of energy</em>:
Potential Energy Lost = Kinetic Energy Gained
where,
v = speed = ?
g = acceleration due to gravity = 9.81 m/s²
h = height lost = halfway = height/2 = 15 m/2 = 7.5 m
Therefore,
<u>v = 12.13 m/s</u>
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(b)
Using the <em>law of conservation of energy</em>:
Potential Energy Lost = Kinetic Energy Gained
where,
v = speed = ?
g = acceleration due to gravity = 9.81 m/s²
h = height lost = three-fourth way down =
Therefore,
<u>v = 14.86 m/s</u>
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Learn more about the <em>law of conservation</em> of energy here:
brainly.com/question/20971995?referrer=searchResults
The attached picture shows the <em>law of conservation of energy</em>.
d = distance between the two point charges = 60 cm = 0.60 m
r = distance of the location of point "a" where the electric field is zero from charge between the two charges.
= magnitude of charge on one charge
= magnitude of charge on other charge
= 3
= Electric field by charge
at point "a"
= Electric field by charge
at point "a"
Electric field by charge at point "a" is given as
= k
/r²
Electric field by charge at point "a" is given as
= k
/(d-r)²
For the electric field to be zero at point "a"
=
k /(d-r)² = k
/r²
/(d-r)² = 3
/r²
1/(0.60 - r)² = 3 /r²
r = 0.38 m
r = 38 cm