Power = I^2 x R
Energy = Power x Time
At point E
- the kinetic energy of the rollercoaster is small compared to the potential energy
- the potential energy is greater than the kinetic energy
- the total energy is a mixture of potential and kinetic energy
<h3>What is the energy of the roller coaster at point E?</h3>
The energy of a roller coaster could either be potential energy, kinetic energy or a combination of both potential and kinetic energy.
Using analogies, the energy of the roller coaster at point E can be compared to a falling fruit from a tree which falls onto a pavement and is the rolling towards the floor. Point E can be compared to the midpoint of the fall of the fruit.
At point E
- the kinetic energy of the rollercoaster is small compared to the potential energy
- the potential energy is greater than the kinetic energy
- the total energy is a mixture of potential and kinetic energy
In conclusion, the energy of the rollercoaster at E is both Kinetic and potential energy,
Learn more about potential and kinetic energy at: brainly.com/question/18963960
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Answer:
Explanation:
Work is defined as the scalar product of force and distance
W=F•d
Given that
F = 8.5i + -8.5j. +×-=-
F=8.5i-8.5j
d = 2.5i + cj
If the work in the practice is zero, then W=0
therefore,
W=F•ds
0=F•ds
0=(8.5i -8.5j)•(2.5i + cj)
Note that
i.i=j.j=k.k=1
i.j=j.i=k.i=i.k=j.k=k.j=0
So applying this
0=(8.5i -8.5j)•(2.5i + cj)
0= (8.5×2.5i.i + 8.5×ci.j -8.5×2.5j.i-8.5×cj.j)
0=21.25-8.5c
Therefore,
8.5c=21.25
c=21.25/8.5
c=2.5
The solution for this problem is through this formula:Ø = w1 t + 1/2 ã t^2
where:Ø - angular displacement w1 - initial angular velocity t - time ã - angular acceleration
128 = w1 x 4 + ½ x 4.5 x 5^2 128 = 4w1 + 56.254w1 = -128 + 56.25 4w1 = 71.75w1 = 71.75/4
w1 = 17.94 or 18 rad s^-1
w1 = wo + ãt
w1 - final angular velocity
wo - initial angular velocity
18 = 0 + 4.5t t = 4 s
Answer:
Explanation:
Work done in lifting the weight once = mgh
= 20 x 9.8 x (1.9+1.7)
= 705.6 J
= 705.6 / 4.2 calorie
= 168 cals
Total energy to be spent = 600 x 10³ cals
No of times weight is required to be lifted
= 600 x 10³ / 168
= 3.57 x 10³ times
Total time to be taken = 2 x 3.57 x 10³
= 7.14 x 10³ s
=119 minutes .
