The potential energy of an object is defined by the equation: PE = mgh, where m = the mass of the object, g = the gravitational acceleration and h = the object's height above the ground.
Answer:
Part a)

Part b)

Part c)

Explanation:
Part a)
As we know that force on the passenger while moving in circle is given as

now variation in force is given as

here speed is constant
Part b)
Now if the variation in force is required such that r is constant then we will have

so we have

Part c)
As we know that time period of the circular motion is given as

so here if radius is constant then variation in time period is given as

Answer:
Energy of Photon = 4.091 MeV
Explanation:
From the conservation of energy principle, we know that total energy of the system must remain conserved. So, the energy or particles before collision must be equal to the energy of photons after collision.
K.E OF electron + Rest Energy of electron + K.E of positron + Rest Energy of positron = 2(Energy of Photon)
where,
K.E OF electron = 3.58 MeV
Rest Energy of electron = 0.511 MeV
Rest Energy of positron = 0.511 MeV
K.E OF positron = 3.58 MeV
Energy of Photon = ?
Therefore,
3.58 MeV + 0.511 MeV + 3.58 MeV + 0.511 MeV = 2(Energy of Photon)
Energy of Photon = 8.182 MeV/2
<u>Energy of Photon = 4.091 MeV</u>
J can get answer on this way:
Ek=m*V*V/2= (24kg*2m/s*2m/s)/2=48 Ј
Hi there!
Recall the equation for centripetal force:

We can rearrange the equation to solve for 'r'.
Multiply both sides by r:

Divide both sides by Fc:
