Answer:
m = mass of the penny
r = distance of the penny from the center of the turntable or axis of rotation
w = angular speed of rotation of turntable
F = centripetal force experienced by the penny
centripetal force "F" experienced by the penny of "m" at distance "r" from axis of rotation is given as
F = m r w²
in the above equation , mass of penny "m" and angular speed "w" of the turntable is same at all places. hence the centripetal force directly depends on the radius .
hence greater the distance from center , greater will be the centripetal force to remain in place.
So at the edge of the turntable , the penny experiences largest centripetal force to remain in place.
Explanation:
Answer: c. 4.56 × 105 J
Explanation:
Given that
mass of lead brick, m= 7.25kg
Temperature T1 = 18.0 °C
Temperature T2 = 328 °C
specific heat capacity of lead, c = 128 J/(kg∙C°)
latent heat of fusion Lfusion =23,200 J/kg
Amount of energy Q =?
Using the formulae
Amount of energy ,Q =mc ( T2-T1)+ mLfusion
7.25kg x 128 J/(kg∙C°) x (328-18°C) + 7.25kg x 23200 J/kg
=455880J
=4.56 x 10^5 J
Answer:
(a) coefficient of friction = 0.451
This was calculated by the application of energy conservation principle (the total sum of energy in a closed system is conserved)
(b) No, it comes to a stop 5.35m short of point B. This is so because the spring on expanding only does a work of 43 J on the block which is not enough to meet up the workdone of 398 J against friction.
Explanation:
The detailed step by step solution to this problems can be found in the attachment below. The solution for part (a) was divided into two: the motion of the body from point A to point B and from point B to point C. The total energy in the system is gotten from the initial gravitational potential energy. This energy becomes transformed into the work done against friction and the work done in compression the spring. A work of 398J was done in overcoming friction over a distance of 6.00m. The energy used in doing so is lost as friction is not a conservative force. This leaves only 43J of energy which compresses the spring. On expansion the spring does a work of 43J back on the block is only enough to push it over a distance of 0.65m stopping short of 5.35m from point B.
Thank you for reading and I hope this is helpful to you.
Answer:
60*12.0= 720 = v/60 * 12.0 squared which is 1,728
Explanation:
Horizontal velocity component: Vx = V * cos(α)