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omeli [17]
3 years ago
12

Potential energy and kinetic energy are forms of what kind of energy? 1. chemical 2. nuclear 3. electromagnetic 4. heat 5. mecha

nical
Physics
1 answer:
blsea [12.9K]3 years ago
4 0
Heat and mechanical energy.
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A disk has 128 tracks of 32 sectors each, on each surface of eight platters. The disk spins at 3600 RPM and takes 15 ms to move
Serga [27]

Answer:

the longest time needed to read an arbitrary sector located anywhere on the disk is 2971.24 ms

Explanation:

 Given the data in the question;

first we determine the rotational latency

Rotational latency = 60/(3600×2) = 0.008333 s = 8.33 ms

To get the longest time, lets assume the sector will be found at the last track.

hence we will access all the track, meaning that 127 transitions will be done;

so the track changing time = 127 × 15 = 1905 ms

also, we will look for the sectors, for every track rotations that will be done;

128 × 8.33 = 1066.24 ms

∴The Total Time = 1066.24 ms + 1905 ms

Total Time = 2971.24 ms

Therefore, the longest time needed to read an arbitrary sector located anywhere on the disk is 2971.24 ms

7 0
3 years ago
On the return home, the cyclist accelerates from 2.0 m/s with a an acceleration of 2.2 m/s 2 . Calculate the time it takes the c
jonny [76]

motion

vt=vo+at

10 m/s = 2 m/s + 2.2 m/s².t

8 = 2.2t

\tt t=\dfrac{8}{2.2}=3.63~s

7 0
3 years ago
In a Broadway performance, an 77.0-kg actor swings from a R = 3.65-m-long cable that is horizontal when he starts. At the bottom
krek1111 [17]

Answer: h =1.22 m

Explanation:

from the question we were given the following

mass of performer ( M1 ) = 77 kg

length of cable ( R ) = 3.65 m

mass of costar ( M2 ) = 55 kg

maximum height (h) = ?

acceleration due to gravity (g) = 9.8 m/s^2  (constant value)

We first have to find the velocity of the performer. From the work energy theorem work done = change in kinetic energy

work done = 1/2 x mass x ( (final velocity)^2 - (initial velocity)^2 )

initial velocity is zero in this case because the performer was at rest before swinging, therefore

work done = 1/2 x 77 x ( v^2 - 0)

work done = 38.5 x ( v^2 ) ......equation 1

work done is also equal to m x g x distance ( the distance in this case is the length of the rope), hence equating the two equations we have

m x g x R =  38.5 x ( v^2 )

77 x 9.8 x 3.65 =  38.5 x ( v^2 )

2754.29 = 38.5 x ( v^2 )

( v^2 ) =  71.54

v = 8.4 m/s  ( velocity of the performer)

After swinging, the performer picks up his costar and they move together, therefore we can apply the conservation of momentum formula which is

initial momentum of performer (P1) + initial momentum of costar (P2) = final momentum of costar and performer after pick up (Pf)  

momentum = mass x velocity therefore the equation above now becomes

(77 x 8.4) + (55 x 0) = (77 +55) x Vf  

take note the the initial velocity of the costar is 0 before pick up because he is at rest

651.3 = 132 x Vf

Vf = 4.9 m/s

the performer and his costar is 4.9 m/s after pickup

to finally get their height we can use the energy conservation equation for from after pickup to their maximum height. Take note that their velocity at maximum height is 0

initial Kinetic energy + Initial potential energy = Final potential energy + Final Kinetic energy

where

kinetic energy = 1/2 x m x v^2

potential energy  = m x g x h

after pickup they both will have kinetic energy and no potential energy, while at maximum height they will have potential energy and no kinetic energy. Therefore the equation now becomes

initial kinetic energy = final potential energy

(1/2 x (55 + 77) x 4.9^2) + 0 = ( (55 + 77) x 9.8 x h) + 0

1584.7 = 1293 x h

h =1.22 m

3 0
4 years ago
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