1. As he climbs... He's gaining potential Energy.
PE=mgh = 60x9.8x3.5=2058J. This is the potential Energy gain.
2. Power = Workdone/Time.
65kW=65,000watts
65000=Workdone/35
Workdone=65,000 x 35=2,275,000J
Recall... Workdone=Force x Distance
given distance =17.5m
Force =Workdone/distance
=2,275,000/17.5
Force=130,000N or 130kN.
3.Potential Energy Increases or decreases as height increases or decreases respectively.
a) PE=mgh =25x9.8x425=104,125J.
b) change in PE=mg(h°-h) =25 x 9.8(225-425)
= -49000J. The Energy is negative because the system loses 49000J.
4. At Constant velocity; d=vt
where d=distance,v=velocity, t=time
v=d/t
1.8km=1800m
10min =10x60=600seconds
v=1800/600
v=3ms-¹
KE=1/2Mv²
=1/2 x 45 x 3² =202.5J.
5.Mary weighs 505N.
Weight=Mass x acceleration due to gravity
W=mg
505=m x 9.8
m=505/9.8 =51.53kg.
Since Mary is coming down... she's decreasing her potential energy cos Potential Energy Decreases as you decrease your height. So Her Initial height was 5.5m... she alighted to the ground which we'll say is the ground(h=0)
Change in PE=mg(h°-h)=51.53 x 9.8(0-5.5) = -2777.5J.
This energy is negative because she lost potential Energy.
6. Since each has a Mass of 45kg... and they're moving together.... Their combined Mass =45+45=90kg.
Velocity given as =10ms-¹
a)Combined KE=1/2MV² = 1/2 x 90 x 10² =4500J.
b)Ratio of combined Mass to Katias mass = CM/KM
Where CM=Combined Mass
KM=Katia's mass
=90/45 =2:1
c) Katia's kinetic E =1/2 x 45 x 10² =2250J.
Ratio of combined to Katia's = CK/KK
CK=Combiner Kinetic Energy
KK=Katia's kinetic energy
= 4500/2250 =2:1.
This answer should be the same cause from the Formula KE=1/2mv²
KInetic Energy varies Directly with Mass
so If the ratio of Masses is 2:1... That of KE should be the same because of their direct Relationship✌.
7.The height at which the Rocket Engine stops is not how far it went. Theres a Minute height that it'll still go before it starts going back down.
from the question... The energy when the engine stops is 1960J
PE=mgh
1960=10x9.8xh
h=1960/98 =20m.
The engine stops at this height
but The additional Minute height that was added before itll start falling cant be calculated from the insufficient info given in this question. If we had the Energy at the additional height-added... We can calculate that height and then subtract 20 from it. That will give the additional added height.
Thats pretty Much Everything✌
Spectro... Out
The energy in the system containing DeAnna and Earth gain from this climb will be 2058J.
From the information, DeAnna, with a mass of 60.0 kg, climbs 3.5 m up a gymnasium rope. The potential energy will be:
= mgh
= 60 × 9.8 × 3.5
= 2058J
When an electric motor develops 65 kW of power as it lifts a loaded elevator 17.5 m in 35s, the force that the motor exerts will be:
= Work done / Time
= 2275000/17.5
= 130000N
The shell-Earth system’s gravitational potential energy when the shell’s height is 425m will be:
= mgh
= 25 × 9.8 × 425
= 104125J
Learn more about energy on:
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