The prop blades of an airplane spin with a linear velocity of 875 m/s and have a centripetal acceleration on the farthest edge o
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<h2>Answer</h2>
The answer is 29495.3 kg
<h2>Explanation</h2>As we know, the Newton's Second Law of Motion
The F exerted by catapult = 9.35 *
The Time is = 2.10 s
Velocity of the catapult = 240 km/h = 66.6 m/s (240*1000 m /3600 s)
The mass of jet = ?
First detrmine the acceleration as
{/tex}Acceleration = Velocity/Time {/tex}
Rearrange the formaula as
The mass will be 29495.3 kg
Answer:
Explanation:
Force of friction = μ N , where μ is coefficient of friction , N is normal force on the body .
a )
Given,
Normal force N = 400 N
Force of friction = 300 N
μ = coefficient of static friction = ?
Putting the values ,
300 = 400 μ
μ = .75
b )
Normal force N = 400 N
Force of friction = 200 N
μ = coefficient of kinetic friction = ?
Putting the values ,
200 = 400 μ
μ = .50
c ) see attached file .
