22,25,26 30m 20m 1m 10m 45m An object is thrown from ground with an angle of 30

Question

22,25,26 30m 20m 1m 10m 45m An object is thrown from ground with an angle of 30 degree with an initial velocity of 60 m/s. How much time is required to reach the maximum height? g = 10m/s^2 5 see 4 sec 3 sec 2 sec 1 sec When you take exit from highway and round a circle with the presented speed limit you will not skid away because road surface has Kinetic friction static friction banking angle both B & C both A & C An object of mass 2 kg starts to move at a constant acceleration of 3 m/s^2. What is its momentum after 10 second? 80 kg-m 60 kg-m/s (C) 64 kg-m/s 40 kg-m's 50 kg-m/s F= 5 N force is working on m 10 kg object at 0=45 degree angle. How much work will be done to move 10 m distance? Ignore force P. 72.1 J. 50.4 J. 98.2 J 60.2 J, 35.35 J The mass m=20 kg is being pulled by a force F=200 N being oriented at theta = 60 degree with ground overcoming a frictional force P=10 N.^ What is acceleration? 4.5 m/s^2 9.5 m/s^2 10.0 m/s^2 15 m/s^2 20 m/s^2

Explanation / Answer

22. Angle = 30 degree

Velocity = 60m/s

Vertical component of initial velocity = 60Sin(30) = 30 m/s

Vertical component of final velocity =0 (At maximum height)

So 0 = 30-gt

t = 30/g = 3 seconds

Time required to reach max height = 3 seconds

25. Force = 5 N

Angle = 45 degree

displacement =10 m

Work done = Force*Displacement*Cos(45)

= 5*10*(1/21/2)  = 35.35 joule.

26.

Net force = Mass*Acceleration

Net force = FCos(60) - Friction = 200*cos60 - 10 = 90 N

Acceleration = Net force / Mass = 90 / 20 = 4.5 m/s2

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