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nadezda [96]
3 years ago
13

How much force is needed to accelerate a 9760 kg airplane at a rate of 4.6 m/s^2

Physics
1 answer:
seropon [69]3 years ago
8 0
Assuming air resistance / friction is negligable:

force = mass × acceleration

f = m × a
f = 9760 × 4.6
f = 44896N

f= 45kN (2 sig fig)

I hope this helps. If you have any questions, feel free to ask
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A small sphere is at rest at the top of a frictionless semicylindrical surface. The sphere is given a slight nudge to the right
V125BC [204]

Answer:

vi = 4.77 ft/s

Explanation:

Given:

- The radius of the surface R = 1.45 ft

- The Angle at which the the sphere leaves

- Initial velocity vi

- Final velocity vf

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Determine the sphere's initial speed.

Solution:

- Newton's second law of motion in centripetal direction is given as:

                         m*g*cos(θ) - N = m*v^2 / R

Where, m: mass of sphere

             g: Gravitational Acceleration

             θ: Angle with the vertical

             N: Normal contact force.

- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:

                         m*g*cos(θ) - 0 = m*vf^2 / R

                         g*cos(θ) = vf^2 / R    

                         vf^2 = R*g*cos(θ)

                         vf^2 = 1.45*32.2*cos(34)

                        vf^2 = 38.708 ft/s

- Using conservation of energy for initial release point and point where sphere leaves cylinder:

                          ΔK.E = ΔP.E

                          0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))

                          ( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))

                          vi^2 =  vf^2 - 2*g*R*( 1 - cos(θ))

                          vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))

                          vi^2 = 22.744

                           vi = 4.77 ft/s

4 0
3 years ago
This is a two part question: What is the force of friction acting on a 100 kg steel slab at rest on a steel floor?
Crazy boy [7]
1) 0N... friction opposes the motion of an object, since the block is at rest there is no motion thus no friction

2) F=ma
= (5.5kg)(30m/s)
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