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Vesnalui [34]
2 years ago
11

1. A U.S. manufacturer wants to sell computer equipment to companies in Honduras. Because the U.S. government limits trade with

Honduras, the manufacturer is required to obtain an export license from the Department of Commerce before selling and shipping the equipment to the purchasers. The export license fee is $8,100, not including a nonrefundable application fee of $2,700. What are the possible effects of these fees?
Physics
1 answer:
xxTIMURxx [149]2 years ago
3 0

If the US government does not allow the manufacturer to export their products to Honduras, the manufacturer will lose the non-refundable application fee of $2,700 and because the U.S. government limits trade with Honduras, the manufacturer is required to obtain an export license from the Department of Commerce, which can result to a shortage of demand for US products.

In addition, the total value of product exports must not less than $10800 for the duration of the license so that the application and license fee will be worthwhile. 

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If a helicopter's mass is 4,500kg and the net force on it is 18,000 N upward, what is its acceleration?
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4 0
1 year ago
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The coefficient of friction between the 2-lb block and the surface is μ=0.2. The block has an initial speed of Vβ =6 ft/s and is
Taya2010 [7]

Answer:

x = 0.0685 m

Explanation:

In this exercise we can use the relationship between work and energy conservation

            W = ΔEm

Where the work is

             W = F x

The energy can be found in two points

Initial. Just when the block with its spring spring touches the other spring

           Em₀ = K = ½ m v²

Final. When the system is at rest

            Em_{f} = K_{e1}b +K_{e2} = ½ k₁ x² + ½ k₂ x²

We can find strength with Newton's second law

            ∑ F = F - fr

Axis y

           N- W = 0

           N = W

The friction force has the equation

          fr = μ N

          fr = μ W

  The job

         W = (F – μ W) x

We substitute in the equation

            (F - μ W) x = ½ m v² - (½ k₁ x² + ½ k₂ x²)

           ½ x² (k₁ + k₂) + (F - μ W) x - ½ m v² = 0

We substitute values ​​and solve

           ½ x² (20 + 40) + (15 -0.2 2) x - ½ (2/32) 6² = 0

         x² 30 + 14.4 x - 1,125 = 0

        x² + 0.48 x - 0.0375 = 0

           

We solve the second degree equation

        x = [-0.48 ±√(0.48 2 + 4 0.0375)] / 2

        x = [-0.48 ± 0.617] / 2

        x₁ = 0.0685 m

        x₂ = -0.549 m

The first result results from compression of the spring and the second torque elongation.

The result of the problem is x = 0.0685 m

4 0
2 years ago
An object of mass 24kg is accelerated up a frictionless place incline at an angle of 37° with horizontal by a constant force, st
RoseWind [281]

a) Average power: 1425 W

b) Instantaneous power at 3.0 sec: 2850 W

Explanation:

a)

The motion of the object along the ramp is a uniformly accelerated motion (because the force applied is constant), so we can use the suvat equation

s=ut+\frac{1}{2}at^2

where

s = 18 m is the displacement along the ramp

u = 0 is the initial velocity

t = 3.0 s is the time taken

a is the acceleration of the object along the ramp

Solving for a,

a=\frac{2s}{t^2}=\frac{2(18)}{(3.0)^2}=4 m/s^2

Now we can apply Newton's second law to find the net force on the object:

F=ma=(24 kg)(4 m/s^2)=96 N

This net force is the resultant of the applied force forward (F_a) and the component of the weight acting backward (mg sin \theta), so we can find what is the applied force:

F=F_a - mg sin \theta\\F_a = F+mg sin \theta = 96+(24)(9.8)(sin 37^{\circ})=237.5 N

where

m = 24 kg is the mass of the object

g=9.8 m/s^2 is the acceleration of gravity

Now we can finally find what is the work done by the applied force, which is parallel to the ramp, therefore:

W=F_a s = (237.6)(18)=4276 J

where s = 18 m is the displacement.

Therefore the average power needed is:

P=\frac{W}{t}=\frac{4276}{3}=1425 W

b)

The instantaneous power at any point of the motion is given by

P=F_av

where

F_a is the force applied

v is the velocity of the object

We already calculated the applied force:

F_a=237.5 N

While since this is a uniformly accelerated motion, we can find the velocity at the end of the 3.0 seconds using the suvat equation:

v=u+at=0+(4)(3.0)=12.0 m/s

And therefore, the instantaeous power at 3.0 sec is:

P=Fv=(237.5)(12)=2850 W

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

8 0
2 years ago
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