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slava [35]
3 years ago
6

8. If the mass of a balloon is 300 kg and the lift force provided by the atmosphere is 3300 N, what

Physics
1 answer:
zmey [24]3 years ago
3 0

Answer:

Toward the north, 300 N

Explanation:

Consider the upward direction as positive and downward direction as negative.

Given:

Mass of the balloon is, m=300\ kg

Lift force on the balloon in the upward direction is, F_L=3300\ N

Take acceleration due to gravity as 10 N/kg.

Now, weight of the balloon acting downward is given as the product of its mass and acceleration due to gravity. Therefore,

Weight is, W=mg=(300\ kg)(10\ N/kg)=-3000\ N

Now, net force is given as the vector sum of both the forces.

F_{net}=F_L+W\\F_{net}=3300-3000=300\ N(\textrm{Upward direction})

The upward direction is the direction of North. So, the net force acting on the balloon is 300 N towards North.

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A box of mass 0.8 kg is placed on an inclined surface that makes an angle 30 above
earnstyle [38]

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a)    a = 17.1 m / s², b)    F = 3.04 N

Explanation:

This is an exercise of Newton's second law, in this case the selection of the reference system is very important, we have two possibilities

* a reference system with the horizontal x axis, for this selection the normal and the friction force have x and y components

* a reference system with the x axis parallel to the plane, in this case the weight and the applied force have x and y components

We are going to select the second possibility, since it is the most used in inclined plane problems, let's analyze the angle of the applied force (F) it has an angle 10º with respect to the x axis, if we rotate this axis 30º the new angle is

                θ = 10 -30 = -20º

The negative sign indicates that it is in the fourth quadrant. Let's use trigonometry to find the components of the applied force

              sin (-20) = F_{y} / F

              cos (-20) = Fₓ / F

              F_{y} = F sin (-20)

              Fₓ = F cos (-20)

              F_y = 18 sin (-20) = -6.16 N

              Fₓ = 18 cos (-20) = 16.9 N

The decomposition of the weight is the customary

               sin 30 = Wₓ / W

               cos 30 = W_y / W

               Wₓ = W sin 30 = mg sin 30

                W_y = W cos 30 = m g cos 30

                Wₓ = 0.8 9.8 sin 30 = 3.92 N

                 W_y = 0.8 9.8 cos 30 = 6.79 N

Notice that in the case  the angle is measured with respect to the axis y perpendicular to the plane

Now we can write Newton's second law for each axis

X axis

      Fₓ - fr = m a

Y Axis  

      N - F_{y} - Wy = 0

      N =F_{y} + Wy

      N = 6.16 + 6.79

     

They both go to the negative side of the axis and

      N = 12.95 N

The friction force has the formula

        fr = μ N

we substitute

        Fₓ - μ N = m a

        a = (Fₓ - μ N) / m

     

we calculate

       a = (16.9 - 0.25 12.95) / 0.8

       a = 17.1 m / s²

b) now the block slides down with constant speed, therefore the acceleration is zero

ask for the value of the applied force, we will suppose that with the same angle, that is, only its modulus was reduced

       Newton's law for the x axis

              Fₓ -fr = 0

              Fₓ = fr

              F cos 20 = μ N

              F = μ N / cos 20

we calculate

              F = 0.25 12.95 / cos 20

              F = 3.04 N

this is the force applied at an angle of 10º to the horizontal

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