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tiny-mole [99]
2 years ago
14

An airplane is flying in a horizontal circle at a speed of 100 m/s. The 80.0 kg pilot does not want the centripetal acceleration

to exceed 7.00 times free-fall acceleration.
Physics
1 answer:
Tema [17]2 years ago
3 0

Answer:

The force will be "5,488". A further solution is provided below.

Explanation:

The given values are:

speed,

v = 100m/s

mass,

m = 80 kg

acceleration,

a = 7

Now,

Radius will be:

⇒ r=\frac{v^2}{ac}

⇒    =\frac{(100)^2}{7\times 9.8}

⇒    =\frac{10000}{68.6}

⇒    =145.77

Force will be:

⇒ F=m\times ac

⇒     =80\times 68.6

⇒     =5488

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A 29.0 kg beam is attached to a wall with a hi.nge while its far end is supported by a cable such that the beam is horizontal.
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The vertical component of the force exerted by the hi.nge on the beam is 114.77 N.

<h3>Tension in the cable</h3>

Apply the principle of moment and calculate the tension in the cable;

Clockwise torque = TL sinθ

Anticlockwise torque = ¹/₂WL

TL sinθ  =  ¹/₂WL

T sinθ  =  ¹/₂W

T = (W)/(2 sinθ)

T = (29 x 9.8)/(2 x sin57)

T = 169.43 N

<h3>Vertical component of the force</h3>

T + F = W

F = W - T

F = (9.8 x 29) - 169.43

F = 114.77 N

Thus, the vertical component of the force exerted by the hi.nge on the beam is 114.77 N.

Learn more about tension here: brainly.com/question/24994188

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A bead slides without friction around a loop the-loop. The bead is released from a height of 17.6 m from the bottom of the loop-
solong [7]

Answer:

 N₁ = 393.96 N   and  N = 197.96 N

Explanation:

In This exercise we must use Newton's second law to find the normal force. Let's use two points the lowest and the highest of the loop

Lowest point, we write Newton's second law n for the y-axis

          N -W = m a

where the acceleration is ccentripeta

          a = v² / r

           

          N = W + m v² / r

          N = mg + mv² / r

         

we can use energy to find the speed at the bottom of the circle

starting point. Highest point where the ball is released

           Em₀ = U = m g h

lowest point. Stop curl down

           Em_{f} = K = ½ m v²

           Emo = Em_{f}

           m g h = ½ m v²

           v² = 2 gh

we substitute

             N = m (g + 2gh / r)

            N = mg (1 + 2h / r)

let's calculate

          N₁ = 5 9.8 (1 + 2 17.6 / 5)

          N₁ = 393.96 N

headed up

we repeat the calculation in the longest part of the loop

          -N -W = - m v₂² / r

            N = m v₂² / r - W

             N = m (v₂²/r  - g)

we seek speed with the conservation of energy

           Em₀ = U = m g h

final point. Top of circle with height 2r

             Em_{f} = K + U = ½ m v₂² + mg (2r)

              Em₀ =   Em_{f}

            mgh = ½ m v₂² + 2mgr

             v₂² = 2 g (h-2r)

we substitute

            N = m (2g (h-2r) / r - g)

            N = mg (2 (h-r) / r 1) = mg (2h/r  -2 -1)

             N = mg (2h/r  - 3)

            N = 5 9.8 (2 17.6 / 5 -3)

            N = 197.96 N

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