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Stolb23
3 months ago
15

When a 85 kg man sits on the stool, by what percent does the length of the legs decrease? assume, for simplicity, that the stool

's legs are vertical and that each bears the same load?
Physics
1 answer:
Yuliya22 [3.3K]3 months ago
7 0

To determine stress, we utilize the formula Stress = Force / Area. In this scenario, the force exerted by a person weighing 85 kg on the stool is calculated as F = mg = 85 kg * 9.8 m/s² = 833 N. Next, we find the total area of the three legs: A = 3 π (D/2)² = 3 π (2.5 x 10^-2 / 2)² = 1.4726 x 10^-3 m². Substituting into the equation gives Stress = 833 N / 1.4726 x 10^-3 m² = 5.656 x 10^5 N/m². To find strain, we apply the formula strain = Stress / Young’s modulus, with wood having Y = 1.3 x 10^10 N/m². Thus, strain = 5.656 x 10^5 N/m² / 1.3 x 10^10 N/m² = 4.35 x 10^-5 = 0.00435%.



This indicates that the legs will shorten by 0.00435%

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A charged particle (q = −8.0 mC), which moves in a region where the only force acting on the particle is an electric force, is r
Yuliya22 [3333]

Answer:

V_B - V_A = 600 Volts

Explanation:

Given that the charge starts from rest and at point B has gained kinetic energy K = 4.8 J

we can apply the principle of conservation of mechanical energy, indicating that the increase in kinetic energy equals the reduction in electrostatic potential energy.

Thus, we can represent this as

U_B + KE = U_A

KE = U_A - U_B

4.8 = (-8 mC)(V_A - V_B)

4.8 = (8 \times 10^{-3})(V_B - V_A)

V_B - V_A = 600 Volts

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3 months ago
What is the factor involved in increasing an object’s inertia?
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2 months ago
The parasailing system shown uses a winch to pull the rider in towards the boat, which is traveling with a constant velocity. Du
ValentinkaMS [3465]

Answer:

The force magnitude is  F_{net}= 1.837 *10^4N

and it is directed at 57.98° from the horizontal in a counterclockwise manner.

Explanation:

The problem states that

At t = 0, \theta = 20^o

The angular rate of increase is w = 2 \ ^o/s

Converting to revolutions per second gives us  \theta ' = 2 \ ^o/s * \frac{\pi}{180} =0.0349\ rps

The rope length is defined by

                      r = 125- \frac{1}{3}t^{\frac{3}{2} }    

    At \theta =30^o, Tension T of the rope is 18 kN.

     The weight of the para-sailor is M_p = 75kg

In analyzing the question, we observe that the equation for length can be represented as a linear displacement equation.

 The derivative of displacement results in velocity.

Hence,

           r' = -\frac{1}{3} [\frac{3}{2} ] t^{\frac{1}{2} }

signifies the velocity, and further differentiation yields acceleration.

Therefore,

         r'' = -\frac{1}{4} t^{-\frac{1}{2} }

Now considering the moment when the rope forms a 30° angle with the water,

typically angular velocity is expressed as

                      w = \frac{\Delta \theta}{\Delta t}

where \theta represents the angular displacement.

Next, evaluating the interval from 20^o \ to \ 30^o gives us

                 2 = \frac{30 -20 }{t -0}

making t the focal point.

             t = \frac{10}{2}

               = 5s

At this time, the displacement measures

             r = 125- \frac{1}{3}(5)^{\frac{3}{2} }  

                = 121.273 m

The linear velocity computes to

             r' = -\frac{1}{3} [\frac{3}{2} ] (5)^{\frac{1}{2} }

                = -1.118 m/s

Whereas linear acceleration calculates as

          r'' = -\frac{1}{4} (5)^{-\frac{1}{2} }

              = -0.112m/s^2

Generally, radial acceleration is given by

         \alpha _R = r'' -r \theta'^2

              = -0.112 - (121.273)[0.0349]^2

              = 0.271 m/s^2

Simultaneously, angular acceleration can be represented as

                 \alpha_t = r \theta'' + 2 r' \theta '

Then \theta '' = \frac{d (0.0349)}{dt} = 0

Thus,

             \alpha _t = 121.273 * 0 + 2 * (-1.118)(0.0349)

                   = -0.07805 m/s^2

The resultant acceleration is mathematically denoted as

                a = \sqrt{\alpha_R^2 + \alpha_t^2 }

                  = \sqrt{(-0.07805)^2 +(-0.027)^2}

                  = 0.272 m/s^2

Now the acceleration's direction is mathematically expressed as

                  tan \theta_a = \frac{\alpha_R }{\alpha_t }

                       \theta_a = tan^{-1} \frac{-0.271}{-0.07805}

                           = 73.26^o

               

The y-axis force acting on the para-sailor is mathematically shown as

               F_y = mg + Tsin 30^o + ma sin(90- \theta )

                    = (75 * 9.8) + (18 *10^3) sin 30 + (75 * 0.272)sin(90-73.26)

                    = 9.74*10^3 N

The x-axis force acting on the para-sailor is represented as

              F_x = mg + Tcos 30^o + ma cos(90- \theta )    

             = (75 * 9.8) + (18 *10^3) cos 30 + (75 * 0.272)cos(90-73.26)

             = 1.557 *10^4 N

The overall force is calculated as

                      F_{net} = \sqrt{F_x^2 + F_y^2}

                             =\sqrt{(1.557 *10^4)^2 + (9.74*10^3)^2}

                            F_{net}= 1.837 *10^4N

The directional force is evaluated as

              tan \theta_f = \frac{F_y}{F_x}

                   \theta_f = tan^{-1} [\frac{1.557*10^4}{9.74*10^3} ]

                       = tan^{-1} (1.599)

                       = 57.98^o

     

                     

7 0
3 months ago
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