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GaryK [48]
4 years ago
13

A 20.3 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The

bottom of the ladder rests on a floor with a rough surface where the coefficient of static friction is 0.23 . The angle between the horizontal and the ladder is θ . The person wants to climb up the ladder a distance of 1.1 m along the ladder from the ladder’s footWhat is the minimum angle θmin (between the horizontal and the ladder) so that the person can reach a distance of 1.1 m without having the ladder slip? The acceleration of gravity is 9.8 m/s 2 . Answer in units of ◦A 20.3 kg person climbs up a uniform ladder with negligible mass. The upper end of the ladder rests on a frictionless wall. The bottom of the ladder rests on a floor with a rough surface where the coefficient of static friction is 0.23 . The angle between the horizontal and the ladder is θ . The person wants to climb up the ladder a distance of 1.1 m along the ladder from the ladder’s footWhat is the minimum angle θmin (between the horizontal and the ladder) so that the person can reach a distance of 1.1 m without having the ladder slip? The acceleration of gravity is 9.8 m/s 2 . Answer in units of ◦
Mathematics
1 answer:
VMariaS [17]4 years ago
4 0

Answer:

Q = arctan(7.1739) = 82.06

Step-by-step explanation:

Given:

- The mass of the person m = 20.3 kg

- The distance traveled up the ladder s = 1.1 m

- The gravitational constant g = 9.8 m/s^2

- The coefficient of static friction u_s = 0.23

- Total length of the ladder

Find:

The minimum angle θ, that would allow the person to climb without ladder slipping

Solution:

- Taking moments about point of ladder and wall contact A to be zero:

                   -F_n,b*cos(Q)*4 - F_f*sin(Q)*4+ m*g*cos(Q)*2.6 = 0

- Taking Sum of vertical forces to be zero:

                    F_n,b - m*g = 0

                    F_n,b = m*g

- The frictional force F_f is given by:

                   F_f = u_s*F_n,b = u_s*m*g

- Plug the values back in:

                  - m*g*cos(Q)*4 + u_s*m*g*sin(Q)*4 - m*g*cos(Q)*2.6 = 0

Simplify:

                  4*cos(Q) + 2.6*cos(Q) = u_s*4*sin(Q)

                        6.6*cos(Q) = 4*u_s*sin(Q)

                               tan(Q) = 6.6 / 4*u_s

- Plug in the values:

                               tan(Q) = 6.6 / 4*0.23

                                    Q = arctan(7.1739) = 82.06

                   

                     

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Step-by-step explanation:

8 - 4y + (-2y) + 5 = <em>13 - 6y</em>

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Andreas93 [3]

Answer:

Proportion of the students recalled more than 15 names is 91.77%.

Step-by-step explanation:

We are given that a researcher was interested in seeing how many names a class of 38 students could remember after playing a name game After playing the name game, the students were asked to recall as many first names of fellow students as possible.

The mean number of names recalled was 19.41 with a standard deviation of 3.17.

<em>Let X = number of names recalled</em>

SO, X ~ N(\mu = 19.41,\sigma^{2} = 3.17^{2})

The z-score probability distribution is given by ;

                  Z = \frac{X-\mu}{\sigma} } } ~ N(0,1)

where, \mu = mean number of names recalled = 19.41

            \sigma = standard deviation = 3.17

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

Now, proportion of the students recalled more than 15 names is given by = P(X > 15 names)

     P(X > 15) = P( \frac{X-\mu}{{\sigma} } } > \frac{15-19.41}{3.17}  } ) = P(Z > -1.39)

                                                      = P(Z < 1.39) = 0.9177  {using z table}

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3 years ago
What is the area of the composite figure? (use 3.14 for pi). Explain how you arrived at your answer. Please round to the nearest
maksim [4K]

The area of the given diagram is 29.817 square centimeters. The given diagram is combination of rectangle and semi circle.

Step-by-step explanation:

The given is,

                 Given diagram is combination of Rectangle and Semi circle.

Step:1

         Res the attachment,

                      Area of given diagram = Area of A + Area of B..............(1)

Step:2

         For A,

         The A section in the given diagram is semi circle,

                     Diameter of semi circle = Total distance - ( 2+3)

                                          ( ∵ 2, 3 are top distance in the given diagram)

                                                             = 10 - 5

                                        Diameter, d = 5 cm

                                             Radius, r = \frac{d}{2}

                                                          r = \frac{5}{2}

                                                         r = 2.5 cm

                                          Area, A_{A}  = \frac {\pi r^{2} }{2}........................(2)

                                                    A_{A}  = \frac {\pi (2.5)^{2} }{2}  

                                                           = \frac {\pi ( 6.25) }{2}

                                                           = \frac{19.63}{2}

                                                     A_{A} = 9.8174 cm^{2}

Step:3

               For B,

               Area of rectangle is,

                                                      A_{B} =lb.....................(3)

              Where, l - Length = 10 cm

                          b - Width = 2 cm

              Equation (3) become,

                                                            = (10)(2)

                                                            = 20

                                    Area of B, A_{B} = 20 cm^{2}

Step:4

               From the equation (1),

                               Area of given diagram = 20+ 9.81747

                                                            Area = 29.817 square centimeters

Result:

            The area of the given diagram is 29.817 square centimeters. The given diagram is combination of rectangle and semi circle.

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