5. An acrobat, starting from rest, swings freely on a trapeze of
length 3.7 m (Figure 6). If the initial angle of the trapeze is
48°, use the law of conservation of energy to determine
(a) the acrobat's speed at the bottom of the swing
(b) the maximum height, relative to the initial position, to
which the acrobat can rise

Answers

Answer 1

The energy conservation and trigonometry we can find the results for the questions about the movement of the acrobat are;

     a) The maximum speed is v = 4.89 m / s

     b) The maximum height is h = 1.22 m

The energy conservation is one of the most fundamental principles of physics, stable that if there are no friction forces the mechanistic energy remains constant. Mechanical energy is the sum of the kinetic energy plus the potential energies.

               Em = K + U

Let's write the energy in two points.

Starting point. Highest part of the oscillation

            Em₀ = U = m g h

Final point. Lower part of the movement

            [tex]Em_f[/tex] = K = ½ m v²

Energy is conserved.

            Emo = [tex]Em_f[/tex]  

            m g h = ½ m v²

            v² = 2 gh

Let's use trigonometry to find the height, see attached.

         h = L - L cos θ

         h = L (1- cos θ)

They indicate that the initial angle is tea = 48º and the length is L = 3.7 m, let's calculate.

         h = 3.7 (1- cos 48)

          h = 1.22 m

this  is the maximum height of the movement.

Let's calculate the velocity.  

          [tex]v= \sqrt{2 \ 9.8 \ 1.22}[/tex]  

          v = 4.89 m / s

In conclusion using the conservation of energy and trigonometry we can find the results for the questions about the movement of the acrobat are;

     a) The maximum speed is v = 4.89 m / s

     b) The maximum height is h = 1.22 m

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5. An Acrobat, Starting From Rest, Swings Freely On A Trapeze Oflength 3.7 M (Figure 6). If The Initial

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Newtons second law of motion states that, the force acting on a body is directly proportion to the mass times the acceleration of the body

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Hi there!

We can begin by calculating the acceleration using the kinematic equation:

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Answer:

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The linear speed of a point located on the 32nd parallel, as a result of Earth’s rotation is 392.91 m/s

We find the linear speed of a point on the 32nd parallel from v = rω where r = radius of 32nd parallel north = Rcos32° (since it is the radius of the small circle at the 32nd parallel) where R = radius of earth = 6,371,000 m and ω = angular speed of the earth = 2π/T where T = period of earth = 24h = 24 × 60 × 60 s = 86400 s.

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Answers

Explanation:

Correct option is

B

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2

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The car accelerates 60 mph in 6.3 seconds. Thus, the acceleration is [tex]\frac{60}{6.3}[/tex]. [tex]f=ma[/tex], so [tex]4106 = m(\frac{60}{6.3})[/tex], so [tex]m=431.13[/tex].

a race car traveling at 44m/s slows at a constant rate to a velocity of 22m/s over 11 seconds , how far does it move during this time

find distance and acceleration

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Answers

Answer:

add 44m/s and 22m/s then multiply it by 11

Explanation:

A race car traveling at 44m/s slows at a constant rate to a velocity of 22m/s over 11 seconds, Acceleration is -2 m/s² (deceleration), and distance is 242 meters.

To solve this problem, it is given:

Initial velocity ([tex]\(v_i\)[/tex] ) = 44 m/s

Final velocity ([tex]\(v_f[/tex] ) = 22 m/s

Time ([tex]\(t\)[/tex] ) = 11 s

We have two tasks: finding the acceleration (a) and then using it to calculate the distance (d).

1. Calculate Acceleration (a):

[tex]\[a = \frac{v_f - v_i}{t} \\\\= \frac{22 \, \text{m/s} - 44 \, \text{m/s}}{11 \, \text{s}} \\\\= -2 \, \text{m/s}^2\][/tex]

The negative sign indicates deceleration (slowing down).

2. Calculate Distance (d):

Using the kinematic equation:

[tex]\[d = v_i \cdot t + \frac{1}{2} \cdot a \cdot t^2\][/tex]

Plugging in the values:

[tex]\[d = (44 \, \text{m/s}) \cdot (11 \, \text{s}) + \frac{1}{2} \cdot (-2 \, \text{m/s}^2) \cdot (11 \, \text{s})^2\][/tex]

Calculating the first part:

[tex]\[d = 484 \, \text{m} - 121 \, \text{m} \\\\= 363 \, \text{m}\][/tex]

Calculating the second part:

[tex]\[d = 363 \, \text{m} + \frac{1}{2} \cdot (-2 \, \text{m/s}^2) \cdot (121 \, \text{s}^2) \\\\= 363 \, \text{m} - 121 \, \text{m} \\\\= 242 \, \text{m}\][/tex]

The total distance the race car moves during this time is 242 meters.

Thus, acceleration is -2 m/s² (deceleration), and distance is 242 meters.

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A 46.5-kg ball has a momentum of 57.2 kg m/s. What is the ball's speed?

Answers

Answer:

1.23 m/s

Explanation:

p=mv

57.2 = 46.5v

v= 57.2/46.5

v= 1.23

If you want to verify your answer, just insert the value of v in the equation.

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