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valentina_108 [34]
3 years ago
13

While punting a football, a kicker rotates his leg about the hip joint. the moment of inertia of the leg is 3.75 kg m2 and its r

otational kinetic energy is 175 j. what is the angular velocity of the leg?
Physics
2 answers:
otez555 [7]3 years ago
7 0
<span>Angular velocity is the rate of the change of angular displacement of a body that is in a circular motion. It is a vector quantity so it consists of a magnitude and direction. It is equal to the linear velocity divided by the radius of the circular motion. However, we are not given the linear velocity and the radius of the motion. We instead use the equation for the rotational kinetic energy. It is expressed as:
</span>
Rotational kinetic energy = Iω^2 / 2

where I is the inertia and ω is the angular velocity

175 = (3.75)ω^2 / 2
ω = 9.66 rad/s

The angular velocity of the motion is about 9.66 rad / s.
Ber [7]3 years ago
7 0

The angular velocity of the kicker’s leg while he kicks the ball is  \boxed{9.66\,{{{\text{rad}}} \mathord{\left/{\vphantom {{{\text{rad}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}}.

Further Explanation:

Given:

The moment of inertia of the leg is 3.75\,{\text{kg}} \cdot {{\text{m}}^{\text{2}}}.

The rotational kinetic energy of the leg is 175\,{\text{J}}.

Concept:

As the kicker rotates his leg to kick the ball, he gains the rotational kinetic energy by rotating it at a particular angular speed and this rotational kinetic energy of the leg is converted into the linear kinetic energy of the football and it moves forward at certain speed.

The rotational kinetic energy of the leg is expressed as:

K{E_R}=\dfrac{1}{2}I{\omega ^2}  

Here, K{E_R} is the rotational kinetic energy, I is the moment of inertia of the leg and \omega  is the angular speed of the leg.

Substitute the values of energy and the moment of inertia in above expression.

\begin{aligned}175&=\frac{1}{2}\times 3.75\times {\omega ^2}\\\omega&=\sqrt {\frac{{2 \times 175}}{{3.75}}}\\&=\sqrt {93.33}\,{{{\text{rad}}} \mathord{\left/{\vphantom {{{\text{rad}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\&=9.66\,{{{\text{rad}}} \mathord{\left/{\vphantom {{{\text{rad}}} {\text{s}}}} \right.\kern-\nulldelimiterspace} {\text{s}}}\\\end{aligned}

Thus, the angular velocity of the kicker’s leg while he kicks the ball is  \boxed{9.66\,{{{\text{rad}}}\mathord{\left/{\vphantom {{{\text{rad}}} {\text{s}}}}\right.\kern-\nulldelimiterspace} {\text{s}}}}.

Learn More:

1. Assume that, at a certain angular speed ω2, the radius r becomes twice l. Find ω2 brainly.com/question/5813257

2. Suppose we replace both hover pucks with pucks that are the same size as the originals but twice as massive brainly.com/question/3923773

3. Mass m, moving at speed 2v, approaches mass 4m, moving at speed v. The two collide elastically head-on brainly.com/question/2097915

Answer Details:

Grade: High School

Subject: Physics

Chapter: Rotational Motion

Keywords:  Punting a football, kicker rotates, about the hip joint, moment of inertia, 3.75 kg m2, rotational kinetic energy, angular velocity, kinetic energy, moves forward.

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7 0
2 years ago
Scenario
Anvisha [2.4K]

Answer:

1) t = 23.26 s,  x = 8527 m, 2)   t = 97.145 s,  v₀ = 6.4 m / s

Explanation:

1) First Scenario.

After reading your extensive problem, we are going to solve it, for this exercise we must use the parabolic motion relationships. Let's carry out an analysis of the situation, for deliveries the planes fly horizontally and we assume that the wind speed is zero or very small.

Before starting, let's reduce the magnitudes to the SI system

         v₀ = 250 miles/h (5280 ft / 1 mile) (1h / 3600s) = 366.67 ft/s

         y = 2650 m

Let's start by looking for the time it takes for the load to reach the ground.

         y = y₀ + v_{oy} t - ½ g t²

in this case when it reaches the ground its height is zero and as the plane flies horizontally the vertical speed is zero

         0 = y₀ + 0 - ½ g t2

          t = \sqrt{ \frac{2y_o}{g} }

          t = √(2 2650/9.8)

          t = 23.26 s

this is the horizontal scrolling time

          x = v₀ t

          x = 366.67  23.26

          x = 8527 m

the speed at the point of arrival is

         v_y = v_{oy} - g t = 0 - gt

         v_y = - 9.8 23.26

         v_y = -227.95 m / s

Module and angle form

        v = \sqrt{v_x^2 + v_y^2}

         v = √(366.67² + 227.95²)

        v = 431.75 m / s

         θ = tan⁻¹ (v_y / vₓ)

         θ = tan⁻¹ (227.95 / 366.67)

         θ = - 31.97º

measured clockwise from x axis

We see that there must be a mechanism to reduce this speed and the merchandise is not damaged.

2) second scenario. A catapult located at the position x₀ = -400m y₀ = -50m with a launch angle of θ = 50º

we look for the components of speed

           cos θ = v₀ₓ / v₀

           sin θ = v_{oy} / v₀

            v₀ₓ = v₀ cos θ

            v_{oy} = v₀ sin θ

we look for the time for the arrival point that has coordinates x = 0, y = 0

            y = y₀ + v_{oy} t - ½ g t²

            0 = y₀ + vo sin θ t - ½ g t²

            0 = -50 + vo sin 50 t - ½ 9.8 t²

            x = x₀ + v₀ₓ t

            0 = x₀ + vo cos θ t

            0 = -400 + vo cos 50 t

podemos ver que tenemos un sistema de dos ecuación con dos incógnitas

          50 = 0,766 vo t – 4,9 t²

          400 =   0,643 vo t

resolved

          50 = 0,766 ( \frac{400}{0.643 \ t}) t – 4,9 t²

          50 = 476,52 t – 4,9 t²

          t² – 97,25 t + 10,2 = 0

we solve the quadratic equation

         t = [97.25 ± \sqrt{97.25^2 - 4 \ 10.2}] / 2

         t = 97.25 ±97.04] 2

         t₁ = 97.145 s

         t₂ = 0.1 s≈0

the correct time is t1 the other time is the time to the launch point,

         t = 97.145 s

let's find the initial velocity

         x = x₀ + v₀ cos 50 t

         0 = -400 + v₀ cos 50 97.145

         v₀ = 400 / 62.44

         v₀ = 6.4 m / s

5 0
3 years ago
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Answer:

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Δ=7.4 mm

Explanation:

Given that

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Diameter of sleeve ,D=32 mm

Length .L=50 mm

E= 5 MPa

n=0.45

As we know that

Lateral strain

\varepsilon _t=\dfrac{D-d}{d}

\varepsilon _t=\dfrac{32-30}{30}

\varepsilon _t=0.0667

We know that

n=-\dfrac{\epsilon _t}{\varepsilon _{long}}

\varepsilon _{long}=-\dfrac{\epsilon _t}{n}

\varepsilon _{long}=-\dfrac{0.0667}{0.45}

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P=E\times \varepsilon _{long}

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\Delta =\varepsilon _{long}\times L

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3 years ago
A tetrahedron has an equilateral triangle base with 25.0-cm-long edges and three equilateral triangle sides. The base is paralle
djverab [1.8K]

Answer:

a. 7.046 Nm²/C

b. 2.348 Nm²/C

Explanation:

Data given:

Base of equilateral triangle = 25.0 cm = 0.25 m

Strength of electric field = 260 N/C

In order to find the electric flux we first have to find out the area of triangle.

Area of triangle = \frac{\sqrt{3} }{4} a^{2}

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Lets find electric flux,

      Electric Flux = E. A

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Now we can find the electric flux through each of the three sides.

Electric flux through three sides = \frac{7.046}{3}

                                                = 2.348 N m²/C

       

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