answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Sergio [31]
2 years ago
9

A gymnast's backflip is considered more difficult to do in the layout (straight body) position than in the tucked position. Why?

- The body's rotational inertia is greater in layout position than in tucked position. Because the body remains airborne for roughly the same time interval in either position, the gymnast must have much greater kinetic energy in tucked position to complete the backflip. - The body's rotational inertia is greater in tucked position than in layout position. Because the body remains airborne for roughly the same time interval in either position, the gymnast must have much greater kinetic energy in layout position to complete the backflip. - The body's rotational inertia is greater in layout position than in tucked position. Because the body remains airborne for roughly the same time interval in either position, the gymnast must have much greater kinetic energy in layout position to complete the backflip. - The body's rotational inertia is greater in tucked position than in layout position. Because the body remains airborne for roughly the same time interval in either position, the gymnast must have much greater kinetic energy in tucked position to complete the backflip.
Physics
1 answer:
spin [16.1K]2 years ago
6 0

Answer:

The body's rotational inertia is greater in layout position than in tucked position. Because the body remains airborne for roughly the same time interval in either position, the gymnast must have much greater kinetic energy in layout position to complete the backflip.

Explanation:

A gymnast's backflip is considered more difficult to do in the layout (straight body) position than in the tucked position.

When the body is straight , its moment of rotational inertia is more than the case when he folds his body round. Hence rotational inertia ( moment of inertia x angular velocity ) is also greater. To achieve that inertia , there is need of greater imput of energy in the form of kinetic energy  which requires greater effort.

So a gymnast's backflip is considered more difficult to do in the layout (straight body) position than in the tucked position.

You might be interested in
A 0.500-kg ball traveling horizontally on a frictionless surface approaches a very massive stone at 20.0 m/s perpendicular to wa
gregori [183]

The magnitude of the change in momentum of the stone is about 18.4 kg.m/s

\texttt{ }

<h3>Further explanation</h3>

Let's recall Impulse formula as follows:

\boxed {I = \Sigma F \times t}

<em>where:</em>

<em>I = impulse on the object ( kg m/s )</em>

<em>∑F = net force acting on object ( kg m /s² = Newton )</em>

<em>t = elapsed time ( s )</em>

Let us now tackle the problem!

\texttt{ }

<u>Given:</u>

mass of ball = m = 0.500 kg

initial speed of ball = vo = 20.0 m/s

final kinetic energy = Ek = 70% Eko

<u>Asked:</u>

magnitude of the change of momentum of the stone = Δp = ?

<u>Solution:</u>

<em>Firstly, we will calculate the final speed of the ball as follows:</em>

Ek = 70\% \ Ek_o

\frac{1}{2} m v^2 = 70\% \ ( \frac{1}{2} m (v_o)^2 )

v^2 = 70 \% \ (v_o)^2

v = - v_o \sqrt{70 \%} → <em>negative sign due to ball rebounds</em>

v = - v_o \sqrt{0.7} \texttt{ m/s}

\texttt{ }

<em>Next, we could find the magnitude of the change of momentum of the stone as follows:</em>

\Delta p_{stone} = - \Delta p_{ball}

\Delta p_{stone} = - [ mv - mv_o ]

\Delta p_{stone} = m[ v_o - v ]

\Delta p_{stone} = m[ v_o + v_o\sqrt{0.7} ]

\Delta p_{stone} = mv_o [ 1 + \sqrt{0.7} ]

\Delta p_{stone} = 0.500 ( 20.0 ) [ 1 + \sqrt{0.7} ]

\Delta p_{stone} \approx 18.4 \texttt{ kg.m/s}

\texttt{ }

<h3>Learn more</h3>
  • Velocity of Runner : brainly.com/question/3813437
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302
  • Average Speed of Plane : brainly.com/question/12826372
  • Impulse : brainly.com/question/12855855
  • Gravity : brainly.com/question/1724648

\texttt{ }

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Dynamics

8 0
2 years ago
1 Ten (10) ml aqueous solutions of drug A (10% w/v) and drug B (25% w/v) are stored in two identical test tubes under identical
Reil [10]

Answer:

YOUR answer is given below:

Explanation:

3 0
1 year ago
2.0 kg of solid gold (Au) at an initial temperature of 1000K is allowed to exchange heat with 1.5 kg of liquid gold at an initia
Elanso [62]

Answer:

Explanation:

The specific heat of gold is 129 J/kgC

It's melting point is 1336 K

It's Heat of fusion is 63000 J/kg

Assuming that the mixture will be solid, the thermal energy to solidify the gold has to be less than that needed to raise the solid gold to the melting point. So,

The first is E1 = 63000 J/kg x 1.5 = 94500 J

the second is E2 = 129 J/kgC x 2 kg x (1336–1000)K = 86688 J

Therefore, all solid is not correct. You will have a mixture of solid and liquid.

For more detail, the difference between E1 and E2 is 7812 J, and that will melt

7812/63000 = 0.124 kg of the solid gold

8 0
2 years ago
1. A particular lever is 90.0% efficient. If 50.0 J of work are done on the lever, then how much work does the lever do on its l
laila [671]

Answer:

Explanation:

Using the efficiency formula;

Efficiency = Work done by the machine (output)/work done on the machine (input) ×100%

Efficiency =w/50 ×100

90 = 100w/50

Cross multiply

90×50 = 100W

4500 = 100W

W = 4500/100

W = 45Joules

Hence the lever does 45Joules of work on its load

2) Mechanical Advantage= Load/Effort

Given

MA = 4

Load = 500N

4 = 500/Effort

Effort = 500/4

Effort =125N

Hence the effort required to lift the load is 125N

8 0
1 year ago
A semi is traveling down the highway at a velocity of v = 26 m/s. The driver observes a wreck ahead, locks his brakes, and begin
Dovator [93]

Answer:

fcosθ + Fbcosθ  =Wtanθ

Explanation:

Consider the diagram shown in attachment

fx= fcosθ (fx: component of friction force in x-direction ; f: frictional force)

Fbx= Fbcosθ ( Fbx: component of braking force in x-direction ; Fb: braking force)

Wx= Wtanθ (Wx: component of weight in x-direction ; W: Weight of semi)

sum of x-direction forces = 0

fx+ Fbx=Wx

fcosθ + Fbcosθ  =Wtanθ

7 0
2 years ago
Other questions:
  • Find τf, the torque about point p due to the force applied by the achilles' tendon.
    11·1 answer
  • How much does it cost to operate a 25-w soldering iron for 8.0 hours if energy costs 8.0¢/kwh?
    15·1 answer
  • Calculate the volume occupied by 25.2 g of co2 at 0.84 atm and 25°c. r = 0.08206 latm/kmol.
    13·1 answer
  • in a hydraulic press the small cylinder has a diameter 10.0cm while the large has 25cm if the force of 600N is applied to the sm
    8·2 answers
  • A small car meshes with a large truck in a head-on collision. Which of the following statements concerning the magnitude of the
    14·1 answer
  • You are driving on the highway, and you come to a steep downhill section. As you roll down the hill, you take your foot off the
    12·2 answers
  • In this problem you are to consider an adiabaticexpansion of an ideal diatomic gas, which means that the gas expands with no add
    6·1 answer
  • A skateboarder is attempting to make a circular arc of radius r = 16 m in a parking lot. The total mass of the skateboard and sk
    10·1 answer
  • A wire loop is suspended from a string that is attached to point P in the drawing. When released, the loop swings downward, from
    11·1 answer
  • Part A
    11·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!